XI. On the Isomorphism of a Group with Itself.

Chapter XI.
On the Isomorphism of a Group with Itself.

I t is shewn in § 24 that, if all the operations of a group are transformed by one of themselves, which is not self-conjugate, a correspondence is thereby established among the operations of the group which exhibits the group as simply isomorphic with itself.

In an Abelian group every operation is self-conjugate, and the only correspondence established in the manner indicated is that in which every operation corresponds to itself. If however in an Abelian group we take, as the operation which corresponds to any given operation $S$ , its inverse $S^{- 1}$ , then to $S T$ will correspond $S^{- 1} T^{- 1}$ or ${(S T)}^{- 1}$ ; and the correspondence exhibits the group as simply isomorphic with itself. For this particular correspondence, a group of order $2$ is the only one in which each operation corresponds to itself.

It is therefore possible for every group, except a group of order $2$ , to establish a correspondence between the operations of the group, which shall exhibit the group as simply isomorphic with itself. Moreover, we shall see that in general there are such correspondences which cannot be established by either of the processes above given. We devote the present Chapter to a discussion of the isomorphism of a group with itself. It will be seen that, for many problems of group-theory, and in particular for the determination of the various types of group which are possible when the factor-groups of the composition-series are given, this discussion is most important.

155. Deﬁnition. A correspondence between the operations of a group, such that to every operation $S$ there corresponds a single operation $S'$ , while to the product $S T$ of two operations there corresponds the product $S' T'$ of the corresponding operations, is said to deﬁne an isomorphism of the group with itself .

That isomorphism in which each operation corresponds to itself is called the identical isomorphism. In every isomorphism of a group with itself, the identical operation corresponds to itself; and the orders of two corresponding operations are the same. For if $1$ and $S$ were corresponding operations, so also would be $1 \cdot 1$ and $S^{2}$ ; and therefore more than one operation would correspond to $1$ . Again, if $S$ and $S'$ , of orders $n$ and $n'$ , are corresponding operations, so also are $S^{n}$ and $S'^{n}$ ; and therefore $n$ must be a multiple of $n'$ . Similarly $n'$ must be a multiple of $n$ ; and therefore $n$ and $n'$ are equal.

If the operations of a group of order $N$ are represented by

1, S_{1}, S_{2}, \dots, S_{N - 1},

and if, for a given isomorphism of the group with itself,

S'_{r}

is the operation that corresponds to

S_{r}

(

r = 1, 2, \dots, N - 1

), the isomorphism will be completely represented by the symbol

[\binom{1, S_{1}, S_{2}, \dots, S_{N - 1}}{1, S', S', \dots, S'}] .

In this symbol, two operations in the same vertical line are corresponding operations. When no risk of confusion is thereby introduced, the simpler symbol

[\binom{S}{S'}]

is used.

156. An isomorphism of a group with itself, thus deﬁned, is not an operation. The symbol of an isomorphism however deﬁnes an operation. It may, in fact, be regarded as a substitution performed upon the $N$ letters which represent the operations of the group. Corresponding to every isomorphism there is thus a deﬁnite operation; and it is obvious that the operations, which correspond to two distinct isomorphisms, are themselves distinct. The totality of these operations form a group. For let

[\binom{S}{S'}] and [\binom{S'}{S''}]

be any two isomorphisms of the group with itself. Then if, as hitherto, we use curved brackets to denote a substitution, we have

(\binom{S}{S'}) (\binom{S'}{S''}) = (\binom{S}{S''}) .

But since $[\binom{S}{S'}]$ is an isomorphism, the relation

S_{p} S_{q} = S_{r}

requires that

S' S' = S' .

And since $[\binom{S'}{S''}]$ is an isomorphism, the relation

S' S' = S'

requires that

S'' S'' = S'' .

Hence if

S_{p} S_{q} = S_{r},

then

S'' S'' = S'';

and therefore

[\binom{S}{S''}]

is an isomorphism.

The product of the substitutions which correspond to two isomorphisms is therefore the substitution which corresponds to some other isomorphism.

The set of substitutions which correspond to the isomorphisms of a given group with itself, therefore form a group.

Deﬁnition. A group, which is simply isomorphic with the group thus derived from a given group, is called the group of isomorphisms of the given group.

It is not, of course, necessary always to regard this group as a group of substitutions performed on the symbols of the operations of the given group. But however the group of isomorphisms may be represented, each one of its operations corresponds to a deﬁnite isomorphism of the given group. To avoid an unnecessarily cumbrous phrase, we may brieﬂy apply the term “isomorphism” to the operations of the group of isomorphisms. So long, at all events, as we are dealing with the properties of a group of isomorphisms, no risk of confusion is thereby introduced. Thus we shall use the phrase “the isomorphism $(\binom{S}{S'})$ ” as equivalent to “the operation of the group of isomorphisms which corresponds to the isomorphism $[\binom{S}{S'}]$ .”

157. If $Σ$ is some operation of a group $G$ , while for $S$ each operation of the group is put in turn, the symbol

(\binom{S}{Σ^{- 1} S Σ})

deﬁnes an isomorphism of the group. For if

S_{p} S_{q} = S_{r},

then

Σ^{- 1} S_{p} Σ \cdot Σ^{- 1} S_{q} Σ = Σ^{- 1} S_{p} S_{q} Σ = Σ^{- 1} S_{r} Σ;

and

Σ_{- 1} S_{r} Σ

is an operation of the group. An isomorphism of a group, which is thus formed on transforming the operations of the group by one of themselves, is called a cogredient isomorphism. All others are called contragredient isomorphisms⁵⁸. If

(\binom{S}{S'})

is any contragredient isomorphism, the isomorphisms

(\binom{S}{S'}) (\binom{S}{Σ^{- 1} S Σ}),

when for

Σ

each operation of the group is taken successively, are said to form a class of contragredient isomorphisms.

THEOREM I. The totality of the cogredient isomorphisms of a group $G$ form a group isomorphic with $G$ ; this group is a self-conjugate sub-group of the group of isomorphisms of $G$ ⁵⁹.

The product of the isomorphisms

(\binom{S}{Σ^{- 1} S Σ}) and (\binom{S}{Σ'^{- 1} S Σ'})

is given by

\begin{array}{l} (\binom{S}{Σ^{- 1} S Σ}) (\binom{S}{{Σ'}^{- 1} S Σ'}) & = (\binom{S}{Σ^{- 1} S Σ}) (\binom{Σ^{- 1} S Σ}{{Σ'}^{- 1} Σ^{- 1} S Σ Σ'}) \\ = (\binom{S}{{Σ'}^{- 1} Σ^{- 1} S Σ Σ'}) \\ = (\binom{S}{{Σ''}^{- 1} S Σ''}), \end{array}

where

Σ Σ' = Σ'' .

The product of two cogredient isomorphisms is therefore another cogredient isomorphism; hence the cogredient isomorphisms form a group. Moreover, if we take the isomorphism

(\binom{S}{Σ^{- 1} S Σ})

as corresponding to the operation

Σ

of the group

G

, then to every operation of

G

there will correspond a deﬁnite cogredient isomorphism, so that to the product of any two operations of

G

there corresponds the product of the two corresponding isomorphisms. The group

G

and its group of cogredient isomorphisms are therefore isomorphic. If

G

contains no self-conjugate operation, identity excepted, no two isomorphisms corresponding to diﬀerent operations of

G

can be identical; and therefore, in this case,

G

is simply isomorphic with its group of cogredient isomorphisms. If however

G

contains self-conjugate operations, forming a self-conjugate sub-group

H

, then to every operation of

H

there corresponds the identical isomorphism; and the group of cogredient isomorphisms is simply isomorphic with

\frac{G}{H}

Let now

(\binom{S}{S'})

be any isomorphism. Then

\begin{array}{l} {(\binom{S}{S'})}^{- 1} (\binom{S}{Σ^{- 1} S Σ}) (\binom{S}{S'}) & = (\binom{S'}{S}) (\binom{S}{Σ^{- 1} S Σ}) (\binom{S}{S'}) \\ = (\binom{S'}{Σ^{- 1} S Σ}) (\binom{Σ^{- 1} S Σ}{{Σ'}^{- 1} S' Σ'}) \\ = (\binom{S'}{{Σ'}^{- 1} S' Σ'}) . \end{array}

The isomorphism $(\binom{S}{S'})$ therefore transforms every cogredient isomorphism into another cogredient isomorphism. It follows that the group of cogredient isomorphisms is self-conjugate within the group of isomorphisms.

158. Let $G$ be a group of order $N$ , whose operations are

1, S_{1}, S_{2}, \dots, S_{N - 1};

and let

L

be the group of isomorphisms of

G

. We have seen in § 20 that

G

may be represented as a transitive group of substitutions performed on the

N

symbols

1, S_{1}, S_{2}, \dots, S_{N - 1};

and that, when it is so represented, the substitution which corresponds to the operation

S_{x}

(\binom{1, S_{1}, S_{2}, \dots, S_{N - 1}}{S_{x}, S_{1} S_{x}, S_{2} S_{x}, \dots, S_{N - 1} S_{x}}),

or more shortly

(\binom{S}{S S_{x}}) .

When $G$ is thus represented, we will denote it by $G'$ . We have already seen that $L$ can be represented as an intransitive substitution group of the same $N$ symbols; a typical substitution of $L$ , when it is so represented, is

(\binom{1, S_{1}, S_{2}, \dots, S_{N - 1}}{1, S'_{1}, S'_{2}, \dots, S'_{N - 1}}),

or more shortly

(\binom{S}{S'}) .

When $L$ is thus represented, we will denote it by $L'$ . It is clear that the two substitution groups $G'$ and $L'$ have no substitution in common except identity. For every substitution of $L'$ leaves the symbol $1$ unchanged; and no substitution of $G'$ , except identity, leaves $1$ unchanged.

Now

\begin{array}{l} {(\binom{S}{S'})}^{- 1} (\binom{S}{S S_{x}}) (\binom{S}{S'}) & = (\binom{S'}{S S_{x}}) (\binom{S S_{x}}{S' S'_{x}}) \\ = (\binom{S'}{S' S'_{x}}) \\ = (\binom{S}{S S'_{x}}) . \end{array}

Every operation of $L'$ is therefore permutable with $G'$ . Hence if $M$ is the order of $L$ , the group ${G', L'}$ , which we will call $K'$ , is a transitive group of degree $N$ and order $N M$ , containing $G'$ self-conjugately. Further $(\binom{S}{S'})$ transforms $(\binom{S}{S S_{x}})$ into $(\binom{S}{S S'_{x}})$ ; and these two substitutions of $G'$ correspond to the operations $S_{x}$ and $S'$ of $G$ . Hence the isomorphism, established on transforming the substitutions of $G'$ by any substitution $(\binom{S}{S'})$ of $L'$ , is the isomorphism denoted by the symbol $(\binom{S}{S'})$ .

Since $(\binom{S}{S_{x} S S_{x}^{- 1}})$ is a substitution of $L'$ , the substitution $(\binom{S}{S_{x} S S_{x}^{- 1}}) (\binom{S}{S S_{x}})$ , or $(\binom{S}{S_{x} S})$ , belongs to $K'$ . Hence $K'$ contains the set of substitutions

(\binom{S}{S_{x} S}), (x = 0, 1, 2, \dots, N - 1) .

These form (§ 107) a transitive group $G''$ , simply isomorphic with $G'$ and such that every substitution of $G''$ is permutable with every substitution of $G'$ . Moreover (l.c.), the substitutions of $G''$ are the only substitutions of the $N$ symbols which are permutable with each of the substitutions of $G'$ .

Suppose now that $Σ$ is any substitution of the $N$ symbols which is permutable with $G'$ . When the substitutions of $G'$ are transformed by $Σ$ , the resulting isomorphism is identical with that given by some substitution, say $(\binom{S}{S'})$ , of $L'$ . Hence $Σ {(\binom{S}{S'})}^{- 1}$ is a substitution of the $N$ symbols which is permutable with every substitution of $G'$ . It therefore belongs to $G''$ ; and hence $Σ$ belongs to $K'$ . It follows that $K'$ contains every substitution of the $N$ symbols which is permutable with $G'$ .

The only substitutions common to $G'$ and $G''$ are the self-conjugate substitutions of either. The factor group $\frac{K'}{{G', G''}}$ is simply isomorphic with $\frac{L}{g}$ , where $g$ is the group of cogredient isomorphisms of $G$ contained in $L$ . The groups $G'$ and $G''$ are identical only when $G'$ is Abelian; in this case, $g$ consists of the identical operation alone.

Deﬁnition. A group $K$ , simply isomorphic with the substitution group $K'$ which has just been constructed, we shall call the holomorph of $G$ .

159. An isomorphism must change any set of operations, which are conjugate to each other, into another set which are conjugate. For if

(\binom{S}{S'})

be the isomorphism, and if

S_{z}^{- 1} S_{x} S_{z} = S_{y},

then

{S'_{z}}^{- 1} S'_{x} S'_{z} = S'_{y},

so that

S'_{x}

and

S'_{y}

are conjugate operations when

S_{x}

and

S_{y}

are conjugate. A cogredient isomorphism changes every set of conjugate operations into itself; and all the members of a class of contragredient isomorphisms permute the conjugate sets in the same way. If

(\binom{S}{S'})

is an isomorphism which changes every conjugate set of operations of

G

into itself, and if

(\binom{S}{S''})

is any isomorphism of

G

, then the isomorphism

{(\binom{S}{S''})}^{- 1} (\binom{S}{S'}) (\binom{S}{S''})

changes every conjugate set into itself. It follows that those isomorphisms, which change every conjugate set of operations into itself, form a self-conjugate sub-group of the complete group of isomorphisms. This sub-group clearly contains the group of cogredient isomorphisms and may be identical with it.

If now $(\binom{S}{S'})$ is any isomorphism of $G$ of order $n$ , the substitutions

(\binom{S}{S'}) and (\binom{S}{S S_{x}}), (x = 0, 1, \dots, N - 1),

generate a group of order

N_{n}

. When

J

is used to represent the isomorphism, this group may be denoted by

{J, G}

; as shewn above, it contains

G

self-conjugately. Suppose that

n

is prime and not a factor of

N

. The operation

J

is not permutable with every operation of

G

; and therefore there must be operations

S

G

which are permutable with no operation of the conjugate set to which

J

belongs. The number of operations which in

{J, G}

are conjugate to such an operation

S

must be a multiple of

n

; and since

n

is not a factor of

N

, this conjugate set of operations must be made up of

n

distinct conjugate sets of operations in

G

. The isomorphism

J

must therefore interchange some of the conjugate sets of

G

The same result is clearly true if the order $n$ of $J$ has any prime factor not contained in $N$ . Hence:—

THEOREM II. An isomorphism of a group $G$ , whose order contains a prime factor which does not occur in the order of $G$ , must interchange some of the conjugate sets of $G$ .

160. If the isomorphism $(\binom{S}{S'})$ or $J$ leaves no operation except identity unchanged, it must in ${J, G}$ be one of $N$ conjugate operations. For if

S_{x}^{- 1} J S_{x} = S_{y}^{- 1} J S_{y},

J

would be permutable with

S_{y} S_{x}^{- 1}

, which is not the case.

These $N$ conjugate operations are

J, J S_{1}, J S_{2}, \dots, J S_{N - 1},

and since the ﬁrst transforms every operation of

G

, except identity, into a diﬀerent one, the same must be true of all the set. If now

J

transformed any operation

S

into a conjugate operation

Σ^{- 1} S Σ, J Σ^{- 1}

would transform

S

into itself; hence

J

must transform every conjugate set of

G

into a diﬀerent conjugate set.

The special case in which the order of $J$ is two may here be considered. Representing the $N$ operations conjugate to $J$ by

J, J_{1}, J_{2}, \dots, J_{N - 1},

the

N

operations of

G

are

J^{2}, J J_{1}, J J_{2}, \dots, J J_{N - 1} .

Now

J^{- 1} \cdot J J_{x} \cdot J = J_{x} J = {(J J_{x})}^{- 1},

so that

J

transforms every operation of

G

into its own inverse. But if

S' = S^{- 1},

and

T' = T^{- 1},

then

S' T' = S^{- 1} T^{- 1} = {(T S)}^{- 1} .

Now as $S' T'$ is the operation into which the isomorphism transforms $S T$ , it must be ${(S T)}^{- 1}$ , and therefore

S T = T S .

The group $G$ is therefore an Abelian group of odd order.

161. Any sub-group $H$ of $G$ is transformed by an isomorphism into a simply isomorphic sub-group $H'$ ; but $H$ and $H'$ are not necessarily conjugate within $G$ . If however the set of conjugate sub-groups

H_{1}, H_{2}, \dots, H_{m},

are the only sub-groups of

G

of a given type, every isomorphism must interchange them among themselves; and if no isomorphism transforms each one of the set into itself, the group of isomorphisms can be represented as a transitive group of degree

m

Suppose now that no operation of $G$ is permutable with each of the conjugate sub-groups

H_{1}, H_{2}, \dots, H_{m},

so that

G

, or what in this case is the same thing (since

G

can have no self-conjugate operation except identity) the group of cogredient isomorphisms of

G

, can be expressed as the transitive group of

m

symbols that arises on transforming the set of

m

sub-groups by each operation of

G

. Let

J

be any operation, of order

μ

, that transforms

G

and each of the set of

m

conjugate sub-groups, into itself. Then

J^{μ}

is the lowest power of

J

that can occur in

G

, since no operation of

G

transforms each of the

m

sub-groups into itself. Now in

{J, G}

, the greatest sub-group that contains

H_{r}

self-conjugately is

{J, I_{r}}

I_{r}

being the greatest sub-group of

G

that contains

H_{r}

self-conjugately. Also, in

{J, G}

the set of sub-groups

{J, I_{r}}

, (

r = 1, 2, \dots, m

), is a complete conjugate set. Now the set of groups

I_{1}, I_{2}, \dots, I_{m}

have by supposition no common operation except identity; and therefore the greatest common sub-group of

{J, I_{1}}, {J, I_{2}}, \dots, {J, I_{m}}

{J}

. Hence

{J}

is a self-conjugate sub-group of

{J, G}

; and since

G

is also a self-conjugate sub-group of

{J, G}

, while

{J}

and

G

have no common operation except identity,

J

must be permutable with every operation of

G

. Every operation therefore which is permutable with

G

, and with each of the sub-groups

H_{1}, H_{2}, \dots, H_{m},

is permutable with every operation of

G

. Thus ﬁnally, no contragredient isomorphism can transform each of the sub-groups

H_{r}

(

r = 1, 2, \dots, m

) into itself. Hence:—

THEOREM III. If the conjugate set of $m$ sub-groups

H_{1}, H_{2}, \dots, H_{m}

contains all the sub-groups of

G

of a given type, and if no operation of

G

is permutable with each sub-group of the set, the group of isomorphisms of

G

can be represented as a transitive group of degree

m

Corollary I. If $G$ contains $k p + 1$ sub-groups of order $p^{α}$ , where $p^{α}$ is the highest power of a prime $p$ that divides the order of $G$ ; and if the greatest sub-group $I$ , that contains a sub-group of order $p^{α}$ self-conjugately, contains no self-conjugate sub-group of $G$ ; then the group of isomorphisms of $G$ can be represented as a transitive group of degree $k p + 1$ .

For it has been seen that the groups of order $p^{α}$ contained in $G$ form a single conjugate set.

Corollary II. If the conjugate set of $m$ sub-groups of $G$

H_{1}, H_{2}, \dots, H_{m}

is changed into itself by every isomorphism of

G

, and if no operation of

G

is permutable with every one of these sub-groups; then the group of isomorphisms of

G

can be expressed as a transitive group of degree

m

In fact, under the conditions stated, the reasoning applied to prove the theorem may be used to shew that no isomorphism of $G$ can transform each of the $m$ sub-groups into itself.

162. Deﬁnition. Any sub-group of a group $G$ which is transformed into itself by every isomorphism of $G$ , is called⁶⁰ a characteristic sub-group of $G$ .

A characteristic sub-group of a group $G$ is necessarily a self-conjugate sub-group of $G$ ; but a self-conjugate sub-group is not necessarily characteristic. A simple group, having no self-conjugate sub-groups, can have no characteristic sub-groups. Let now $G$ be any group, and let $K$ be the holomorph of $G$ . A characteristic sub-group of $G$ is then a self-conjugate sub-group of $K$ ; and conversely, every self-conjugate sub-group of $K$ which is contained in $G$ is a characteristic sub-group of $G$ .

Suppose now a chief-series of $K$ formed which contains $G$ . If $G$ has no characteristic sub-group, it must be the last term but one of this series, the last term being identity. It follows by § 94 that $G$ must be the direct product of a number of simply isomorphic simple groups. Hence:—

THEOREM IV. A group, which has no characteristic sub-group, must be either a simple group or the direct product of simply isomorphic simple groups.

The converse of this theorem is clearly true.

163. Suppose now that $G$ is a group which has characteristic sub-groups; and let

G, G_{1}, \dots, G_{r}, G_{r + 1}, \dots, 1

be a series of such sub-groups, each containing all that follow it and chosen so that, for each consecutive pair

G_{r}

and

G_{r + 1}

, there is no characteristic sub-group of

G

contained in

G_{r}

and containing

G_{r + 1}

, except

G_{r + 1}

itself. Such a series is called⁶¹ a characteristic series of

G

It may clearly be possible to choose such a series in more than one way. If

G, G', \dots, G', G', \dots, 1

be a second characteristic series of

G

, then

\begin{array}{l} K, J, \dots, H, G, G_{1}, \dots, G_{r}, G_{r + 1}, \dots, 1 \\ a n d & K, J, \dots, H, G, G', \dots, G', G', \dots, 1 \end{array}

are two chief-series of $K$ . In fact, if $K$ had a self-conjugate sub-group contained in $G_{r}$ and containing $G_{r + 1}$ , then $G$ would have a characteristic sub-group contained in $G_{r}$ and containing $G_{r + 1}$ . The two chief-series of $K$ coincide in the terms from $K$ to $G$ inclusive. Hence the two sets of factor-groups

\frac{G}{G_{1}}, \frac{G_{1}}{G_{2}}, \dots, \frac{G_{r}}{G_{r + 1}}, \dots

and

\frac{G}{G'}, \frac{G'}{G'}, \dots, \frac{G'}{G'}, \dots

must be equal in number and, except possibly as regards the sequence in which they occur, identical in type. Moreover, each factor-group must be either a simple group or the direct product of simply isomorphic simple groups.

164. We will now shew how to determine a characteristic series for a group whose order is the power of a prime⁶².

First, let the group $G$ be Abelian; and suppose that it is generated by a set of independent operations, of which $n_{s}$ are of the order $p^{m}$ , ( $s = 1, 2, \dots, r$ ), while

m_{1} > m_{2} > \dots > m_{r} .

The sub-group $G_{μ}$ (§ 42), formed of the operations of $G$ which satisfy the relation

S^{p^{μ}} = 1,

is clearly a characteristic sub-group. As a ﬁrst step towards forming the characteristic series, we may take the set of groups

G_{m_{1}} (= G), G_{m_{1} - 1}, G_{m_{1} - 2}, \dots, G_{2}, G_{1}, 1;

for this is a set of characteristic sub-groups such that each contains the one that follows it.

Now the sub-group $H_{ν}$ (§ 45), formed of the distinct operations that remain when every operation of $G$ is raised to the power $p^{ν}$ , is also a characteristic sub-group; and since the operations common to two characteristic sub-groups also form a characteristic sub-group, the sub-group $K_{μ, ν}$ (common to $G_{μ}$ and $H_{ν}$ ) is characteristic. It follows from this that $G_{1}$ will be a characteristic sub-group only when $r = 1$ . If $r > 1$ , $G_{1}$ is not contained in $H_{m_{r - 1} - 1}$ , and the common sub-group $K_{1, m_{r - 1} - 1}$ of these two is characteristic. If $r > 2$ , this sub-group again is not contained in $H_{m_{r - 2} - 1}$ ; and the common sub-group $K_{1, m_{r - 2} - 1}$ of $G_{1}$ and $H_{m_{r - 2} - 1}$ is a characteristic sub-group contained in $K_{1, m_{r - 1} - 1}$ . Continuing thus, we form between $G_{1}$ and $1$ the series

G_{1}, K_{1, m_{r - 1} - 1}, K_{1, m_{r - 2} - 1}, \dots, K_{1, m_{1} - 1}, 1 .

In a similar way, between $G_{α}$ and $G_{α - 1}$ we introduce such of the series

{G_{α - 1}, K_{α, m_{r - 1} - α}}, {G_{α - 1}, K_{α, m_{r - 2} - α}}, \dots, {G_{α - 1}, K_{α, m_{1} - α}}

as are distinct, the symbol

m_{s} - α

being replaced by zero where it is negative.

From the original series we thus form a new one, in which again each group is characteristic and contains the following. This series may be shewn to be a characteristic series.

Let

P_{m_{s}, 1}, P_{m_{s}, 2}, \dots, P_{m_{s}, n_{s}}

be the

n_{s}

generating operations of

G

, whose orders are

p^{m_{s}}

. Then if

{G_{α - 1}, K_{α, m_{s} - α}}

and

{G_{α - 1}, K_{α, m_{s - 1} - α}}

are distinct, the generating operations of the latter diﬀer only from those of the former in containing the set

P_{m_{s}, x}^{p^{m_{s} - α + 1}}, (x = 1, 2, \dots, n_{s}),

in the place of

P_{m_{s}, x}^{p^{m_{s} - α}}, (x = 1, 2, \dots, n_{s}) .

Now any permutation of the $n_{s}$ generating operations

P_{m_{s}, x}, (x = 1, 2, \dots, n_{s}),

among themselves, the remaining generating operations being unaltered, must clearly give an isomorphism of

G

with itself; and therefore no sub-group of

G

, contained in

{G_{α - 1}, K_{α, m_{s} - α}}

and containing

{G_{α - 1}, K_{α, m_{s - 1} - α}}

, can be a characteristic sub-group. This result being true for every pair of distinct groups which succeed each other in the series that has been formed, it follows that the series is a characteristic series. It may be noticed that, if

Γ

and

Γ'

are any two consecutive sub-groups in a characteristic series of

G

, the order of

\frac{Γ}{Γ'}

must be

p^{ν}

, where

ν

is one of the

r

numbers

n_{s}

Secondly, suppose that $G$ is not Abelian. We may ﬁrst consider the series of sub-groups

G, H_{n}, H_{n - 1}, \dots, H_{1}, 1

of § 53. Each of these is clearly a characteristic sub-group, and each contains the succeeding. Moreover,

\frac{H_{r + 1}}{H_{r}}

is an Abelian group; and, by the process that we have just investigated, a characteristic series may be formed for it. To each group in this series will correspond a characteristic sub-group of

H_{r + 1}

containing

H_{r}

; and the set of groups so obtained forms part of a complete characteristic series of

G

. When between each consecutive pair of groups in the above series the groups thus formed are interpolated, the resulting series of groups is a characteristic series for

G

165. The isomorphisms of a given group with itself are closely connected with the composition of every composite group in which the given group enters as a self-conjugate sub-group. Let $G$ be any composite group and $H$ a self-conjugate sub-group of $G$ . Then since every operation of $G$ transforms $H$ into itself, to every such operation will correspond an isomorphism of $H$ with itself. If $S$ is an operation of $G$ not contained in $H$ , and if the isomorphism of $H$ arising on transforming its operations by $S$ is contragredient, so also is the isomorphism arising from each of the set of operations $S H$ . In this case, no one of this set of operations is permutable with every operation of $H$ . If however the isomorphism arising from $S$ is cogredient, there must be some operation $h$ of $H$ which gives the same isomorphism as $S$ ; and then $S h^{- 1}$ is permutable with every operation of $H$ . In this case, the set of operations $S H$ will give all the cogredient isomorphisms of $H$ .

Suppose now that $H_{1}$ is that sub-group of $G$ which is formed of all the operations of $G$ that are permutable with every operation of $H$ . Then to every operation of $G$ , not contained in ${H, H_{1}}$ , must correspond a contragredient isomorphism of $H$ ; and to every operation of the factor group $\frac{G}{{H, H_{1}}}$ corresponds a class of contragredient isomorphisms. If then $L$ is the group of isomorphisms of $H$ , and if $L_{1}$ is that self-conjugate sub-group of $L$ which gives the cogredient isomorphisms of $H$ , $\frac{G}{{H, H_{1}}}$ must be simply isomorphic with a sub-group of $\frac{L}{L_{1}}$ .

If now $H$ contains no self-conjugate operation except identity, $H$ and $H_{1}$ can contain no common operation except identity; and since each of them is a self-conjugate sub-group of $G$ , every operation of $H$ is permutable with every operation of $H_{1}$ . In this case, ${H, H_{1}}$ is the direct product of $H$ and $H_{1}$ .

If, further, $L$ coincides with $H$ , so that $H$ admits of no contragredient isomorphisms, $\frac{G}{{H, H_{1}}}$ must reduce to identity. In this case, $G$ is the direct product of $H$ and $H_{1}$ .

Deﬁnition. A group, which contains no self-conjugate operation except identity and which admits of no contragredient isomorphism, is called⁶³ a complete group.

The result of the present paragraph may be expressed in the form:—

THEOREM V. A group, which contains a complete group as a self-conjugate sub-group, must be the direct product of the complete group and some other group⁶⁴.

166. THEOREM VI. If $G$ is a group with no self-conjugate operations except identity; and if the group of cogredient isomorphisms of $G$ is a characteristic sub-group of $L$ , the group of isomorphisms of $G$ ; then $L$ is a complete group⁶⁵.

With the notation of § 158, the operations of $L$ may be represented by the substitutions

(\binom{S}{S'}) .

The group of cogredient isomorphisms, which we will call

G'

, is given by the substitutions

(\binom{S}{S_{x}^{- 1} S S_{x}}), (x = 0, 1, \dots, N - 1);

it is simply isomorphic with

G

Now

\begin{array}{l} {(\binom{S}{S'})}^{- 1} (\binom{S}{S_{x}^{- 1} S S_{x}}) (\binom{S}{S'}) & = (\binom{S'}{{S'}^{- 1} S' S'}), \\ = (\binom{S}{{S'}^{- 1} S S'}) : \end{array}

and therefore no operation of $L$ is permutable with every operation of $G'$ . Hence every isomorphism of $G'$ is given on transforming its operations by those of $L$ . Suppose now that $J$ is an operation which transforms $L$ into itself. Since $G'$ is by supposition a characteristic sub-group of $L$ , the operation $J$ transforms $G'$ into itself. If $J$ does not belong to $L$ , we may assume that $J$ is permutable with every operation of $G'$ . For if it is not, it must give the same isomorphism of $G'$ as some operation $S$ of $L$ ; and then $J S^{- 1}$ is permutable with every operation of $G'$ , and is not contained in $L$ . Now $J$ being permutable with every operation of $G'$ , we have

J^{- 1} s^{- 1} g s J = s^{- 1} g s,

where

s

is any operation of

L

, and

g

any operation of

G'

Moreover

J g J^{- 1} = g,

and therefore

J^{- 1} s^{- 1} J g J^{- 1} s J = s^{- 1} g s .

Hence $s$ and $J^{- 1} s J$ give the same isomorphism of $G'$ . Now no two distinct operations of $L$ give the same isomorphism of $G'$ , so that $s$ and $J^{- 1} s J$ must be identical; in other words, $J$ is permutable with every operation of $L$ . Hence $L$ admits of no contragredient isomorphisms. Moreover, $G'$ has no self-conjugate operations, and no operation of $L$ is permutable with every operation of $G'$ ; hence $L$ has no self-conjugate operations. It is therefore a complete group.

Corollary. If $G$ is a simple group of composite order, or if it is the direct product of a number of isomorphic simple groups of composite order, the group of isomorphisms $L$ of $G$ is a complete group.

For suppose, if possible, in this case that $G'$ is not a characteristic sub-group of $L$ ; and that, by a contragredient isomorphism of $L$ , $G$ is transformed into $G''$ . Then $G''$ is a self-conjugate sub-group of $L$ , and each of the groups $G'$ and $G''$ transforms the other into itself. Hence (§ 34) either every operation of $G'$ is permutable with every operation of $G''$ , or $G'$ and $G''$ must have a common sub-group. The former supposition is impossible since no operation of $L$ is permutable with every operation of $G$ . On the other hand, if $G'$ and $G''$ have a common sub-group, it is a self-conjugate sub-group of $L$ and it therefore is a characteristic sub-group of $G'$ . Now (§ 162) $G$ has no characteristic sub-groups, and therefore the second supposition is also impossible. It follows that, in this case, $G'$ is a characteristic sub-group of $L$ , and that $L$ is a complete group.

167. THEOREM VII. If $G$ is an Abelian group of odd order, and if $K$ is the holomorph of $G$ ; then when $G$ is a characteristic sub-group of $K$ , the latter group is a complete group.

If $N$ is the order of $G$ , then $K$ can be expressed (§ 158) as a transitive group of degree $N$ . When $K$ is so expressed, those operations of $K$ which leave one symbol unchanged form a sub-group $H$ , which is simply isomorphic with the group of isomorphisms of $G$ . Now (§ 160) an Abelian group of odd order admits of a single isomorphism of order two, which changes every operation into its own inverse. The corresponding substitution of $H$ is a self-conjugate substitution in $H$ , and is one of $N$ conjugate substitutions in $K$ . These are the only substitutions of $K$ which transform every substitution of $G$ into its inverse. If $G$ is a characteristic sub-group of $K$ , every isomorphism of $K$ must transform $G$ , and therefore also the set of $N$ conjugate substitutions of order two, into itself. Also, no substitution of $K$ can be permutable with each one of these $N$ substitutions, since each of them keeps just one symbol unchanged. Hence (Theorem III, Cor. II, § 161) the group of isomorphisms of $K$ can be expressed as a transitive group of degree $N$ , which contains $G$ as a transitive self-conjugate sub-group. But when the group of isomorphisms of $K$ is so expressed, $K$ itself consists of all the substitutions of the $N$ symbols which are permutable with $G$ ; and at the same time, every isomorphism of $K$ transforms $G$ into itself. Hence the group of isomorphisms of $K$ must coincide with $K$ itself; i.e. $K$ admits of no contragredient isomorphisms. Also $K$ obviously contains no self-conjugate operation except identity; hence it is a complete group.

Corollary. The holomorph of an Abelian group of order $p^{m}$ , where $p$ is an odd prime, and type $(1, 1, \dots, to m units)$ , is a complete group.

For if, in this case, $G$ is not a characteristic sub-group of $K$ , let $G'$ be a sub-group of $K$ which, in the group of isomorphisms of $K$ , is conjugate to $G$ . Then $G$ and $G'$ being both self-conjugate in $K$ must have a common sub-group, since $K$ cannot contain their direct product. But the common sub-group of $G$ and $G'$ , being self-conjugate in $K$ , is a characteristic sub-group of $G$ . This is impossible (§ 162); hence $G$ is a characteristic sub-group of $K$ .

Ex. Shew that the holomorph of an Abelian group of degree $2^{m}$ and type $(1, 1, \dots, to m units)$ is a complete group.

168. We shall now discuss the groups of isomorphisms of certain special groups; and we shall begin with the case of a cyclical group $G$ , of prime order $p$ , generated by an operation $P$ . Every isomorphism of such a group must interchange among themselves the $p - 1$ operations

P, P^{2}, \dots, P^{p - 1};

and if any isomorphism replaces

P

P^{α}

, it must replace

P^{2}

P^{2 α}

, and so on. Moreover, the symbol

(\binom{P, P^{2}, \dots, P^{p - 1}}{P^{α}, P^{2 α}, \dots, P^{(p - 1) α}})

does actually represent an isomorphism whatever number

α

may be from

1

p - 1

; for each operation occurs once in the second line, and the change indicated leaves the multiplication-table of the group unaltered. If

α = p

, the symbol does not represent an isomorphism: if

α = p + α'

, the symbol represents the same isomorphism as

(\binom{P, P^{2}, \dots, P^{p - 1}}{P^{α'}, P^{2 α'}, \dots, P^{(p - 1) α'}}) .

The group of isomorphisms of a group of prime order $p$ is therefore a group of order $p - 1$ . Now the $n$ th power of the isomorphism

(\binom{P}{P^{α}})

(\binom{P}{P^{α^{n}}}) .

Hence if $α$ is a primitive root of the congruence

α^{p - 1} - 1 \equiv 0, (mod p),

the group of isomorphisms is a cyclical group generated by the isomorphism

(\binom{P}{P^{α}}) .

Finally, if $S$ is an operation satisfying the relations

S^{p - 1} = 1, S^{- 1} P S = P^{α},

where

α

is a primitive root of

p

{S, P}

is the holomorph of

G

The reader will at once observe that this group of order $p (p - 1)$ is identical with the doubly transitive group of § 112. It is a complete group.

169. We shall consider next the case of any cyclical group.

Suppose, ﬁrst, that $G$ is a cyclical group of order $p^{n}$ , where $p$ is an odd prime; and let it be generated by an operation $S$ . The group contains $p^{n - 1} (p - 1)$ operations of order $p^{n}$ ; if $S'$ is any one of these,

(\binom{S}{S'})

deﬁnes a distinct isomorphism. The group of isomorphisms is therefore a group of order

p^{n - 1} (p - 1)

. Moreover, since the congruence

α^{p^{n - 1} (p - 1)} - 1 \equiv 0, (mod p^{n}),

has primitive roots, the group of isomorphisms is a cyclical group. The holomorph of

G

is deﬁned by

S^{p^{n}} = 1, J^{p^{n - 1} (p - 1)} = 1, J^{- 1} S J = S^{α},

where

α

is a primitive root of the congruence

α^{p^{n - 1} (p - 1)} - 1 \equiv 0, (mod p^{n}) .

If $G$ is a cyclical group of order $2^{n}$ , it follows, in the same way, that the group of isomorphisms is an Abelian group of order $2^{n - 1}$ . In this case, however, the congruence

α^{2^{n - 1}} - 1 \equiv 0, (mod 2^{n}), n > 2,

has no primitive root, and therefore the group of isomorphisms is not cyclical. The congruence

α^{2^{n - 2}} - 1 \equiv 0, (mod 2^{n})

however has primitive roots, and a primitive root

α

of this congruence can always be found to satisfy the condition

α^{2^{n - 3}} \equiv 1 + 2^{n - 1}, (mod 2^{n}) .

The powers of the isomorphism

(\binom{S}{S^{α}})

then form a cyclical group of order

2^{n - 2}

; and the only isomorphism of order

2

contained in it is

(\binom{S}{S^{1 + 2^{n - 1}}}) .

Hence

(\binom{S}{S^{α}}) and (\binom{S}{S^{- 1}}),

the latter not being contained in the sub-group generated by the former, are two permutable and independent isomorphisms of orders

2^{n - 2}

and

2

. They generate an Abelian group of order

2^{n - 1}

which is the group of isomorphisms of

G

. The corresponding holomorph is given by

\begin{matrix} S^{2^{n}} = 1, J_{1}^{2^{n - 2}} = 1, J_{2}^{2} = 1, J_{1} J_{2} = J_{2} J_{1}, \\ J_{1}^{- 1} S J_{1} = S^{α}, J_{2} S J_{2} = S^{- 1}, \end{matrix}

where $α$ satisﬁes the conditions given above.

If $G$ is a cyclical group of order $4$ , its group of isomorphisms is clearly a group of order $2$ .

170. It is now easy to construct the group of isomorphisms of any cyclical group $G$ , and the corresponding holomorph. If the order of $G$ is $2^{n} p_{1}^{m_{1}} p_{2}^{m_{2}} \dots$ , where $p_{1}$ , $p_{2}$ , … are odd primes, $G$ is the direct product of cyclical groups of orders $2^{n}$ , $p_{1}^{m_{1}}$ , $p_{2}^{m_{2}}$ , …; and every isomorphism of $G$ transforms each of these groups into itself. Hence if the groups of isomorphisms of these cyclical groups be formed, and their direct product be then constructed, every operation of the group so formed will give a distinct isomorphism of the group $G$ . Moreover, the order

2^{n - 1} p_{1}^{m_{1} - 1} (p_{1} - 1) p_{2}^{m_{2} - 1} (p_{2} - 1) \dots

of this group is equal to the number of operations of

G

whose order is

2^{n} p_{1}^{m_{1}} p_{2}^{m_{2}} \dots

, or in other words to the number of isomorphisms of which

G

is capable. The group thus formed is therefore the group of isomorphisms of

G

. The corresponding holomorph is clearly the direct product of the holomorphs of the cyclical groups of orders

2^{n}

p_{1}^{m_{1}}

, etc.

When the order of $G$ is odd, the holomorph $K$ is easily shewn to be a complete group. Suppose it to be expressed transitively, as in § 158, in $N$ symbols, where $N$ is the order of $G$ ; if $G$ is not a characteristic sub-group of $K$ , let $G'$ be a group into which $G$ is transformed by a contragredient isomorphism of $K$ . Then $G'$ is a self-conjugate sub-group of $K$ ; and since $G'$ is cyclical, every sub-group of $G$ ’ is a self-conjugate sub-group of $K$ . Hence a generating operation of $G'$ must be a circular substitution of $N$ symbols. The $N$ operations of order $2$ which transform each operation of $G'$ into its own inverse therefore each keep one symbol ﬁxed; hence each of them must transform every operation of $G$ into its inverse. But there is only one such set of operations of order $2$ , and therefore $G'$ cannot diﬀer from $G$ . It follows by Theorem VII, § 167 that, as $G$ is a characteristic sub-group of $K$ , the group $K$ itself is complete.

If the order of $G$ is even, $K$ must contain a self-conjugate operation other than identity, namely the operation of order $2$ contained in $G$ . Moreover, $K$ admits of a contragredient isomorphism whose square is cogredient. From the mode of formation of $K$ , it is clearly suﬃcient to verify this when the order of $G$ is a power of $2$ . The holomorph $K$ is given by

\begin{matrix} S^{2^{n}} = 1, J_{1}^{2^{n - 2}} = 1, J_{2}^{2} = 1, J_{1} J_{2} = J_{2} J_{1}, \\ J_{1}^{- 1} S J_{1} = S^{α}, J_{2} S J_{2} = S^{- 1}; \end{matrix}

where $α$ is a primitive root of

α^{2^{n - 2}} - 1 \equiv 0 (mod 2^{n}),

such that

α^{2^{n - 3}} + 1 ≢ 0 (mod 2^{n}) .

In this group, $J_{2}$ is one of $2^{n - 1}$ conjugate operations

J_{2}, J_{2} S^{2}, J_{2} S^{4}, \dots .

Now it may be directly veriﬁed that $K$ admits the isomorphism represented by

(\binom{S, J_{1}, J_{2}}{S, J_{1} S^{\frac{1}{2} (1 - α)}, J_{2} S}),

Moreover, since this isomorphism changes

J_{2}

into

J_{2} S

, it cannot be a cogredient isomorphism. Finally, the square of this isomorphism is

(\binom{S, J_{1}, J_{2}}{S, J_{1} S^{1 - α}, J_{2} S^{2}}),

(\binom{S, J_{1}, J_{2}}{S, S^{- 1} J_{1} S, S^{- 1} J_{2} S}),

which is a cogredient isomorphism.

171. We shall next consider the group of isomorphisms of an Abelian group of order $p^{n}$ and type $(1, 1, \dots, to n units)$ . Such a group is generated by $n$ independent permutable operations of order $p$ , say

P_{1}, P_{2}, \dots, P_{n} .

Since every operation of the group is self-conjugate and of order $p$ , while the group contains no characteristic sub-group, there must be isomorphisms transforming any one operation of the group into any other. We may therefore begin by determining under what conditions the symbol

(\binom{P_{r}}{P_{1}^{a_{1 r}} P_{2}^{a_{2 r}} \dots P_{n}^{a_{n r}}}) (r = 1, 2, \dots, n),

deﬁnes an isomorphism. This symbol replaces the operation

P_{1}^{x_{1}} P_{2}^{x_{2}} \dots P_{n}^{x_{n}}

P_{1}^{y_{1}} P_{2}^{y_{2}} \dots P_{n}^{y_{n}}

, where

\begin{aligned} y_{1} & \equiv a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n}, \\ y_{2} & \equiv a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n}, \\ . . . . . . . . . . . . . \\ y_{n} & \equiv a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{n n} x_{n}, \end{aligned} (mod p) .

Unless the $p^{n}$ operations $P_{1}^{y_{1}} P_{2}^{y_{2}} \dots P_{n}^{y_{n}}$ thus formed are all distinct, when for $P_{1}^{x_{1}} P_{2}^{x_{2}} \dots P_{n}^{x_{n}}$ is put successively each of the $p^{n}$ operations of the group, the symbol does not represent an isomorphism. On the other hand, when this condition is satisﬁed, the symbol represents a permutation of the operations among themselves which leaves the multiplication table of the group unchanged; it is therefore an isomorphism.

If this condition is satisﬁed, $x_{1}$ , $x_{2}$ , …, $x_{n}$ must be deﬁnite numbers $(mod p)$ , when $y_{1}$ , $y_{2}$ , …, $y_{n}$ are given; and therefore the above set of $n$ simultaneous congruences must be capable of deﬁnite solution with respect to the $x$ ’s. The necessary and suﬃcient condition for this is that the determinant

|\begin{matrix} a_{11}, & a_{12}, & \dots, & a_{1 n} \\ a_{21}, & a_{22}, & \dots, & a_{2 n} \\ . . . . . . . . . . . . . . . . . . . . \\ a_{n 1}, & a_{n 2}, & \dots, & a_{n n} \end{matrix}|

should not be congruent to zero

(mod p)

Every distinct set of congruences of the above form, for which this condition is satisﬁed, represents a distinct isomorphism of the group, two sets being regarded as distinct if the congruence

a_{r s} \equiv a'_{r s} (mod p)

does not hold for each corresponding pair of coeﬃcients. Moreover, to the product of two isomorphisms will correspond the set of congruences which results from carrying out successively the operations indicated by the two sets that correspond to the two isomorphisms.

The group of isomorphisms is therefore simply isomorphic with the group of operations deﬁned by all sets of congruences

\begin{aligned} y_{1} & \equiv a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n}, \\ y_{2} & \equiv a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n}, \\ . . . . . . . . . . . . . \\ y_{n} & \equiv a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{n n} x_{n}, \end{aligned} (mod p)

for which the relation

|\begin{matrix} a_{11}, & a_{12}, & \dots, & a_{1 n} \\ a_{21}, & a_{22}, & \dots, & a_{2 n} \\ . . . . . . . . . . . . . . . . . . . . \\ a_{n 1}, & a_{n 2}, & \dots, & a_{n n} \end{matrix}| ≢ 0 (mod p)

is satisﬁed.

172. The group thus deﬁned is of great importance in many branches of analysis. It is known as the linear homogeneous group. In a subsequent Chapter we shall consider some of its more important properties; we may here conveniently determine its order. Let this be represented by $N_{n}$ , so that $N_{r}$ represents the number of distinct solutions of the congruences

|\begin{matrix} a_{11}, & a_{12}, & \dots, & a_{1 r} \\ a_{21}, & a_{22}, & \dots, & a_{2 r} \\ . . . . . . . . . . . . . . . . . . . \\ a_{r 1}, & a_{r 2}, & \dots, & a_{r r} \end{matrix}| ≢ 0 (mod p)

Since the group of isomorphisms of the Abelian group of order $p^{n}$ transforms every one of its operations (identity excepted) into every other, it can be represented as a transitive substitution group of $p^{n} - 1$ symbols, and therefore, if $M$ is the order of the sub-group that keeps $P_{1}$ unchanged,

N_{n} = (p^{n} - 1) M .

Now, in the congruences corresponding to an operation that keeps $P_{1}$ unchanged, we have

a_{11} \equiv 1, a_{12} \equiv a_{13} \equiv \dots \equiv a_{1 n} \equiv 0, (mod p) .

Hence $M$ is equal to the number of solutions of the congruences

|\begin{matrix} 1, & 0, & \dots, & 0 \\ a_{21}, & a_{22}, & \dots, & a_{2 n} \\ . . . . . . . . . . . . . . . . . . . \\ a_{n 1}, & a_{n 2}, & \dots, & a_{n n} \end{matrix}| ≢ 0 (mod p)

The value of the determinant does not depend on the values of $a_{21}$ , $a_{31}$ , …, $a_{n 1}$ , and therefore

M = p^{n - 1} N_{n - 1} .

Hence

N_{n} = (p^{n} - 1) p^{n - 1} N_{n - 1},

and therefore immediately

N_{n} = (p^{n} - 1) (p^{n} - p) \dots (p^{n} - p^{n - 1}) .

The reader will notice that an independent proof of this result has already been obtained in § 48. The discussion there given of the number of distinct ways, in which independent generating operations of an Abelian group of type $(1, 1, \dots, 1)$ may be chosen, is clearly equivalent to a determination of the order of the corresponding group of isomorphisms.

The holomorph of an Abelian group, of order $p^{n}$ and type $(1, 1, \dots, to n units)$ , can similarly be represented as a group of linear transformations to the prime modulus $p$ . Consider, in fact, the set of transformations

\begin{aligned} y_{1} & \equiv a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} + b_{1}, \\ y_{2} & \equiv a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} + b_{2}, \\ . . . . . . . . . . . . . . . . . . \\ y_{n} & \equiv a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{n n} x_{n} + b_{n}, \end{aligned} (mod p);

where the coeﬃcients take all integral values

(mod p)

consistent with

|\begin{matrix} a_{11}, & a_{12}, & \dots, & a_{1 n} \\ a_{21}, & a_{22}, & \dots, & a_{2 n} \\ . . . . . . . . . . . . . . . . . . . . \\ a_{n 1}, & a_{n 2}, & \dots, & a_{n n} \end{matrix}| ≢ 0 .

The set of transformations clearly forms a group whose order is $N_{n} p^{n}$ . The sub-group formed by all the transformations

y_{1} \equiv x_{1} + b_{1}, y_{2} \equiv x_{2} + b_{2}, \dots, y_{n} \equiv x_{n} + b_{n}, (mod p),

is an Abelian group of order

p^{n}

and type

(1, 1, \dots, to n units)

, and it is a self-conjugate sub-group. Moreover, the only operations of the group, which are permutable with every operation of this self-conjugate sub-group, are the operations of the sub-group itself; and, since the order of the group is equal to the order of the holomorph of the Abelian group, it follows that the group of transformations must be simply isomorphic with the holomorph of the Abelian group. In the simplest instance, where

p^{n}

2^{2}

, the holomorph is simply isomorphic with the alternating group of four symbols. In any case the holomorph, when expressed as in § 158, is a doubly transitive group of degree

p^{n}

173. It has been seen in § 142 that, except when $n = 6$ , the symmetric group of degree $n$ has $n$ and only $n$ sub-groups of order $n - 1!$ , which form a conjugate set. Hence by Theorem III, § 161, the group of isomorphisms of the symmetric group of degree $n$ can be expressed, except when $n = 6$ , as a transitive group of degree $n$ . The symmetric group of $n$ symbols however consists of all possible substitutions that can be performed on the $n$ symbols, and therefore it must coincide with its group of isomorphisms. Hence⁶⁶:—

THEOREM VIII. The symmetric group of $n$ symbols is a complete group, except when $n = 6$ .

Corollary. Except when $n = 6$ , the alternating group of $n$ symbols admits of one and only one class of contragredient isomorphisms.

For with this exception, the alternating group of degree $n$ has just $n$ sub-groups of order $\frac{1}{2} (n - 1)!$ .

174. The alternating group of degree $6$ occurs as a special case of another class of groups of which we will determine the groups of isomorphisms. These are the doubly and the triply transitive groups that have been deﬁned in §§ 112, 113.

The doubly transitive group of degree $p^{m}$ and order $p^{m} (p^{m} - 1)$ there considered has a single set of $p^{m}$ conjugate cyclical sub-groups of order $p^{m} - 1$ . Its group of isomorphisms can therefore be expressed as a transitive group of degree $p^{m}$ . Let $p^{m} (p^{m} - 1) μ$ be the order of the group of isomorphisms. The order of a sub-group that keeps two symbols ﬁxed is $μ$ ; and every operation of this sub-group must transform a cyclical sub-group of order $p^{m} - 1$ into itself. With the notation of § 112, we will consider the sub-group which keeps $0$ and $i$ ﬁxed. Every operation of this sub-group must transform the cyclical sub-group generated by the congruence⁶⁷

x' \equiv i x

into itself; and no operation of the sub-group can be permutable with the given operation. If

S

is an operation of the sub-group which transforms

x' \equiv i x

into

x' \equiv i^{α} x,

then

S

, when represented as a substitution, is given by

(\binom{i^{2}, i^{3}, i^{4}, \dots}{i^{α + 1}, i^{2 α + 1}, i^{3 α + 1}, \dots}) .

Now $S$ must transform the sub-group of order $p^{m}$ into itself; it must therefore be permutable with that operation of this sub-group which changes $0$ into $i$ .

This operation is given by

x' \equiv x + i,

and if we denote it by

T

, then

S^{- 1} T S

changes

i^{y + 1}

into

i^{α z + 1}

, where

i^{\frac{y}{α} + 1} + i \equiv i^{z + 1} .

Now $T$ changes $i^{y + 1}$ into $i^{y + 1} + i$ . Hence, since $S$ and $T$ are permutable, we must for all values of $y$ have the simultaneous congruences

i^{\frac{y}{α} + 1} + i \equiv i^{z + 1},

and

i^{y + 1} + i \equiv i^{α z + 1} .

Eliminating $z$ , the congruence

1 + i^{y} \equiv {(1 + i^{\frac{y}{α}})}^{α}

must hold for all values of

y

from

0

p^{m} - 1

. If

α

is not a power of

p

, this congruence involves an identity of the form

a_{1} i^{α_{1}} + a_{2} i^{α_{2}} + \dots \equiv 0,

where all the indices are less than

p^{m} - 1

; and this is impossible. Hence the only possible values of

α

are

p

p^{2}

p^{3}

, …; and the greatest possible value of

μ

m

Now the congruence

x' \equiv x^{p}

deﬁnes a substitution performed on the

p^{m}

symbols permuted by the group, and this substitution is permutable with the group. For if we denote this operation by

J

, and any operation

x' \equiv α x + β

of the group by

Σ

, then

J^{- 1} Σ J

x' \equiv α^{p} x + β^{p},

another operation of the group. Moreover,

J

clearly transforms

x' \equiv i x

into its

p

th power.

The group of isomorphisms of the doubly transitive group of order $p^{m}$ is therefore the group deﬁned by

x' \equiv i x, x' \equiv x + i, x' \equiv x^{p};

where

i

is a primitive root of the congruence

i^{p^{m - 1}} \equiv 1 .

From this it immediately follows that the group of isomorphisms of the triply transitive group, of degree $p^{m} + 1$ and order

(p^{m} + 1) p^{m} (p^{m} - 1),

deﬁned by the congruences

x' \equiv \frac{α x + β}{γ x + δ},

where

α

β

γ

δ

satisfy the conditions of § 113, is the group of order

(p^{m} + 1) p^{m} (p^{m} - 1) m

obtained by combining the previous congruences with

x' \equiv x^{p} .

In fact, it may be immediately veriﬁed that the operation given by this congruence is permutable with the group, and does not give a cogredient isomorphism of the group. Moreover, by Theorem III, § 161, the group of isomorphisms of the given group can be expressed as a transitive group of degree $p^{m} + 1$ ; and therefore, among a class of contragredient isomorphisms, there must be some transforming into itself a sub-group which keeps one symbol ﬁxed. Hence the order of the group of isomorphisms cannot exceed

(p^{m} + 1) p^{m} (p^{m} - 1) m .

When $p$ is an odd prime, the triply transitive group of degree $p^{m} + 1$ , which may be deﬁned by

x' \equiv - \frac{1}{x}, x' \equiv i x, x' \equiv x + i,

contains as a self-conjugate sub-group a doubly transitive group deﬁned by

x' \equiv - \frac{1}{x}, x' \equiv i^{2} x, x' \equiv x + i .

It will be shewn in Chapter XIV that this is a simple group. When $p^{m}$ is equal to $3^{2}$ , it is easy to verify that this group is simply isomorphic with the alternating group of degree $6$ .

The group of isomorphisms of this simple group can be expressed as a transitive group of degree $p^{m} + 1$ , and must clearly contain the triply transitive group of order $(p^{m} + 1) p^{m} (p^{m} - 1)$ self-conjugately. It therefore coincides with the group of order $(p^{m} + 1) p^{m} (p^{m} - 1) m$ , which has just been determined. The latter group is therefore (Theorem VI, Cor. § 166) a complete group. Further, if the simple group be denoted by $G$ and the group of isomorphisms by $L$ , the factor group $\frac{L}{G}$ will be determined by

x' \equiv i x, x' \equiv x^{p},

when all operations of

G

are treated as the identical operation. Denoting these operations by

I

and

J

, then

J^{- 1} I J

I^{p}

, which is the same as

I

multiplied by an operation of

G

. Hence

\frac{L}{G}

is an Abelian group generated by two independent operations of orders

2

and

m

The group of isomorphisms of the alternating group of degree $6$ is therefore a group of order $1440$ ; and the symmetric group of degree $6$ admits a single class of contragredient isomorphisms⁶⁸.

175. Let $P$ be any group whose order is the power of a prime, and let

P, P_{1}, P_{2}, \dots, P_{n}, 1

of orders

p^{α}, p^{α_{1}}, p^{α_{2}}, \dots, P^{α_{n}}, 1

be a characteristic series (§ 163) of

P

. Every isomorphism of

P

must transform

P_{r}

and

P_{r + 1}

into themselves, and therefore also

\frac{P_{r}}{P_{r + 1}}

into itself. Suppose now that an isomorphism

I

P

transforms every operation of

P_{r + 1}

into itself and every operation of

\frac{P_{r}}{P_{r + 1}}

into itself. If

S

is any operation of

P_{r}

, not contained in

P_{r + 1}

I

must transform

S

into

S A

, where

A

is some operation of

P_{r + 1}

; so that, if

p^{μ}

is the order of

A

I

transforms the operations of the set

S P_{r + 1}

cyclically among themselves in sets of

p^{μ}

. Similarly, if

S'

is an operation of

P_{r}

not contained in the set

S P_{r + 1}

I

will transform the operations of the set

S' P_{r + 1}

cyclically among themselves in sets of

p^{μ'}

. Hence the order of the isomorphism

I

is a multiple of

p

; and any isomorphism, that transforms every operation of

P_{r + 1}

into itself and every operation of

\frac{P_{r}}{P_{r + 1}}

into itself and is of order prime to

p

, will transform every operation of

P_{r}

into itself. Therefore, the only isomorphism of

P

, that transforms every operation of each of the groups

\frac{P}{P_{1}}, \frac{P_{1}}{P_{2}}, \dots, \frac{P_{n - 1}}{P_{n}}, P_{n}

into itself and is of order prime to

p

, is the identical isomorphism. Now each of these groups is an Abelian group, whose operations are all of order

p

; and it has been shewn that, if

p^{m}

is the order of such a group, the order of its group of isomorphisms is

(p^{m} - 1) (p^{m} - p) \dots (p^{m} - p^{m - 1}) .

Every isomorphism of $P$ , whose order is relatively prime to $p$ , must therefore be such that its order is a factor of one of the expressions of the above form, obtained by writing

α - α_{1}, α_{1} - α_{2}, \dots, α_{n},

in succession for

m

. If then

k

is the greatest of these numbers, the order of any isomorphism of

P

, whose order is relatively prime to

p

, is a factor of

(p - 1) (p^{2} - 1) \dots (p^{k} - 1) .

Herr Frobenius⁶⁹ has introduced the symbol $𝜗 (P)$ to denote this product. If $G$ is a group of order $p_{1}^{α_{1}} p_{2}^{α_{2}} \dots p_{n}^{α_{n}}$ , where $p_{1}$ , $p_{2}$ , …, $p_{n}$ are distinct primes, and if $P_{1}$ , $P_{2}$ , …, $P_{n}$ are groups of orders $p_{1}^{α_{1}}$ , $p_{2}^{α_{2}}$ , …, $p_{n}^{α_{n}}$ contained in $G$ , we shall use the symbol $𝜗 (G)$ to denote the least common multiple of

𝜗 (P_{1}), 𝜗 (P_{2}), \dots, 𝜗 (P_{n}) .

176. If $P'$ is a sub-group of a group $P$ of order $p^{α}$ , $𝜗 (P')$ is not necessarily a factor of $𝜗 (P)$ . For instance, the group of order $p^{4}$ , generated by the four operations $A$ , $B$ , $C$ , $D$ of order $p$ , of which $A$ and $B$ are self-conjugate while

D^{- 1} C D = C A,

has a characteristic sub-group

{A, B}

of order

p^{2}

, and it has no characteristic sub-group of order

p^{3}

. Hence

𝜗 (P) = (p - 1) (p^{2} - 1) .

The sub-group $P'$ however, which is generated by $A$ , $B$ and $C$ , is an Abelian group whose operations are all of order $p$ ; and therefore

𝜗 (P') = (p - 1) (p^{2} - 1) (p^{3} - 1) .

So again, for the Abelian group $P$ of order $p^{3}$ , generated by $A$ and $B$ , where

A^{p^{2}} = 1, B^{p} = 1, A B = B A,

we have

𝜗 (P) = p - 1;

while its sub-group

P'

, generated by

A^{p}

and

B

, is an Abelian group whose operations are all of order

p

, so that

𝜗 (P') = (p - 1) (p^{2} - 1) .

Suppose now that $P$ is an Abelian group of order $p^{α}$ , generated by $ρ$ independent and permutable operations. For such a group, we deﬁne⁷⁰ a new symbol $𝜃 (P)$ by the equation

𝜃 (P) = (p - 1) (p^{2} - 1) \dots (p^{ρ} - 1) .

In forming the characteristic series of $P$ in § 164, we commenced with the series of groups $G_{r}$ ( $r = 1, 2, \dots$ ), such that $G_{r}$ consists of all the operations of $P$ satisfying the relation

S^{p^{r}} = 1 .

Since $P$ is generated by $ρ$ independent operations, the order of $G_{1}$ is $p^{ρ}$ , and the order of $\frac{G_{r + 1}}{G_{r}}$ cannot be greater than $p^{ρ}$ . If the generating operations of $P$ are all of the same order $β$ , the series

P, G_{β - 1}, G_{β - 2}, \dots, G_{1}, 1

is a complete characteristic series of

P

, and each factor-group

\frac{G_{r + 1}}{G_{r}}

is of order

p^{ρ}

. In this case, therefore,

𝜗 (P) = 𝜃 (P) .

If however the generating operations are not all of the same order, $G_{1}$ will not be the last term of the complete characteristic series; nor will $G_{r + 1}$ and $G_{r}$ be consecutive terms in the series, if the order of $\frac{G_{r + 1}}{G_{r}}$ is $p^{ρ}$ . Hence, in this case, $𝜗 (P)$ will be a factor of $𝜃 (P)$ .

If now $P'$ is any sub-group of $P$ , then it has been seen (§ 46) that the number of independent generating operations of $P'$ is equal to or less than $ρ$ . Hence $𝜃 (P')$ is equal to or is a factor of $𝜃 (P)$ .

177. THEOREM IX. If a group $G$ , of order $N$ , is transformed into itself by an operation $S$ , whose order is relatively prime to $N 𝜗 (G)$ , every operation of $G$ is permutable with $S$ ⁷¹.

Let $p_{1}^{α_{1}}$ be the highest power of a prime $p_{1}$ which divides $N$ . The number of sub-groups of $G$ , and therefore also of ${S, G}$ , whose order is $p_{1}^{α_{1}}$ is a factor of $N$ ; and since the order of $S$ is relatively prime to $N$ , one at least of them, say $P_{1}$ , must be transformed into itself by $S$ . Now the order of $S$ is relatively prime both to $p_{1}$ and to $𝜗 (P_{1})$ ; and therefore the isomorphism of $P$ given on transforming its operations by $S$ must be the identical isomorphism. In other words, $S$ is permutable with every operation of $P_{1}$ . In the same way it follows that, if $p_{2}^{α_{2}}$ , $p_{3}^{α_{3}}$ , … are the highest powers of $p_{2}$ , $p_{3}$ , … that divide $N$ , there must be sub-groups of orders $p_{2}^{α_{2}}$ , $p_{3}^{α_{3}}$ , … with every operation of each of which $S$ is permutable. But these groups of orders $p_{1}^{α_{1}}$ , $p_{2}^{α_{2}}$ , $p_{3}^{α_{3}}$ , … generate $G$ ; therefore $S$ is permutable with every operation of $G$ .

Corollary. If the order of an isomorphism of $G$ is relatively prime to $N$ , it must be a factor of $𝜗 (G)$ .

It follows immediately, from the theorem, that there is no isomorphism of $G$ whose order contains a prime factor not occurring in $N 𝜗 (G)$ . Suppose, if possible, that the group of isomorphisms of $G$ contains an operation $S$ of order $q^{m}$ , where $q$ is a prime which does not divide $N$ , and that the highest power of $q$ that divides $𝜗 (G)$ is $q^{m'}$ , where $m' < m$ . If $p_{r}^{α_{r}}$ is the highest power of any prime $p_{r}$ which divides $N$ , $S$ must transform some sub-group $P_{r}$ of order $p_{r}^{α_{r}}$ into itself. Since $q^{m}$ is not a factor of $𝜗 (P_{r})$ , some power of $S$ must be permutable with every operation of $P_{r}$ . Hence, as in the theorem, it follows that some power of $S$ , certainly $S^{q^{m - 1}}$ , must be permutable with every operation of $G$ . But no operation of the group of isomorphisms of $G$ , except identity, is permutable with every operation of $G$ . Hence the group of isomorphisms cannot contain an operation of order $q^{m}$ , if $m > m'$ ; and therefore there is no isomorphism of $G$ whose order contains a higher power of $q$ than $q^{m'}$ . If then $q^{r} {q'}^{r'} \dots$ , a number relatively prime to $N$ , is the order of an isomorphism of $G$ , all the numbers $q^{r}$ , ${q'}^{r'}$ , … are factors of $𝜗 (G)$ , and so also is their product.

178. The method, by which it has been shewn in § 175 that the order of any isomorphism of $P_{r}$ which transforms every operation of each of the groups $\frac{P_{r}}{P_{r + 1}}$ and $P_{r + 1}$ into itself is a power of $p$ , may be used to obtain the following more general result.

THEOREM X. If $H$ is a self-conjugate sub-group of $G$ , the order of an isomorphism of $G$ , which transforms every operation of each of the groups $\frac{G}{H}$ and $H$ into itself, is a factor of the order of $H$ .

If $S$ is any operation of $G$ not contained in $H$ , the isomorphism will change $S$ into $S h$ , where $h$ is some operation of $H$ . If then $m$ is the order of $h$ , the isomorphism transforms

S, S h, S h^{2}, \dots, S h^{m - 1}

cyclically; and therefore it transforms all the operations of the set

S H

in cycles of

m

each. If

S'

is any operation of

G

not contained in

S H

, the isomorphism will interchange the operations of the set

S' H

among themselves in cycles of

m'

each, where

m'

again is the order of some operation of

H

. The isomorphism, when expressed as a substitution performed on the operations of

G

, will consist of a number of cycles of

m

m'

, … symbols; and its order is therefore the least common multiple of

m

m'

, …. Now if

q

is any prime that divides the order of

H

, and

q^{n}

the highest power of

q

that occurs as the order of a cyclical operation of

H

, no power of

q

higher than

q^{n}

can occur in any of the numbers

m

m'

, …; and

q^{n}

is therefore the highest power of

q

that can occur in their least common multiple. This least common multiple, which is the order of the isomorphism, must therefore divide the order of

H

179. Ex. 1. Shew that, for the group of order $p^{3}$ deﬁned by

\begin{matrix} P^{p} = 1, Q^{p} = 1, R^{p} = 1, Q^{- 1} P Q = P R, \\ R P = P R, R Q = Q R, \end{matrix}

the symbol

(\binom{P, Q, R}{P^{x} Q^{y} R^{z}, P^{α} Q^{β} R^{γ}, R^{n}})

gives an isomorphism if

β x - α y \equiv n, (mod p) .

Hence determine the order of the group of isomorphisms.

Ex. 2. Shew that, for the group of order $p^{3}$ deﬁned by

P^{p^{2}} = 1, Q^{p} = 1, Q^{- 1} P Q = P^{1 + p},

the symbol

(\binom{P, Q}{P^{x} Q^{y}, P^{α p} Q})

gives an isomorphism if

x

is not a multiple of

p

; and determine the order of the group of isomorphisms.

Ex. 3. Shew that the group of isomorphisms of the group of order $2^{n}$ , deﬁned by (§ 63)

P^{2^{n - 1}} = 1, Q^{2} = P^{2^{n - 2}}, Q^{- 1} P Q = P^{- 1},

is of order

2^{2 n - 3}

, when

n > 3

. If

n = 3

, its order is

24

and it is simply isomorphic with the last type but one of § 84.

Ex. 4. If $G$ is a complete group of order $N$ , shew that the order of $K$ , the holomorph of $G$ , is $N_{2}$ ; and that the order of the holomorph of $K$ is $2 N^{4}$ .

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Chapter XI.On the Isomorphism of a Group with Itself.

Chapter XI.
On the Isomorphism of a Group with Itself.