XIV. On the Linear Group86.

Chapter XIV.
On the Linear Group⁸⁷.

W e shall now, in illustration of the general principles that have been developed in the preceding chapters, proceed to discuss and give an analysis of certain special groups. The ﬁrst that we choose for this purpose is the group of isomorphisms of an Abelian group of order $p^{n}$ and type $(1, 1, \dots, to n units)$ . This group has been deﬁned and its order determined in §§ 171, 172. It is there shewn that the group is simply isomorphic with the homogeneous linear group 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),

whose determinants are not congruent to zero; and that its order is

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

The operation given by the above set of congruences will be denoted in future by the symbol

(a_{11} x_{1} + \dots + a_{1 n} x_{n}, a_{21} x_{1} + \dots + a_{2 n} x_{n}, \dots, a_{n 1} x_{1} + \dots + a_{n n} x_{n}) .

216. It may be easily veriﬁed by direct calculation that, if $D$ and $D'$ are the determinants of two operations $S$ and $S'$ of the group, then $D D'$ is the determinant of the operation $S S'$ , all the numbers involved being reduced, $(mod p)$ . Hence it immediately follows that those operations of the group, whose determinant is unity, form a sub-group. If this sub-group is denoted by $Γ$ , the group itself being $G$ , then $Γ$ is a self-conjugate sub-group of $G$ . For if $Σ$ is any operation of $Γ$ and $S$ any operation of $G$ whose determinant is $D$ , the determinant of $S^{- 1} Σ S$ is $D^{- 1} D$ or unity; and therefore $S^{- 1} Σ S$ belongs to $Γ$ . Suppose now that $S$ is an operation⁸⁸ of $G$ whose determinant is $z$ , a primitive root of the congruence

z^{p - 1} \equiv 1 (mod p) .

Then the determinant of every operation of the set

S^{r} Γ

z^{r}

; and therefore, if

r

and

s

are not congruent

(mod p)

, the two sets

S^{r} Γ and S^{s} Γ

can have no operation in common. Moreover, if

S'

is any operation of

G

whose determinant is

z^{r}

, then

S^{- r} S'

belongs to

Γ

, and therefore

S'

belongs to the set

S^{r} Γ

. Hence ﬁnally, the sets

Γ, S Γ, S^{2} Γ, \dots, S^{p - 2} Γ

are all distinct, and they include every operation of

G

; so that

G = {S, Γ} .

The factor-group $\frac{G}{Γ}$ is therefore cyclical and of order $p - 1$ .

217. It may be very readily veriﬁed that the operations of the cyclical sub-group generated by

(z x_{1}, z x_{2}, \dots, z x_{n})

are self-conjugate operations of

G

. To prove that these are the only self-conjugate operations of

G

, we will deal with the case

n = 3

: it will be seen that the method is perfectly general. Suppose then that

T = (α x_{1} + β x_{2} + γ x_{3}, α' x_{1} + β' x_{2} + γ' x_{3}, α'' x_{1} + β'' x_{2} + γ'' x_{3})

is a self-conjugate operation of

G

, while

S = (a x_{1} + b x_{2} + c x_{3}, a' x_{1} + b' x_{2} + c' x_{3}, a'' x_{1} + b'' x_{2} + c'' x_{3})

is any operation. The relation

S T = T S

involves the nine simultaneous congruences⁸⁹

\begin{array}{l} a α + b α' + c α'' & \equiv α a + β a' + γ a'', \\ a β + b β' + c β'' & \equiv α b + β b' + γ b'', \\ a γ + b γ' + c γ'' & \equiv α c + β c' + γ c'', \\ etc., & etc.; \end{array}

and these must be satisﬁed for all possible values of the coeﬃcients of $S$ . Now

b \equiv c \equiv a' \equiv 0

is a possible relation between the coeﬃcients of

S

, whether regarded as an operation of

G

Γ

; and therefore

γ \equiv 0 .

In the same way, it may be shewn that

β \equiv α' \equiv γ' \equiv α'' \equiv β'' \equiv 0,

and that

α \equiv β' \equiv γ'';

so that

T

is a power of the operation

(z x_{1}, z x_{2}, z x_{3}) .

The only self-conjugate operations of

G

are therefore the powers of

A

, where

A

denotes

(z x_{1}, z x_{2}, \dots, z x_{n});

and the only self-conjugate operations of

Γ

are those operations of this cyclical sub-group which are contained in

Γ

. Now the order of

A

p - 1

and its determinant is

z^{n}

. Hence the self-conjugate operations of

Γ

form a cyclical sub-group

D

of order

d

, where

d

is the greatest common factor of

p - 1

and

n

; and this sub-group is generated by

A^{\frac{p - 1}{d}}

218. To determine completely the composition-series of $G$ , it is necessary to ﬁnd whether $Γ$ has a self-conjugate sub-group greater than and containing $D$ . A simple calculation will shew that, from

(x_{1} + x_{s}, x_{2}, x_{3}, \dots, x_{n})

and its conjugate operations, all the operations of

Γ

may be generated; and hence no self-conjugate sub-group of

Γ

which is diﬀerent from

Γ

itself can contain an operation of this form. If then it is shewn that any self-conjugate sub-group of

Γ

, distinct from

D

, necessarily contains operations of this form, it follows that

D

is a maximum self-conjugate sub-group of

Γ

We shall ﬁrst deal with the case $n = 2$ .

If $p = 2$ , the orders of $G$ , $Γ$ and $D$ are $6$ , $6$ and $1$ . In this case, $Γ$ is simply isomorphic with the symmetric group of three symbols, which has a self-conjugate sub-group of order $3$ . The successive factor-groups of the composition-series of $G$ are therefore cyclical groups of orders $2$ and $3$ .

If $p = 3$ , the orders of $G$ , $Γ$ and $D$ are $48$ , $24$ and $2$ . The factor-group $\frac{Γ}{D}$ has $12$ for its order, and cannot therefore be a simple group. The reader will have no diﬃculty in verifying that, in this case, the successive factor-groups of $G$ have orders $2$ , $3$ , $2$ , $2$ and $2$ . We may therefore, in dealing with the case $n = 2$ , assume that $p$ is not less than $5$ .

Let us suppose now that $Γ$ has a self-conjugate sub-group $I$ that contains $D$ ; and let $S$ or

(a x_{1} + b x_{2}, a' x_{1} + b' x_{2})

be one of its operations, not contained in

D

If $b$ is diﬀerent from zero, $Γ$ contains $Σ$ , where $Σ$ denotes

(α a x_{1} + α b x_{2}, - \frac{1 + α^{2} a^{2}}{α b} x_{1} - α a x_{2}),

and therefore

I

contains

Σ^{- 1} S Σ S

, which is

(- α^{- 2} x_{1}, - \frac{(1 + α^{2}) (b' + a α^{2})}{b α^{2}} x_{1} - α^{2} x_{2}) .

If $b$ is zero, $b'$ is congruent with $a^{- 1}$ ; therefore, in any case, $I$ contains an operation $S'$ of the form

(c x_{1}, d x_{1} + c^{- 1} x_{2}) .

Again, $Γ$ contains the operation $T$ , where $T$ denotes

(x_{1}, x_{1} + x_{2});

and

I

therefore contains

S' T^{- 1} {S'}^{- 1} T

, which is

(x_{1}, (1 - c^{2}) x_{1} + x_{2}) .

Hence unless $1 - c^{2} \equiv 0$ , $I$ must coincide with $Γ$ . Now, when $p > 5$ , $c$ can always be chosen so that this congruence is not satisﬁed. If $p = 5$ , the square of the above operation $Σ^{- 1} S Σ S$ , when unity is written for $α$ , is

(x_{1}, \frac{4}{b} (b' + a) x_{1} + x_{2});

unless

b' + a \equiv 0

, this again requires that

I

coincides with

Γ

. If ﬁnally, the condition

b' + a \equiv 0

is satisﬁed in

S

, it is not satisﬁed in

S T^{- 1} S^{- 1} T

, another operation belonging to

I

; and therefore again, in this case,

I

coincides with

Γ

Hence ﬁnally, if $n = 2$ , the factor-group $\frac{Γ}{D}$ is simple, except when $p$ is $2$ or $3$ .

219. When $n$ is greater than $2$ , it will be found that it is suﬃcient to deal in detail with the case $n = 3$ , as the method will apply equally well for any greater value of $n$ . Suppose here again that $Γ$ has a self-conjugate sub-group $I$ which contains $D$ ; and let $S$ , denoting

(a x_{1} + b x_{2} + c x_{3}, a' x_{1} + b' x_{2} + c' x_{3}, a'' x_{1} + b'' x_{2} + c'' x_{3}),

be one of the operations of

I

which is not contained in

D

S

cannot be permutable with all operations of the form

(x_{1}, x_{2}, x_{3} + x_{1})

, as it would then be permutable with every operation of

Γ

. We may therefore suppose without loss of generality that

S

and

T

are not permutable,

T

denoting

(x_{1}, x_{2}, x_{3} + x_{1})

. Then

T^{- 1} S T S^{- 1}

is an operation, distinct from identity, belonging to

I

. Now a simple calculation shews that this operation, say

U

, is of the form

(x_{1} - c X, x_{2} - c' X, A x_{1} + B x_{2} + C x_{3}),

where

X

is the symbol with which

S^{- 1}

replaces

x_{1}

If $c$ and $c'$ are both diﬀerent from zero, $Γ$ will contain an operation $V$ of the form

(x_{1} - \frac{c}{c'} x_{2}, x_{2}, x_{3});

and

I

contains

V^{- 1} U V

, which is of the form

(x_{1}, α x_{1} + β x_{2} + γ x_{3}, α' x_{1} + β' x_{2} + γ' x_{3}) .

Moreover, if either $c$ or $c'$ is zero, the operation $U$ itself leaves one symbol unaltered. Hence $I$ always contains operations by which one symbol is unaltered.

This process may now be repeated to shew that $I$ necessarily contains operations of the form

(x_{1}, x_{2}, α'' x_{1} + β'' x_{2} + γ'' x_{3});

and, since the determinant of the operation is unity,

γ''

is necessarily congruent to unity. But it has been seen that the group

Γ

is generated from the last operation and the operations conjugate to it. Hence ﬁnally, if

n

is greater than

2

, the factor-group

\frac{Γ}{D}

is simple for all values of

p

220. The composition-series of $G$ is now, except as regards the constitution of the simple group $\frac{Γ}{D}$ , perfectly deﬁnite. It has, in fact, been seen that $\frac{G}{Γ}$ and $D$ are cyclical groups of orders $p - 1$ and $d$ ; and therefore if $α$ , $β$ , $γ$ , … are distinct primes whose product is $p - 1$ , and if $α'$ , $β'$ , $γ'$ , … are distinct primes whose product is $d$ ; the successive factor-groups of $G$ are ﬁrst, a series of simple groups of prime orders $α$ , $β$ , $γ$ , …: then a simple group of composite order $\frac{N}{(p - 1) d}$ : and lastly, a series of simple groups of prime orders $α'$ , $β'$ , $γ'$ , ….

The sequence in which the set of simple groups of orders $α$ , $β$ , $γ$ , … are taken in the composition-series may be clearly any whatever, and the same is true of the set of factor-groups of orders $α'$ , $β'$ , $γ'$ , …; but it is to be noticed that, when $d$ is not equal to $p - 1$ , the composition-series is capable of further modiﬁcations. In this case, ${A, Γ}$ is a self-conjugate sub-group of $G$ of order $\frac{N}{d}$ , which has a maximum self-conjugate sub-group ${A}$ of order $p - 1$ . The successive composition-factors of $G$ may therefore be taken in the sequence

α', β', γ', \dots : \frac{N}{(p - 1) d} : α, β, γ, \dots :

and their arrangement may be yet further changed by considering the self-conjugate sub-group

{A^{m}, Γ}

, where

m

is a factor of

p - 1

less than

\frac{p - 1}{d}

221. For every value of $p^{n}$ , except $2^{2}$ and $3^{2}$ , it thus appears that the linear group may be regarded as deﬁning a simple group of composite order. We shall now proceed to a discussion of the constitution of the simple groups thus deﬁned when $n = 2$ , $p$ being greater than $3$ ⁹⁰. In this case, the group $Γ$ is deﬁned by the congruences

\begin{aligned} y_{1} & \equiv α x_{1} + β x_{2}, \\ y_{2} & \equiv γ x_{1} + δ x_{2}, \\ α δ - β γ & \equiv 1, \end{aligned} (mod p);

and since

p - 1

is divisible by

2

when

p

is an odd prime,

d

is equal to

2

. Hence the self-conjugate operations of

Γ

are

(x_{1}, x_{2}) and (- x_{1}, - x_{2}) .

The order of $Γ$ is $p (p^{2} - 1)$ , and therefore the order of the simple group, $H$ , which it deﬁnes is $\frac{1}{2} p (p^{2} - 1)$ . Suppose now, if possible, that $Γ$ contains a sub-group $g$ simply isomorphic with $H$ . If $S$ is any operation of $Γ$ , not contained in $g$ , the whole of the operations of $Γ$ are contained in the two sets

g, S g .

Now $(x_{2}, - x_{1})$ , whose square $(- x_{1}, - x_{2})$ is a self-conjugate operation, cannot be contained in the simple group $g$ . Hence both $(x_{2}, - x_{1})$ and $(- x_{1}, - x_{2})$ are contained in $S g$ , an obvious contradiction. Therefore $Γ$ contains no sub-group simply isomorphic with $H$ .

For a discussion of the properties of $H$ , some concrete representation of the group itself is necessary; this may be obtained in the following way. Instead of the pair of homogeneous congruences that deﬁne each operation of $Γ$ , let us, as in § 113, consider the single non-homogeneous congruence

y \equiv \frac{α x + β}{γ x + δ}, (mod p),

where

α δ - β γ \equiv 1 .

Corresponding to every operation

(α x_{1} + β x_{2}, γ x_{1} + δ x_{2})

Γ

, there will be a single operation of this new set; namely that in which

α

β

γ

δ

have respectively the same values. But since the operations

y \equiv \frac{α x + β}{γ x + δ} and y \equiv \frac{- α x - β}{- γ x - δ}

are identical, two operations

(α x_{1} + β x_{2}, γ x_{1} + δ x_{2}) and (- α x_{1} - β x_{2}, - γ x_{1} - δ x_{2})

Γ

will correspond to each operation

y \equiv \frac{α x + β}{γ x + δ}

of the new set; the two self-conjugate operations

(x_{1}, x_{2}) and (- x_{1}, - x_{2}),

in particular, corresponding to the identical operation of the new set. Moreover, direct calculation immediately veriﬁes that, to the product of any two operations of

Γ

, corresponds the product of the two corresponding operations of the new set. Hence the new set of operations forms a group of order

\frac{1}{2} p (p^{2} - 1)

, with which

Γ

is multiply isomorphic; the group of order

2

formed by the self-conjugate operations of

Γ

corresponding to the identical operation of the new group.

The simple group $H$ , of order $\frac{1}{2} p (p^{2} - 1)$ , which we propose to discuss, can therefore be represented by the set of operations

y \equiv \frac{α x + β}{γ x + δ}, (mod p);

where

α δ - β γ \equiv 1,

α

β

γ

δ

being integers reduced to modulus

p

222. Since the order of $H$ is divisible by $p$ and not by $p^{2}$ , the group must contain a single conjugate set of sub-groups of order $p$ . Now the operation

y \equiv x + 1,

(x + 1)

as we will write it in future, is clearly an operation of order

p

: for its

n

th power is

(x + n)

, and

p

is the smallest value of

n

for which this is the identical operation. If

(x + 1)

and

(\frac{α x + β}{γ x + δ})

are represented by

P

and

S

, then

S^{- 1} P S = (\frac{(1 - α γ) x + α^{2}}{- γ^{2} x + 1 + α γ}) .

This is identical with $P$ , only if

γ \equiv 0, α^{2} \equiv 1;

and therefore

P

is permutable with no operations except its own powers. On the other hand, if

γ \equiv 0,

then

S^{- 1} P S = P^{a^{2}};

and therefore every operation, for which

γ \equiv 0

, transforms the sub-group

{P}

into itself. These operations therefore form a sub-group: a result that may also be easily veriﬁed directly. The order of this sub-group is the number of distinct operations

(\frac{α x + β}{δ})

for which

α δ \equiv 1

. The ratio

\frac{α}{δ}

must be a quadratic residue, while

β

may have any value whatever. Hence the order of the sub-group is

\frac{1}{2} (p - 1) p

; and

H

therefore contains

p + 1

sub-groups of order

p

. Since

H

is a simple group, it follows (§ 125) that it can be represented as a transitive substitution group of degree

p + 1

This representation of the group can be directly derived, as in § 113, from the congruences already used to deﬁne it. Thus if, in

y \equiv \frac{α x + β}{γ x + δ},

we write for

x

successively

0

1

2

, …,

p - 1

\infty

, the

p + 1

values obtained for

y

, when reduced

(mod p)

, will be the same

p + 1

symbols in some other sequence. For if

\frac{α x_{1} + β}{γ x_{1} + δ} \equiv \frac{α x_{2} + β}{γ x_{2} + δ},

then

(α δ - β γ) (x_{1} - x_{2}) \equiv 0,

and therefore

x_{1} \equiv x_{2}

Each operation of $H$ gives therefore a distinct substitution performed on the symbols $0$ , $1$ , …, $p - 1$ , $\infty$ ; and the complete set of substitutions thus obtained gives the representation of $H$ as a transitive substitution group of degree $p + 1$ . Since $H$ contains operations of order $p$ , this substitution group must be doubly transitive. That this is the case may also be shewn directly. Thus

\frac{y - a}{y - b} \equiv m \frac{x - a'}{x - b'}

is an operation changing

a'

into

a

and

b'

into

b

. This operation may be written

y \equiv \frac{k (b m - a) x + k (a b' - m a' b)}{k (m - 1) x + k (b' - m a')},

and its determinant is

k^{2} m (b - a) (b' - a') .

If now $(b - a) (b' - a')$ is a quadratic residue (or non-residue) $(mod p)$ , $m$ may be any quadratic residue (or non-residue); and $k$ can always be chosen so that the determinant is unity. There are therefore $\frac{1}{2} (p - 1)$ substitutions in the group, changing any two symbols $a'$ , $b'$ into any other two given symbols $a$ , $b$ . Further, if the operation $(\frac{α x + β}{γ x + δ})$ keeps $x$ unchanged in the substitution group, $x$ must satisfy the congruence

x \equiv \frac{α x + β}{γ x + δ},

that is

γ x^{2} + (δ - a) x - β \equiv 0 .

Such a congruence cannot have more than two roots; and therefore every substitution displaces all, all but one, or all but two, of the

p + 1

symbols.

223. The substitutions, which keep either one or two symbols ﬁxed, must therefore be regular in the remaining $p$ or $p - 1$ symbols. Hence the order of every substitution which keeps just one symbol ﬁxed must be $p$ ; and the order of every substitution that keeps two symbols ﬁxed must be equal to or be a factor of $p - 1$ . Now it was seen in the last paragraph that the order of the sub-group that keeps two symbols ﬁxed is $\frac{1}{2} (p - 1)$ . Moreover, if $z$ is a primitive root $(mod p)$ , the sub-group that keeps $a$ and $b$ ﬁxed contains the operation

\frac{y - a}{y - b} = z^{2} \frac{x - a}{x - b},

and the order of this operation is

\frac{1}{2} (p - 1)

. Hence, the sub-group that keeps any two symbols ﬁxed is a cyclical group of order

\frac{1}{2} (p - 1)

; and every operation that keeps two symbols ﬁxed is some power of an operation of order

\frac{1}{2} (p - 1)

. Since the group is a doubly transitive group of degree

p + 1

, there must be

\frac{1}{2} (p + 1) p

sub-groups which keep two symbols ﬁxed; and these must form a conjugate set. Each is therefore self-conjugate in a sub-group of order

p - 1

. To determine the type of this sub-group, we may consider the sub-group keeping

0

and

\infty

ﬁxed: this is generated by

Q

, where

Q

denotes

(\frac{z x}{z^{- 1}})

. If

(\frac{α x + β}{γ x + δ})

is represented by

S

, then

S^{- 1} Q S = (\frac{(z α δ - z^{- 1} β γ) x - (z - z^{- 1}) α β}{(z - z^{- 1}) γ δ x + z^{- 1} α δ - z β γ}),

which can be a power of

Q

only if

α β \equiv 0, γ δ \equiv 0 .

Hence either

β \equiv γ \equiv 0,

in which case

S

is a power of

Q

: or

α \equiv δ \equiv 0, γ \equiv - β^{- 1} .

In the latter case, we have

S = (\frac{- β}{β^{- 1} x}),

which is an operation of order

2

; and then

S^{- 1} Q S = (\frac{z^{- 1} x}{z}) = Q^{- 1} .

The group of order $p - 1$ , which contains self-conjugately a cyclical sub-group of order $\frac{1}{2} (p - 1)$ that keeps two symbols ﬁxed, is therefore a group of dihedral (§ 202) type. Moreover, if $t$ is any factor of $p - 1$ , this investigation shews that ${S, Q}$ is the greatest sub-group that contains ${Q^{t}}$ self-conjugately.

224. A substitution that changes all the symbols must either be regular in the $p + 1$ symbols, or must be such that one of its powers keeps two symbols ﬁxed. The latter case however cannot occur; for we have just seen that, if $Q$ is an operation, of order $\frac{1}{2} (p - 1)$ , which keeps two symbols ﬁxed, the only operations permutable with $Q^{r}$ are the powers of $Q$ . Hence the substitutions that change all the symbols must be regular in the $p + 1$ symbols, and their orders must be equal to or be factors of $p + 1$ .

Suppose now that $i$ is a primitive root of the congruence

i^{p^{2} - 1} - 1 \equiv 0 (mod p),

so that

i

and

i^{p}

are the roots of a quadratic congruence with real coeﬃcients; and consider the operation

K

, denoting

\frac{y - i^{k}}{y - i^{k p}} \equiv i^{2 (p - 1)} \frac{x - i^{k}}{x - i^{k p}},

where

k

is not a multiple of

p + 1

. On solving with respect to

y

K

is expressed in the form

y \equiv \frac{\frac{i^{(\frac{k}{2} + 1) (p - 1)} - i^{- (\frac{k}{2} + 1) (p - 1)}}{i^{\frac{k}{2} (p - 1)} - i^{- \frac{k}{2} (p - 1)}} x - \frac{i^{(p - 1)} - i^{- (p - 1)}}{i^{\frac{k}{2} (p - 1)} - i^{- \frac{k}{2} (p - 1)}} i^{\frac{k}{2} (p + 1)}}{i^{(p - 1)} - i^{- (p - 1)} i^{k 2 (p - 1)} - i^{- k 2 (p - 1)} x i^{- k 2 (p + 1)} + i^{(k 2 - 1) (p - 1)} - i^{- (k 2 - 1) (p - 1)} i^{k 2 (p - 1)} - i^{- k 2 (p - 1)}},

an operation of determinant unity. It will be found, on writing

i^{p}

for

i

in the coeﬃcients of this operation, that they remain unaltered; therefore, since they are symmetric functions of

i

and

i^{p}

, they must be real numbers. The operation therefore belongs to

H

. The

n

th power of this operation is given by

\frac{y - i^{k}}{y - i^{k p}} \equiv i^{2 n (p - 1)} \frac{x - i^{k}}{x - i^{k p}},

and therefore, since the ﬁrst power of

i

which is congruent to unity,

(mod p)

, is the

(p^{2} - 1)

th, the order of the operation is

\frac{1}{2} (p + 1)

. If we write

k p

for

k

in the operation

K

, the new operation is

K^{- 1}

; but if

k

is replaced by any other number

k'

, which is not a multiple of

p + 1

, the new operation

K'

, given by

\frac{y - i^{k'}}{y - i^{k' p}} \equiv i^{2 (p - 1)} \frac{x - i^{k'}}{x - i^{k' p}},

generates a new sub-group of order

\frac{1}{2} (p + 1)

, which has no operation except identity in common with

{K}

. Now there are

p^{2} - p

numbers less than

p^{2} - 1

which are not multiples of

p + 1

; therefore

H

contains

\frac{1}{2} (p^{2} - p)

cyclical sub-groups of order

\frac{1}{2} (p + 1)

, no two of which have a common operation except identity. The corresponding substitutions displace all the symbols.

225. A simple enumeration shews that the operations of the cyclical sub-groups of orders $\frac{1}{2} (p - 1)$ , $p$ and $\frac{1}{2} (p + 1)$ , exhaust all the operations of the group. Thus there are, omitting identity from each sub-group:

(i) $\frac{1}{2} p (p + 1)$ sub-groups of order $\frac{1}{2} (p - 1)$ , containing $\frac{1}{4} p (p^{2} - 1) - \frac{1}{2} p^{2} - \frac{1}{2} p$ distinct operations;

(ii) $\frac{1}{2} p (p - 1)$ sub-groups of order $\frac{1}{2} (p + 1)$ , containing $\frac{1}{4} p (p^{2} - 1) - \frac{1}{2} p^{2} + \frac{1}{2} p$ distinct operations;

(iii) $p + 1$ sub-groups of order $p$ , containing $p^{2} - 1$ distinct operations;
and the sum of these numbers, with $1$ for the identical operation, gives $\frac{1}{2} p (p^{2} - 1)$ , which is the order of the group.

Every operation that displaces all the symbols is therefore the power of an operation of order $\frac{1}{2} (p + 1)$ .

226. We shall now further shew that the $\frac{1}{2} p (p - 1)$ sub-groups of order $\frac{1}{2} (p + 1)$ form a single conjugate set, and that each is contained self-conjugately in a dihedral group of order $p + 1$ . Let $S$ be any operation of $H$ , which is permutable with ${K}$ and replaces $i^{k}$ by some other symbol $j$ . Then $S^{- 1} K S$ is an operation which leaves $j$ unaltered; it may therefore be expressed in the form

\frac{y - j}{y - j'} = m \frac{x - j}{x - j'} .

This can belong to the sub-group generated by $K$ , only if $j$ and $j'$ are the same pair as $i^{k}$ and $i^{k p}$ . Hence $j$ must be either $i^{k}$ or $i^{k p}$ ; and similarly, if $S$ replaces $i^{k p}$ by $j'$ , the latter must be either $i^{k p}$ or $i^{k}$ . Hence either $S$ must keep both the symbols $i^{k}$ and $i^{k p}$ unchanged or it must interchange them; and conversely, every operation which either keeps both the symbols unchanged or interchanges them, must transform ${K}$ into itself. If $S$ keeps both of them unchanged, it is a power of $K$ . If $S$ interchanges them, it is of the form

\frac{y - i^{k}}{y - i^{k p}} \equiv m \frac{x - i^{k p}}{x - i^{k}};

and a simple calculation shews that

S^{- 1} K S = K^{- 1} .

If we take

m = 1

S

becomes

x + y \equiv i^{k} + i^{k p},

an operation belonging to

H

. Hence the cyclical sub-group

{K}

is contained self-conjugately in the sub-group of

{S, K}

which is of dihedral type. If there were any other operation

S'

, not contained in

{S, K}

, which transformed

K

into its inverse, then

S S'

would be an operation permutable with

K

and not contained in

{K}

. It has just been seen that no such operation exists. Hence

{S, K}

, of order

p + 1

, is the greatest sub-group that contains

{K}

self-conjugately; and

{K}

must be one of

\frac{1}{2} p (p - 1)

conjugate sub-groups.

227. The distribution of the operations of $H$ in conjugate sets is now known. A sub-group of order $p$ is contained self-conjugately in a group of order $\frac{1}{2} p (p - 1)$ , while an operation of order $p$ is permutable only with its own powers. There are therefore two conjugate sets of operations of order $p$ , each set containing $\frac{1}{2} (p^{2} - 1)$ operations. Again, each of the operations of a cyclical sub-group of order $\frac{1}{2} (p - 1)$ or $\frac{1}{2} (p + 1)$ is conjugate to its own inverse and to no other of its powers. Hence if $\frac{1}{2} (p + 1)$ is even and therefore $\frac{1}{2} (p - 1)$ odd, there are $\frac{1}{4} (p - 3)$ conjugate sets of operations whose orders are factors of $\frac{1}{2} (p - 1)$ , each set containing $p^{2} + p$ operations; $\frac{1}{4} (p - 3)$ conjugate sets of operations whose orders are factors of $\frac{1}{2} (p + 1)$ , other than the factor $2$ , each set containing $p^{2} - p$ operations; and a single set of operations of order $2$ , containing $\frac{1}{2} (p^{2} - p)$ operations. If $\frac{1}{2} (p - 1)$ is even and $\frac{1}{2} (p + 1)$ odd, there are $\frac{1}{4} (p - 1)$ conjugate sets of operations whose orders are factors of $\frac{1}{2} (p + 1)$ , each containing $p^{2} - p$ operations; $\frac{1}{4} (p - 5)$ conjugate sets whose orders are factors of $\frac{1}{2} (p - 1)$ , other than the factor $2$ , each set containing $p^{2} + p$ operations; and a single set of $\frac{1}{2} (p^{2} + p)$ conjugate operations of order $2$ . In either case, the group contains, exclusive of identity, $\frac{1}{2} (p + 3)$ conjugate sets of operations.

228. Since $p - 1$ and $p + 1$ can have no common factor except $2$ , it follows that, if $q^{m}$ denote the highest power of an odd prime, other than $p$ , which divides the order of $H$ , $q^{m}$ must be a factor of $\frac{1}{2} (p - 1)$ or of $\frac{1}{2} (p + 1)$ ; and the sub-groups of order $q^{m}$ must be cyclical. Moreover, since no two cyclical sub-groups of order $\frac{1}{2} (p - 1)$ , or $\frac{1}{2} (p + 1)$ , have a common operation except identity, the same must be true of the sub-groups of order $q^{m}$ .

If $2^{m}$ is the highest power of $2$ that divides $\frac{1}{2} (p - 1)$ or $\frac{1}{2} (p + 1)$ , $2^{m + 1}$ will be the highest power of $2$ that divides the order of $H$ . Moreover, a sub-group of order $2^{m + 1}$ must contain a cyclical sub-group of order $2^{m}$ self-conjugately, and it must contain an operation of order $2$ that transforms every operation of this cyclical sub-group into its own inverse; in other words, the sub-groups of order $2^{m + 1}$ are of dihedral type.

Suppose now that two sub-groups of order $2^{m + 1}$ ( $m > 1$ ) have a common sub-group of order $2^{r}$ ( $r > 2$ ). Such a sub-group must be either cyclical or dihedral: in the latter case, it contains self-conjugately a single cyclical sub-group of order $2^{r - 1}$ . Hence, on the supposition made, a cyclical sub-group of order $4$ at least would be contained self-conjugately in two distinct cyclical sub-groups of order $2^{m}$ . It has been seen that this is not the case; and therefore the greatest sub-group, that two sub-groups of order $2^{m + 1}$ can have in common, must be a sub-group of order $4$ , whose operations, except identity, are all of order $2$ . Now every group of order $2^{m + 1}$ contains one self-conjugate operation of order $2$ , and $2^{m}$ operations of order $2$ falling into $2$ conjugate sets of $2^{m - 1}$ each. Moreover, the group of order $p \pm 1$ , which has a cyclical sub-group of order $2^{m}$ and contains the operation $A$ of order $2$ of this cyclical group self-conjugately, has $\frac{1}{2} (p \pm 1)$ other operations of order $2$ ; and therefore it contains $\frac{p \pm 1}{2^{m + 1}}$ sub-groups of order $2^{m + 1}$ , each of which has $A$ for its self-conjugate operation. If now $B$ is any operation of order $2$ of this sub-group of order $p \pm 1$ , and if it is distinct from $A$ , then $B$ enters into a sub-group of order $2^{m + 1}$ that contains $A$ self-conjugately. But since $A$ is permutable with $B$ , $A$ must belong to the sub-group of order $p \pm 1$ , which contains $B$ self-conjugately; hence $A$ enters into a sub-group of order $2^{m + 1}$ which contains $B$ self-conjugately. The sub-group ${A, B}$ is therefore common to two distinct sub-groups of order $2^{m + 1}$ . Now no group of order $2^{r}$ ( $r > 2$ ) can be common to two sub-groups of order $2^{m + 1}$ ; and therefore ${A, B}$ must (§ 80) be permutable with some operation $S$ whose order $s$ is prime to $2$ . If $s$ is not $3$ , $S$ must be permutable with $A$ and $B$ : and then ${A, S}$ and ${B, S}$ would be two distinct sub-groups of orders $2 s$ , whose operations are permutable with each other. It has been seen that $H$ does not contain such sub-groups. Hence $s = 3$ ; and $S$ transforms $A$ , $B$ and $A B$ cyclically, or ${S, A, B}$ is a sub-group of tetrahedral type (§ 202).

The number of quadratic⁹¹ sub-groups contained in $H$ may be directly enumerated. A group of order $2^{m + 1}$ contains $2^{m - 1}$ such sub-groups, which fall into $2$ conjugate sets of $2^{m - 2}$ each; a single group of order $8$ containing each quadratic group self-conjugately. The quadratic groups, contained in the $\frac{p \pm 1}{2^{m + 1}}$ sub-groups of order $2^{m + 1}$ of a sub-group of order $p \pm 1$ , are clearly all distinct, and each quadratic group belongs to just $3$ groups of order $p \pm 1$ ; thus ${A, B}$ belongs to the $3$ groups which contain $A$ , $B$ and $A B$ respectively as self-conjugate operations. Hence the total number of quadratic groups contained in $H$ is

\frac{1}{2} p (p \mp 1) \frac{p \pm 1}{2^{m + 1}} 2^{m - 1} \frac{1}{3} = \frac{\frac{1}{2} p (p^{2} - 1)}{12} .

229. The greatest sub-group of a group of order $2^{m + 1}$ , that contains a quadratic group self-conjugately, is a group of order $8$ and dihedral type; and it has been shewn that $3$ is the only factor, prime to $2$ , that occurs in the order of the sub-group containing a quadratic group self-conjugately. Hence ﬁnally, the order of the greatest group containing a quadratic group self-conjugately is $24$ , and the $\frac{\frac{1}{2} p (p^{2} - 1)}{12}$ quadratic groups fall into two conjugate sets of $\frac{\frac{1}{2} p (p^{2} - 1)}{24}$ each. The group of order $24$ , that contains a quadratic group self-conjugately, contains also a self-conjugate tetrahedral sub-group, while the sub-groups of order $8$ are dihedral. Hence (§ 84) this group must be of octohedral type.

Since every tetrahedral sub-group of $H$ contains a quadratic sub-group self-conjugately, and every octohedral sub-group contains a tetrahedral sub-group self-conjugately, there must also be two conjugate sets of tetrahedral sub-groups and two conjugate sets of octohedral sub-groups, the number in each set being $\frac{\frac{1}{2} p (p^{2} - 1)}{24}$ .

230. We have hitherto supposed $m > 1$ , or what is the same thing, $p \equiv \pm 1 (mod 8)$ . If now $m = 1$ , so that $p \equiv \pm 3 (mod 8)$ , the highest power of $2$ that divides the order of $H$ is $2^{2}$ ; and, since $2^{2}$ is not a factor of $\frac{1}{2} (p \pm 1)$ , the sub-groups of order $2^{2}$ are quadratic. Moreover, since $2^{2}$ is the highest power of $2$ dividing the order of $H$ , the quadratic sub-groups form a single conjugate set. Each sub-group of order $p \pm 1$ , which has a self-conjugate operation of order $2$ , contains $\frac{1}{4} (p \pm 1)$ sub-groups of order $4$ , and each of the latter belongs to $3$ of the former. The total number is as before

\frac{\frac{1}{2} p (p^{2} - 1)}{12},

and since they form a single conjugate set, each quadratic group is self-conjugate in a group of order

12

. Also, for the same reason as in the previous case, this sub-group is of tetrahedral type.

Finally, since every sub-group of $H$ of tetrahedral type must contain a quadratic sub-group self-conjugately, $H$ must contain a single conjugate set of $\frac{\frac{1}{2} p (p^{2} - 1)}{12}$ tetrahedral sub-groups. In this case, the order of $H$ is not divisible by $24$ , and therefore the question of octohedral sub-groups does not arise.

231. The group $H$ always contains tetrahedral sub-groups; when its order is divisible by $24$ , it contains also octohedral sub-groups. Now if $p \equiv \pm 1 (mod 5)$ , the order of $H$ is divisible by $60$ : it may be shewn as follows that, in these cases, $H$ contains sub-groups of icosahedral type.

Let us suppose, ﬁrst, that $p \equiv 1 (mod 5)$ ; and let $j$ be a primitive root of the congruence

j^{5} \equiv 1 (mod p) .

Then

(\frac{j x}{j^{- 1}})

, which we will denote by

A

, is an operation of order

5

. The operations of order

2

H

are all of the form

B

, where

B

denotes

(\frac{α x + β}{γ x - α})

, since each is its own inverse. Now

A B = (\frac{α j x + β j^{- 1}}{γ j x - α j^{- 1}});

and (§ 203) if

A

and

B

generate an icosahedral group,

{(A B)}^{3} = 1 .

A simple calculation shews that, if this condition is satisﬁed, then

α^{2} {(j - j^{- 1})}^{2} \equiv 1 .

Also, since the determinant of

B

is unity,

α^{2} + β γ \equiv - 1 .

These two congruences have just

p - 1

distinct solutions, the solutions

α

β

γ

and

- α

- β

- γ

being regarded as identical. There are therefore

p - 1

operations of order two in

H

, namely the operations

(\frac{\frac{1}{j - j^{- 1}} x + β}{γ x - 1 j - j^{- 1}}),

where

β γ \equiv - 1 - \frac{1}{{(j - j^{- 1})}^{2}},

which with

A

generate an icosahedral sub-group.

The group generated by

(\frac{j x}{j^{- 1}}) and (\frac{\frac{1}{j - j^{- 1}} x + β_{0}}{γ_{0} x - 1 j - j^{- 1}}),

contains

5

of the

p - 1

operations of order

2

of the form

(\frac{\frac{1}{j - j^{- 1}} x + β}{γ x - 1 j - j^{- 1}}),

viz. those for which

β \equiv β_{0} j^{n}, γ \equiv γ_{0} j^{- n}, (n = 0, 1, 2, 3, 4) .

Hence the sub-group

{A}

, of order

5

, belongs to

\frac{1}{5} (p - 1)

distinct icosahedral sub-groups. Now each icosahedral sub-group has

6

sub-groups of order

5

; and

H

contains

\frac{1}{2} p (p + 1)

sub-groups of order

5

forming a single conjugate set. The number of icosahedral sub-groups in

H

is therefore

\frac{1}{6} \frac{p - 1}{5} \frac{1}{2} p (p + 1) = \frac{\frac{1}{2} p (p^{2} - 1)}{30} .

The group of isomorphisms of the icosahedral group is the symmetric group of degree

5

(§ 173). Now

H

can contain no sub-group simply isomorphic with the symmetric group of degree

5

. For if it contained such a sub-group, an operation of order

5

would be conjugate to its own square; and this is not the case.

Hence (§ 165), if an icosahedral sub-group $K$ of $H$ is contained self-conjugately in a greater sub-group $L$ , then $L$ must be the direct product of $K$ and some other sub-group. This also is impossible; for the greatest sub-group of $H$ in which any cyclical sub-group, except those of order $p$ , is contained self-conjugately, is of dihedral type. Hence $L$ must coincide with $K$ , and $K$ must be one of $\frac{\frac{1}{2} p (p^{2} - 1)}{60}$ conjugate sub-groups. The icosahedral sub-groups of $H$ therefore fall into two conjugate sets of $\frac{\frac{1}{2} p (p^{2} - 1)}{60}$ each.

In a similar manner, when $p \equiv - 1 (mod 5)$ , we may take as a typical operation $A$ , of order $5$ ,

\frac{y - i}{y - i^{p}} \equiv i^{\frac{p^{2} - 1}{5}} \frac{x - i}{x - i^{p}};

and it may be shewn, the calculation being rather more cumbrous than in the previous case, that there are just

p + 1

operations

B

, of the form

(\frac{α x + β}{γ x - α})

, such that

{(A B)}^{3} = 1,

and that ﬁve of these belong to the icosahedral group generated by

A

and any one of them. It follows, exactly as in the previous case, that

H

contains

\frac{\frac{1}{2} p (p^{2} - 1)}{30}

icosahedral sub-groups, which fall into two conjugate sets, each set containing

\frac{\frac{1}{2} p (p^{2} - 1)}{60}

groups.

232. Finally, we proceed to shew that $H$ has no other sub-groups than those which have been already determined. Suppose, ﬁrst, that a sub-group $h$ of $H$ contains two distinct sub-groups of order $p$ . These must, by Sylow’s theorem, form part of a set of $k p + 1$ sub-groups of order $p$ conjugate within $h$ . Now $H$ contains only $p + 1$ sub-groups of order $p$ , and therefore $k$ must be unity and $h$ must contain all the sub-groups of order $p$ ; or since $H$ is simple, $h$ must contain and therefore coincide with $H$ . Hence the only sub-groups of $H$ , whose orders are divisible by $p$ , are those that contain a sub-group of order $p$ self-conjugately. They are of known types.

Suppose next that $g$ is a sub-group of $H$ , whose order $n$ is not divisible by $p$ , and let $S_{1}$ be an operation of $g$ whose order $q_{1}$ is not less than the order of any other operation of $g$ . In $H$ the sub-group ${S_{1}}$ is self-conjugate in a dihedral group of order $p \pm 1$ ; and the greatest sub-group of this group, which contains no operation of order greater than $q_{1}$ , is a dihedral group of order $2 q_{1}$ . Hence in $g$ the sub-group ${S_{1}}$ is self-conjugate in a group of order $q_{1}$ or $2 q_{1}$ , and therefore it forms one of $\frac{n}{q_{1}}$ or of $\frac{n}{2 q_{1}}$ conjugate sub-groups. Moreover, no two of these sub-groups contain a common operation except identity; and they therefore contain, excluding identity, $\frac{n (q_{1} - 1)}{𝜖_{1} q_{1}}$ distinct operations, where $𝜖_{1}$ is either $1$ or $2$ .

Of the remaining operations of $g$ , let $S_{2}$ be one whose order $q_{2}$ is not less than that of any of the others. The operation $S_{2}$ cannot be permutable with any of the $\frac{n (q_{1} - 1)}{𝜖_{1} q_{1}}$ operations already accounted for, since $S_{2}$ is not a power of any one of these operations. Hence, exactly as before, ${S_{2}}$ must form one of $\frac{n}{𝜖_{2} q_{2}}$ conjugate sub-groups in $g$ , $𝜖_{2}$ being either $1$ or $2$ ; and these sub-groups contain $\frac{n (q_{2} - 1)}{𝜖_{2} q_{2}}$ operations which are distinct from identity, from each other, and from those of the previous set. This process may be continued till the identical operation only remains. Hence, ﬁnally, $n$ being the total number of operations of $g$ , we must have

n = 1 + \sum_{ν} \frac{n (q_{ν} - 1)}{𝜖_{ν} q_{ν}}

\frac{1}{n} = 1 - \sum_{ν} \frac{q_{ν} - 1}{𝜖_{ν} q_{ν}} .

233. In this equation, let $r$ of the $𝜖$ ’s be $1$ and $s$ of them be $2$ , so that $r + s$ is their total number, say $m$ . Then

\begin{array}{l} \frac{1}{n} & = 1 - \sum (1 - \frac{1}{q_{λ}}) - \sum (\frac{1}{2} - \frac{1}{2 q_{μ}}) \\ = 1 - r + \sum \frac{1}{q_{λ}} - \frac{1}{2} s + \sum \frac{1}{2 q_{μ}} \\ \leq 1 - r + \frac{1}{2} r - \frac{1}{2} s + \frac{1}{4} s \\ \leq 1 - \frac{1}{2} r - \frac{1}{4} s \\ \leq 1 - \frac{1}{4} m . \end{array}

Hence, since $n$ is a positive integer, there cannot be more than three terms under the sign of summation. Moreover, since

\frac{1}{n} \leq 1 - \frac{1}{2} r - \frac{1}{4} s,

r

cannot be greater than

1

, and therefore not more than one of the

𝜖

’s can be unity. Also, when one of the

𝜖

’s is unity, we have

\begin{array}{l} \frac{1}{n} & = \frac{1}{q} - \frac{1}{2} s + \sum \frac{1}{2 q_{μ}} \\ \leq \frac{1}{2} - \frac{1}{2} s + \frac{1}{4} s \\ \leq \frac{1}{4} (2 - s), \end{array}

so that, in this case, $s$ cannot be greater than unity. The solutions are now easily obtained by trial.

(i) For one term in the sum, the only possible solution is

𝜖_{1} = 1, n = q_{1},

and the corresponding group is cyclical.

(ii) For two terms in the sum, the solutions are

\begin{array}{l} (α) 𝜖_{1} = 𝜖_{2} = 2, n = \frac{2 q_{1} q_{2}}{q_{1} + q_{2}}; \\ (β) 𝜖_{1} = 2, 𝜖_{2} = 1, q_{2} = 2, n = 2 q_{1}; \\ (γ) 𝜖_{1} = 1, 𝜖_{2} = 2, q_{1} = 3, q_{2} = 2, n = 12 . \end{array}

To the solution ( $α$ ) there corresponds no sub-group; for $n < 2 q_{1}$ , and the values $q_{1} = q_{1}$ , $𝜖_{1} = 2$ imply that $g$ has a sub-group of order $2 q_{1}$ .

To the solution ( $β$ ) correspond the sub-groups of order $2 q_{1}$ of dihedral type, for which $q_{1}$ is odd, so that the operations of order $2$ form a single conjugate set.

To the solution ( $γ$ ) corresponds a sub-group of order $12$ containing $8$ operations of order $3$ and $3$ operations of order $2$ , i.e. a tetrahedral sub-group.

(iii) For three terms in the sum, the solutions are

\begin{array}{l} (α) & 𝜖_{1} = 𝜖_{2} = 𝜖_{3} = 2, & q_{2} & = 2, & q_{3} & = 2, & n & = 2 q_{1}; \\ (β) & ” ” & q_{1} & = 3, & q_{2} & = 3, & q_{3} & = 2, & n & = 12; \\ (γ) & ” ” & q_{1} & = 4, & q_{2} & = 3, & q_{3} & = 2, & n & = 24; \\ (δ) & ” ” & q_{1} & = 5, & q_{2} & = 3, & q_{3} & = 2, & n & = 60 . \end{array}

To the solution ( $α$ ) correspond the sub-groups of order $2 q_{1}$ of dihedral type, in which $q_{1}$ is even, so that the operations of order $2$ , which do not belong to the cyclical sub-group of order $q_{1}$ , fall into two distinct conjugate sets.

To the solution ( $β$ ) would correspond a group of order $12$ containing $3$ operations of order $2$ and $4$ sub-groups of order $3$ which fall into two conjugate sets of $2$ each. Sylow’s theorem shews that such a group cannot exist; and therefore there is no sub-group of $H$ corresponding to this solution.

Solution ( $γ$ ) gives a group of order $24$ , with $3$ conjugate cyclical sub-groups of order $4$ , $4$ conjugate cyclical sub-groups of order $3$ , and $6$ other operations of order $2$ forming a single conjugate set. No operation of this group is permutable with each of the $4$ sub-groups of order $3$ ; and therefore, if the group exists, it can be represented as a transitive group of $4$ symbols. On the other hand, the order of the symmetric group of $4$ symbols, which (§ 203) is simply isomorphic with the octohedral group, is $24$ ; and its cyclical sub-groups are distributed as above. Hence to this solution there correspond the octohedral sub-groups of $H$ .

Solution ( $δ$ ) gives a group of order $60$ , with $6$ conjugate sub-groups of order $5$ , $10$ conjugate sub-groups of order $3$ , and a conjugate set of $15$ operations of order $2$ . It has been shewn, in § 85, that there is only one type of group of order $60$ that has $6$ sub-groups of order $5$ ; viz. the alternating group of degree $5$ : and that, in this group, the distribution of sub-groups in conjugate sets agrees with that just given. Moreover, the alternating group of degree $5$ is simply isomorphic with the icosahedral group. Hence to this solution there correspond the icosahedral sub-groups of $H$ .

234. When $p > 11$ , then $\frac{1}{2} p (p - 1) > 60$ ; and, when $p > 3$ , $\frac{1}{2} p (p - 1) > p + 1$ . Hence when $p > 11$ , the order of the greatest sub-group of $H$ is $\frac{1}{2} p (p - 1)$ , and the least number of symbols in which $H$ can be expressed as a transitive substitution group is $p + 1$ .

When $p$ is $5$ , $7$ or $11$ , however, $H$ can be expressed as a transitive substitution group of $p$ symbols⁹².

For, when $p = 5$ , $H$ contains a tetrahedral sub-group of order $12$ , forming one of $5$ conjugate sub-groups; therefore $H$ can be expressed as a transitive group of $5$ symbols. It is to be noticed that in this case $H$ is an icosahedral group.

When $p = 7$ , $H$ contains an octohedral sub-group of order $24$ , which is one of $7$ conjugate sub-groups; and $H$ can therefore be expressed as a transitive group of $7$ symbols. Similarly, when $p = 11$ , $H$ contains an icosahedral sub-group of order $60$ , which is one of $11$ conjugate sub-groups; and the group can be expressed transitively in $11$ symbols.

235. The simple groups, of the class we have been discussing in the foregoing sections, are self-conjugate sub-groups of the triply transitive groups of degree $p + 1$ , deﬁned by

y \equiv \frac{α x + β}{γ x + δ}, (mod p),

the existence of which was demonstrated in § 113. In fact, since

(\frac{α x + β}{γ x + δ})

and

(\frac{k α x + k β}{k γ x + k δ})

represent the same transformation, the determinant,

α δ - β γ

, of any transformation may always be taken as either unity or a given non-residue; and it follows at once that the transformations of determinant unity form a self-conjugate sub-group of the whole group of transformations.

If, as in § 113, $α$ , $β$ , $γ$ , $δ$ , are powers of $i$ , where $i$ is a primitive root of the congruence

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

the triply transitive group

G

of degree

p^{n} + 1

, which is deﬁned by the transformations, has again, when

p

is an odd prime, a self-conjugate sub-group

H

of order

\frac{1}{2} p^{n} (p^{2 n} - 1)

, which is given by the transformations of determinant unity. It follows from Theorem IX, § 134, that

G

, being a triply transitive group of degree

p^{n} + 1

, must have, as a self-conjugate sub-group, a doubly transitive simple group; and it is easy to shew that

H

is this sub-group.

In fact, if a simple group $h$ is a self-conjugate sub-group of $G$ it must be contained in $H$ . Also, since $h$ is a doubly transitive group of degree $p^{n} + 1$ , it must contain every operation of order $p$ that occurs in $G$ . Now we may shew that these operations generate $H$ . Thus $(\frac{2 x + 1}{- x})$ and $(x + 2 - i - i^{- 1})$ are operations of order $p$ belonging to $G$ . Therefore $(\frac{(i + i^{- 1}) x + 1}{- x})$ belongs to $h$ . But this operation is transformed into $(\frac{i x}{i^{- 1}})$ by $(\frac{x + i^{- 1}}{x + i})$ . Hence $(\frac{i x}{i^{- 1}})$ belongs to $h$ ; and a sub-group of $h$ which keeps one symbol unchanged is the group of order $\frac{1}{2} p^{n} (p^{n} - 1)$ generated by $(x + 1)$ and $(\frac{i x}{i^{- 1}})$ . The order of $h$ therefore is not less than $\frac{1}{2} p^{n} (p^{2 n} - 1)$ ; in other words $h$ is identical with $H$ .

When $p = 2$ , every power of $i$ is a quadratic residue, and the determinant of every transformation is unity. In this case it may be shewn, by an argument similar to the above, that the group $G$ of order $2^{n} (2^{2 n} - 1)$ is itself a simple group.

We are thus led to recognize the existence of a doubly-inﬁnite series of simple groups of orders $2^{n} (2^{2 n} - 1)$ and $\frac{1}{2} p^{n} (p^{2 n} - 1)$ , which are closely analogous to the groups of order $\frac{1}{2} p (p^{2} - 1)$ already discussed. For an independent proof of the existence of these simple groups and for an investigation of their properties, the reader is referred to the memoirs mentioned below⁹³.

236. We will now return to the linear homogeneous group $G$ of transformations of $n$ symbols, taken to a prime modulus $p$ ; and consider it more directly as the group of isomorphisms of an Abelian group of order $p^{n}$ and type $(1, 1, \dots, to n units)$ . As in § 156, it may be expressed in the form of a substitution group performed on the $p^{n} - 1$ symbols of the operations, other than identity, of the Abelian group. In this form it is clearly transitive, since there are isomorphisms changing any operation of the Abelian group into any other operation. If $P$ is any operation of the Abelian group, an isomorphism which changes any one of the $p - 1$ operations

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

into any other, will certainly interchange the set among themselves. Hence, when expressed as a group of degree

p^{n} - 1

G

is imprimitive; and the symbols forming an imprimitive system are those of the operations, other than identity, of any sub-group of order

p

of the Abelian group. If

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

are a set of generating operations of the Abelian group, an isomorphism, which changes each of the sub-groups

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

into itself, must be of the form

(\binom{P_{1}, P_{2}, \dots, P_{n}}{P_{1}^{α_{1}}, P_{2}^{α_{2}}, \dots, P_{n}^{α_{n}}}) .

This isomorphism changes

P_{1} P_{2}

into

{(P_{1} P_{2}^{\frac{α_{2}}{α_{1}}})}^{α_{1}}

; therefore it will only transform the sub-group

{P_{1} P_{2}}

into itself when

α_{1} \equiv α_{2}, (mod p)

. If then the given isomorphism changes every sub-group of order

p

into itself, we must have

α_{1} \equiv α_{2} \equiv \dots \equiv α_{n}, (mod p) .

Hence the only operations of $G$ , which interchange the symbols of each imprimitive system among themselves, are those given by the powers of

(\binom{P_{1}, P_{2}, \dots, P_{n}}{P_{1}^{α}, P_{2}^{α}, \dots, P_{n}^{α}}),

where

α

is a primitive root of

p

. This operation is the same as that denoted by

A

in § 217. It follows immediately that the factor-group

\frac{G}{{A}}

can be represented as a transitive group in

\frac{p^{n} - 1}{p - 1}

symbols. In fact, the operations of

{A}

are the only operations of

G

which transform each of the

\frac{p^{n} - 1}{p - 1}

sub-groups of order

p

in itself; and these

\frac{p^{n} - 1}{p - 1}

sub-groups must be permuted among themselves by every operation of

G

. The substitution group thus obtained is doubly transitive; for if

P

and

P'

are any two operations of the Abelian group such that

P'

is not a power of

P

, and if

Q

and

Q'

are any other two operations of the Abelian group subject to the same condition, there certainly exists an isomorphism of the form

(\binom{P, P', \dots}{Q, Q', \dots}),

and this isomorphism changes the sub-groups

{P}

and

{P'}

into the sub-groups

{Q}

and

{Q'}

These results will still hold if, instead of considering $G$ the total group of isomorphisms, we take $Γ$ the group of isomorphisms of determinant unity. Thus the determinant of

(\binom{P, P', \dots}{Q^{α}, Q', \dots})

α

times the determinant of

(\binom{P, P', \dots}{Q, Q', \dots}) .

It is therefore possible always to choose

α

so that the determinant of

(\binom{P, P', \dots}{Q^{α}, Q', \dots})

shall be unity; and this isomorphism still changes the sub-groups

{P}

and

{P'}

into

{Q}

and

{Q'}

respectively.

The lowest power of $A$ contained in $Γ$ is (§ 217) $A^{\frac{p - 1}{d}}$ . Hence the group $\frac{Γ}{{A^{\frac{p - 1}{d}}}}$ can be represented as a doubly transitive group of degree $\frac{p^{n} - 1}{p - 1}$ . This group is (§ 220) simply isomorphic with the simple group of order $\frac{N}{(p - 1) d}$ , which is deﬁned by the composition-series of $G$ .

We may sum up these results as follows:—

THEOREM. The homogeneous linear group of order

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

when

p^{n}

is neither

2^{2}

nor

3^{2}

, deﬁnes, by its composition-series, a simple group of order

\frac{N}{(p - 1) d}

, where

d

is the greatest common factor of

p - 1

and

n

. This simple group can be represented as a doubly transitive group of degree

p^{n - 1} + p^{n - 2} + \dots + p + 1

237. The $\frac{p^{n} - 1}{p - 1}$ symbols, permuted by one of these doubly transitive simple groups, may be regarded as the sub-groups of order $p$ of an Abelian group of order $p^{n}$ and type $(1, 1, \dots, to n units)$ . Now every pair of sub-groups of such an Abelian group enters in one, and only in one, sub-group of order $p^{2}$ ; and every sub-group of order $p^{2}$ contains $p + 1$ sub-groups of order $p$ . Hence from the $\frac{p^{n} - 1}{p - 1}$ symbols permuted by the doubly transitive group, $\frac{p^{n} - 1 \cdot p^{n - 1} - 1}{p - 1 \cdot p^{2} - 1}$ sets of $p + 1$ symbols each may be formed, such that every pair of symbols occurs in one set and no pair in more than one set, while the sets are permuted transitively by the operations of the group. These groups therefore belong to the class of groups referred to in § 148. The sub-group, that leaves two symbols unchanged, permutes the remaining symbols in two transitive systems of $p - 1$ and $p^{n - 1} + p^{n - 2} + \dots + p^{2}$ ; and the sub-group, that leaves unchanged each of the symbols of one of the sets of $p + 1$ , is contained self-conjugately in a sub-group whose order is $(p + 1) p$ times that of a sub-group leaving two symbols unchanged. This latter sub-group permutes the symbols in two transitive systems of $p + 1$ and $p^{n - 1} + p^{n - 2} + \dots + p^{2}$ . It may be pointed out that, when $n$ is $3$ , such a sub-group is simply isomorphic with, but is not conjugate to, the sub-groups that leave one symbol unchanged: this may be seen at once by noticing that an Abelian group, of order $p^{3}$ and type $(1, 1, 1)$ , has the same number of sub-groups of orders $p$ and $p^{2}$ .

238. Some special cases may be noticed. First, when $p = 2$ , both $p - 1$ and $d$ are unity, and the homogeneous linear group is itself a simple group.

If $n = 3$ , then $N = 168$ ; so that the group of isomorphisms of a group of order $8$ , whose operations are all of order $2$ , is the simple group of order $168$ (§ 146).

If $n = 4$ , then $N = 2^{6} \cdot 3^{2} \cdot 5 \cdot 7$ . This is the order of the alternating group of $8$ symbols; and it may be shewn that this group is simply isomorphic with the group of isomorphisms.

The Abelian group of order $16$ contains $35$ sub-groups of order $4$ ; and it may be shewn that, from these $35$ sub-groups, sets of $5$ can be formed in $56$ distinct ways, so that each set of $5$ contains every operation of order $2$ of the Abelian group once, and once only. If

P_{1}, P_{2}, P_{3}, P_{4}

are a set of generating operations of the Abelian group, such a set of

5

sub-groups of order

4

is given by

{P_{1}, P_{2}}, {P_{3}, P_{4}}, {P_{1} P_{3}, P_{2} P_{4}}, {P_{1} P_{4}, P_{1} P_{2} P_{3}}, {P_{2} P_{3}, P_{1} P_{3} P_{4}} .

Now

(\binom{P_{1}, P_{2}, P_{3}, P_{4}}{P_{1}, P_{2} P_{3}, P_{4}, P_{2} P_{4}})

is an isomorphism of order

7

of the Abelian group, and

P_{1}

is the only operation of the group, except identity, which is left unchanged by this isomorphism. It may be directly veriﬁed that the

7

sets of

5

groups of order

4

, into which the given set is transformed by the powers of this isomorphism, contain every sub-group of order

4

of the Abelian group. The

7

sets, being interchanged among themselves by this isomorphism of order

7

which leaves only

P_{1}

unchanged, must be interchanged among themselves by isomorphisms of order

7

which leave any other single operation of the Abelian group unchanged. There are therefore at least

15

isomorphisms of order

7

which interchange the

7

sets among themselves. Now the isomorphisms, which interchange the

7

sets among themselves, form a sub-group of the group of isomorphisms, which is isomorphic with a group of degree

7

; and the only groups of degree

7

, which contain at least

15

operations of order

7

, are the symmetric and the alternating groups. The group of isomorphisms must therefore contain a sub-group which is isomorphic with the symmetric or with the alternating group of degree

7

. Hence at once, since the group of isomorphisms is simple, it must contain a sub-group which is simply isomorphic with the alternating group of degree

7

. Since this must be one of

8

conjugate sub-groups, the group of substitutions itself is simply isomorphic with the alternating group of degree

8

If $p^{n} = 3^{3}$ , then $p^{n - 1} + \dots + p + 1 = 13$ , $d = 1$ , and $N = 2^{4} \cdot 3^{3} \cdot 13$ . There is therefore a doubly transitive simple group of degree $13$ and order $2^{4} \cdot 3^{3} \cdot 13$ (§§ 145, 149).

239. The homogeneous linear group may be generalized by taking for the coeﬃcients powers of a primitive root of

i^{p^{ν} - 1} \equiv 1, (mod p),

instead of powers of a primitive root of

i^{p - 1} \equiv 1, (mod p) .

When the coeﬃcients are thus chosen, the order of the group

G_{p, n, ν}

, deﬁned by all sets of transformations

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

whose determinant diﬀers from zero, may be shewn, as in § 172, to be

N = (p^{n ν} - 1) (p^{n ν} - p^{ν}) (p^{n ν} - p^{2 ν}) \dots (p^{n ν} - p^{(n - 1) ν});

and the order of the sub-group

Γ

, formed of those transformations whose determinant is unity, is

\frac{N}{p^{ν} - 1}

. The only self-conjugate operations of

Γ

are the operations of the sub-group generated by

(i x_{1}, i x_{2}, \dots, i x_{n})

, which are contained in

Γ

. If

δ

is the greatest common factor of

p^{ν} - 1

and

n

, these self-conjugate operations of

Γ

form a cyclical sub-group

γ

of order

δ

. Finally, the argument of § 219 maybe repeated to shew that

\frac{Γ}{γ}

is a simple group.

The homogeneous linear group $G_{p, n, ν}$ , when values of $ν$ greater than unity are admitted, thus deﬁnes a triply inﬁnite system of simple groups; it may be proved that these groups can, for all values of $ν$ , be expressed as doubly transitive groups of degree $\frac{p^{n ν} - 1}{p^{ν} - 1}$ .

240. We may shew, in conclusion, that the group $G_{p, n, ν}$ is simply isomorphic with a sub-group of $G_{p, n ν, 1}$ . For this purpose, we consider the group deﬁned by

\begin{matrix} x_{1}^{1} \equiv x_{1} + i^{r_{1}}, x_{2}^{1} \equiv x_{2} + i^{r_{2}}, \dots, x_{n}^{1} \equiv x_{n} + i^{r_{n}}, \\ (r_{1}, r_{2}, \dots, r_{n} = 0, 1, 2, \dots, p^{ν} - 1); \end{matrix}

the congruences being taken to modulus $p$ . This is an Abelian group of order $p^{n ν}$ and type $(1, 1, \dots, to n ν units)$ . Moreover, the operation

\begin{array}{l} x_{1}^{1} & \equiv a_{11} x_{1} + \dots + a_{1 n} x_{n}, \\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \\ x_{n}^{1} & \equiv a_{n 1} x_{1} + \dots + a_{n n} x_{n}, \end{array}

of $G_{p, n, ν}$ transforms the given operation of the Abelian group into

x_{1}^{1} \equiv x_{1} + i^{s_{1}}, x_{2}^{1} \equiv x_{2} + i^{s_{2}}, \dots, x_{n}^{1} \equiv x_{n} + i^{s_{n}};

where

\begin{array}{l} i^{s_{1}} & \equiv a_{11} i^{r_{1}} + a_{12} i^{r_{2}} + \dots + a_{1 n} i^{r_{n}}, \\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \\ i^{s_{n}} & \equiv a_{n 1} i^{r_{1}} + a_{n 2} i^{r_{2}} + \dots + a_{n n} i^{r_{n}} . \end{array}

Every operation of $G_{p, n, ν}$ as deﬁned in § 239, is therefore permutable with the Abelian group, and gives a distinct isomorphism of it; or in other words, as stated above, $G_{p, n, ν}$ is simply isomorphic with a sub-group of $G_{p, n ν, 1}$ .

Further, the sub-group

\begin{matrix} x_{1}^{1} \equiv x_{1} + i^{r}, x_{2}^{1} \equiv x_{2}, \dots, x_{n}^{1} \equiv x_{n}, \\ (r = 0, 1, 2, \dots, p^{ν} - 1), \end{matrix}

is transformed by the given operation of $G_{p, n, ν}$ into the sub-group

\begin{matrix} x_{1}^{1} \equiv x_{1} + a_{11} i^{r}, x_{2}^{1} \equiv x_{2} + a_{21} i^{r}, \dots, x_{n}^{1} \equiv x_{n} + a_{n 1} i^{r}, \\ (r = 0, 1, 2, \dots, p^{ν} - 1) . \end{matrix}

a_{21} \equiv a_{31} \equiv \dots \equiv a_{n 1} \equiv 0,

the two sub-groups are identical; but if these conditions are not satisﬁed, they have no operation in common except identity. Moreover,

a_{11}, a_{21}, \dots, a_{n 1}

may each have any value from

0

i^{p^{ν} - 1}

, simultaneous zero values alone excluded. Hence the sub-group of order

p^{ν}

deﬁned by

\begin{matrix} x_{1}^{1} \equiv x_{1} + i^{r}, x_{2}^{1} \equiv x_{2}, \dots, x_{n}^{1} \equiv x_{n}, \\ (r = 0, 1, \dots, p^{ν} - 1), \end{matrix}

is one of $\frac{p^{n ν} - 1}{p^{ν} - 1}$ conjugate sub-groups in the group formed by combining the Abelian group with $G_{p, n, ν}$ ; and no two of these conjugate sub-groups have a common operation except identity.

The $p^{n ν} - 1$ operations, other than identity, of an Abelian group of order $p^{n ν}$ and type $(1, 1, \dots, to n ν units)$ , can therefore be divided into $\frac{p^{n ν} - 1}{p^{ν} - 1}$ sets of $p^{ν} - 1$ each, such that each set, with identity, forms a sub-group of order $p^{ν}$ ; and the group $G_{p, n, ν}$ is isomorphic with a group of isomorphisms of the Abelian group which permutes among themselves such a set of $\frac{p^{n ν} - 1}{p^{ν} - 1}$ sub-groups of order $p^{ν}$ .

Ex. 1. Shew that the $\frac{p^{n ν} - 1 \cdot p^{n ν} - p \dots p^{n ν} - p^{ν}}{p - 1 \cdot p^{2} - 1 \dots p^{ν} - 1}$ sub-groups of order $p^{ν}$ of an Abelian group of order $p^{n ν}$ and type $(1, 1, \dots, to n ν units)$ can be divided into sets of $\frac{p^{n ν} - 1}{p^{ν} - 1}$ each, such that each set contains every operation of the group, other than identity, once and once only; and discuss in how many distinct ways such a division may be carried out.

Ex. 2. Shew that the simple group, deﬁned by the group of isomorphisms of an Abelian group of order $p^{n}$ and type $(1, 1, \dots, 1)$ , admits a class of contragredient isomorphisms, which change the operations of the simple group, that correspond to isomorphisms leaving a sub-group of order $p$ of the Abelian group unaltered, into operations that correspond to isomorphisms leaving a sub-group of order $p^{n - 1}$ of the Abelian group unaltered.

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Chapter XIV.On the Linear Group87.

Chapter XIV.
On the Linear Group⁸⁷.