IV. On Abelian Groups.

Chapter IV.
On Abelian Groups.

W e shall now apply the general results, that have been obtained in the last chapter, to the study of two special classes of groups; in the present chapter we shall deal particularly with those groups whose operations are all permutable with each other.

Deﬁnition. A group, whose operations are all permutable with each other, is called an Abelian¹³ group.

It is to be expected (and it will be found) that the theory of Abelian groups is much simpler than that of groups in general; for the process of multiplication of the operations of such groups is commutative as well as associative.

Every sub-group of an Abelian group is itself an Abelian group, since its operations are necessarily all permutable. For the same reason, every operation and every sub-group of an Abelian group is self-conjugate both in the group itself and in any sub-group in which it is contained.

If $G$ is an Abelian group and $H$ any sub-group of $G$ , then since $H$ is necessarily self-conjugate, there exists a factor-group $\frac{G}{H}$ , and this again must be an Abelian group. (The reader must not however infer that, if $H$ and $\frac{G}{H}$ are both Abelian, then $G$ is also Abelian. It is indeed clear that this is not necessarily the case.)

37. Let now $G$ be any Abelian group, and let $p^{m}$ be the highest power of a prime $p$ that divides its order. We shall ﬁrst shew that $G$ has a single sub-group of order $p^{m}$ , consisting of all the operations of $G$ whose orders are powers of $p$ .

If $S_{1}$ , $S_{2}$ are any two operations of $G$ , it follows from § 33, that, because $S_{1}$ and $S_{2}$ are permutable, the order of ${S_{1}, S_{2}}$ is equal to or is a factor of the product of the orders of $S_{1}$ and $S_{2}$ . So again, if $S_{3}$ is an operation of $G$ not contained in ${S_{1}, S_{2}}$ , the order of ${S_{1}, S_{2}, S_{3}}$ is equal to or is a factor of the product of the orders of ${S_{1}, S_{2}}$ and $S_{3}$ . Now by continually including a fresh operation, not contained in the group already arrived at, we must in this way after a ﬁnite number of steps arrive at the group $G$ , whose order is divisible by $p$ . Hence, among the operations $S_{1}$ , $S_{2}$ , $S_{3}$ , …, there must be at least one whose order is divisible by $p$ , and some power of this, say $S$ , will be an operation of order $p$ . Now, if $m$ is greater than unity, the order of the factor-group $\frac{G}{{S}}$ , which is also Abelian, is divisible by $p$ , and therefore this factor-group must have an operation of order $p$ . Hence $G$ will (§ 28) contain a sub-group of order $p^{2}$ . If $m$ is greater than $2$ , the same reasoning may be repeated to shew that $G$ has a sub-group of order $p^{3}$ , and so on. Hence, ﬁnally, $G$ has a sub-group of order $p^{m}$ . Let $H$ be this sub-group; and suppose if possible that $G$ contains an operation $T$ , whose order is a power of $p$ , which does not belong to $H$ . Then ${H, T}$ is a sub-group of $G$ , whose order is a power of $p$ , greater than $p^{m}$ ; and this is impossible (§ 22). The sub-group $H$ must therefore contain all the operations of $G$ whose orders are powers of $p$ . Hence:—

THEOREM I. If $p^{m}$ is the highest power of a prime $p$ that divides the order of an Abelian group $G$ , then $G$ contains a single sub-group of order $p^{m}$ , which consists of all the operations of $G$ whose orders are powers of $p$ .

38. Let the order of $G$ be

N = p_{1}^{m_{1}} p_{2}^{m_{2}} \dots p_{n}^{m_{n}},

where

p_{1}

p_{2}

, …,

p_{n}

are distinct primes; and let

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

be the sub-groups of

G

of orders

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

Since the orders of $H_{1}$ and $H_{2}$ are relatively prime, they can have no common operation except identity; and therefore the order of ${H_{1}, H_{2}}$ is $p_{1}^{m_{1}} p_{2}^{m_{2}}$ . This sub-group contains all the operations of $G$ whose orders are relatively prime to $\frac{N}{p_{1}^{m_{1}} p_{2}^{m_{2}}}$ . For if $T$ were an operation of $G$ of order $p_{1}^{α} p_{2}^{β}$ , not contained in ${H_{1}, H_{2}}$ , then ${H_{1}, H_{2}, T}$ would be a sub-group of $G$ , of order $p_{1}^{n_{1}} p_{2}^{n_{2}}$ , where $n_{1} > m_{1}$ if $α > 0$ and $n_{2} > m_{2}$ if $β > 0$ ; and this is impossible (§ 22).

This process may clearly be continued to shew that, if $N = μ ν$ , where $μ$ and $ν$ are relatively prime, then $G$ contains a single sub-group of order $μ$ , consisting of all the operations of $G$ whose orders are relatively prime to $ν$ . Moreover $G$ itself is the direct product (§ 31) of $H_{1}$ , $H_{2}$ , …, $H_{n}$ .

39. The ﬁrst problem of pure group-theory that presents itself in connection with Abelian groups is the determination of all distinct Abelian groups of given order $N$ . Let $H_{1}$ and $H'$ be two distinct Abelian groups of order $p_{1}^{m_{1}}$ , i.e. two groups which are not simply isomorphic. Then two Abelian groups of order $N$ , whose sub-groups of order $p_{1}^{m_{1}}$ are simply isomorphic with $H_{1}$ and $H'$ respectively, are necessarily distinct. Since then $G$ is the direct product of $H_{1}$ , $H_{2}$ , …, $H_{n}$ , the general problem for any composite order $N$ will be completely solved when we have determined all distinct types of Abelian groups of order $p_{1}^{m_{1}}$ , $p_{2}^{m_{2}}$ , …, $p_{n}^{m_{n}}$ . We may therefore, for the purpose of this problem, conﬁne our attention to those Abelian groups whose orders are powers of primes.

40. Suppose then that $G$ is an Abelian group whose order is the power of a prime. If among the operations of $G$ we choose at random a set

P, P', P'', \dots

from which the group can be generated, they will not in general be independent of each other.

As an instance, we may take the group whose multiplication table is:—

	$1$	$P_{1}$	$P_{2}$	$P_{3}$	$P_{4}$	$P_{5}$	$P_{6}$	$P_{7}$


$1$	$1$	$P_{1}$	$P_{2}$	$P_{3}$	$P_{4}$	$P_{5}$	$P_{6}$	$P_{7}$
$P_{1}$	$P_{1}$	$P_{2}$	$P_{3}$	$1$	$P_{5}$	$P_{6}$	$P_{7}$	$P_{4}$
$P_{2}$	$P_{2}$	$P_{3}$	$1$	$P_{1}$	$P_{6}$	$P_{7}$	$P_{4}$	$P_{5}$
$P_{3}$	$P_{3}$	$1$	$P_{1}$	$P_{2}$	$P_{7}$	$P_{4}$	$P_{5}$	$P_{6}$
$P_{4}$	$P_{4}$	$P_{5}$	$P_{6}$	$P_{7}$	$1$	$P_{1}$	$P_{2}$	$P_{3}$
$P_{5}$	$P_{5}$	$P_{6}$	$P_{7}$	$P_{4}$	$P_{1}$	$P_{2}$	$P_{3}$	$1$
$P_{6}$	$P_{6}$	$P_{7}$	$P_{4}$	$P_{5}$	$P_{2}$	$P_{3}$	$1$	$P_{1}$
$P_{7}$	$P_{7}$	$P_{4}$	$P_{5}$	$P_{6}$	$P_{3}$	$1$	$P_{1}$	$P_{2}$

Here $P_{1}$ and $P_{5}$ are two operations from which every operation of the group may be generated: an inspection of the table will shew that they are connected by the relation

P_{1}^{2} P_{5}^{2} = 1 .

On the other hand, if we choose $P_{1}$ and $P_{4}$ as generating operations, we ﬁnd that every operation of the group can be expressed in the form

P_{1}^{α} P_{4}^{β}, (α = 0, 1, 2, 3; β = 0, 1),

while the only conditions to which the permutable operations

P_{1}

and

P_{4}

are submitted are

P_{1}^{4} = 1, P_{4}^{2} = 1 .

The question then arises as to whether $G$ can be generated by a set of permutable and independent operations, i.e. by a set of permutable operations which are connected by no relations except those that give their orders. That this question is always to be answered in the aﬃrmative may be proved in the following manner.

41. Let the order of $G$ be $p^{m}$ ; and let $P_{1}$ be an operation of $G$ whose order $p^{m_{1}}$ is not less than that of any other operation of the group. Then every operation of the group satisﬁes the equation

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

If $m_{1} = m$ , the order of ${P_{1}}$ is equal to the order of $G$ ; the latter is then a cyclical group generated by the operation $P_{1}$ .

If $m_{1} < m$ , $G$ must contain other operations besides those of ${P_{1}}$ . Denoting ${P_{1}}$ by $G_{1}$ , let $Q$ be any operation of $G$ not contained in $G_{1}$ , and let $Q^{a}$ be the lowest power of $Q$ that is contained in $G_{1}$ . Then (§ 33) $a$ must be a power of $p$ ; and when for $Q$ each operation of $G$ that is not contained in $G_{1}$ is taken in turn, $a$ must have some maximum value, say $p^{m_{2}}$ . Since no operation of $G$ is of greater order than $p^{m_{1}}$ it follows that $m_{2} \leq m_{1}$ . We may suppose then $Q$ to be an operation of $G$ , such that $Q^{p^{m_{2}}}$ is the lowest power of $Q$ which is contained in $G_{1}$ .

Then

Q^{p^{m_{2}}} = P_{1}^{λ_{1}},

and

Q^{p^{m_{1}}} = 1 = P_{1}^{λ_{1} p^{m_{1} - m_{2}}},

so that

λ_{1}

is divisible by

p^{m_{2}}

and may be expressed in the form

x p^{m_{2}}

. The case

λ_{1} = 0

forms no exception to this statement, since

λ_{1}

is congruent to zero,

(mod p^{m_{1}})

; but we actually take

x = 0

in this case.

If now we write

Q P_{1}^{- x} = P_{2},

then

P_{2}^{p^{m_{2}}} = 1,

and

P_{2}^{p^{m_{2}}}

is the lowest power of

P_{2}

which is contained in

G_{1}

. Let the sub-group

{P_{1}, Q}

be denoted by

G_{2}

. Then

G_{2}

is generated by the two independent operations

P_{1}

and

P_{2}

of orders

p^{m_{1}}

and

p^{m_{2}}

; and every operation

S

G

is such that

S^{p^{m_{2}}} = P_{1}^{α_{1}},

where

α_{1}

is divisible by

p^{m_{2}}

This process may now be continued step by step. That it will ultimately lead to a representation of $G$ as generated by a set of independent operations may be shewn by induction. To this end, we will suppose that a sub-group $G_{n}$ has been arrived at which is generated by the $n$ independent operations

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

of orders

p^{m_{1}}, p^{m_{2}}, \dots, p^{m_{n}},

where

m_{1} \geq m_{2} \geq \dots \geq m_{n} .

We will also suppose that, if $T$ is any operation of $G$ ,

T^{p^{m_{n}}} = P_{1}^{α_{1}} P_{2}^{α_{2}} \dots P_{n - 1}^{α_{n - 1}},

where

α_{1}

α_{2}

, …,

α_{n - 1}

are all divisible by

p^{m_{n}}

. In the special case

n = 2

these suppositions have been justiﬁed.

Let $Q'$ be any operation of $G$ , and let $Q'^{a}$ be the lowest power of $Q'$ that occurs in $G_{n}$ . Then when for $Q'$ each operation of $G$ that is not contained in $G_{n}$ is taken in turn, $a$ will have a certain maximum value, say $p^{m_{n + 1}}$ ; and from the suppositions made above, $m_{n + 1}$ is not greater than $m_{n}$ . We may therefore write

Q'^{p^{m_{n + 1}}} = P_{1}^{β_{1}} P_{2}^{β_{2}} \dots P_{n - 1}^{β_{n - 1}} P_{n}^{β_{n}},

and hence

Q'^{p^{m_{n}}} = P_{1}^{β_{1} p^{m_{n} - m_{n + 1}}} P_{2}^{β_{2} p^{m_{n} - m_{n + 1}}} \dots P_{n - 1}^{β_{n - 1} p^{m_{n} - m_{n + 1}}} P_{n}^{β_{n} p^{m_{n} - m_{n + 1}}} .

Now $P_{1}$ , $P_{2}$ , …, $P_{n}$ are independent; it follows therefore from the second assumption made above, that the exponents of $P_{1}$ , $P_{2}$ , …, $P_{n}$ in this last equation are divisible by $p^{m_{n}}$ ; and hence that $β_{1}$ , $β_{2}$ , …, $β_{n}$ are divisible by $p^{m_{n + 1}}$ .

If we now suppose that $Q'$ itself is chosen so that $Q'^{p^{m_{n + 1}}}$ is the lowest power of $Q'$ that occurs in $G_{n}$ , then

Q'^{p^{m_{n + 1}}} = P_{1}^{x_{1} p^{m_{n + 1}}} P_{2}^{x_{2} p^{m_{n + 1}}} \dots P_{n}^{x_{n} p^{m_{n + 1}}};

hence, if

Q' P_{1}^{- x_{1}} P_{2}^{- x_{2}} \dots P_{n}^{- x_{n}} = P_{n + 1},

then

P_{n + 1}^{p^{m_{n + 1}}} = 1;

and

P_{n + 1}^{p^{m_{n + 1}}}

is the lowest power of

P

that occurs in

G_{n}

. Hence ﬁnally

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

is generated by the

n + 1

independent operations

P_{1}

P_{2}

, …,

P_{n + 1}

of orders

p^{m_{1}}

p^{m_{2}}

, …,

p^{m_{n + 1}}

(

m_{1} \geq m_{2} \geq \dots \geq m_{n + 1}

); and

S

being any operation of

G

S^{p^{m_{n + 1}}} = P_{1}^{γ_{1}} P_{2}^{γ_{2}} \dots P_{n}^{γ_{n}},

where

γ_{1}

γ_{2}

, …,

γ_{n}

are divisible by

p^{m_{n + 1}}

This completes the proof by induction, and we may now state the result in the form of the following theorem:—

THEOREM II. An Abelian group of order $p^{m}$ , where $p$ is a prime, can be generated by $r$ $≤ m$

independent operations $P_{1}$ , $P_{2}$ , …, $P_{r}$ , of orders $p^{m_{1}}$ , $p^{m_{2}}$ , …, $p^{m_{r}}$ , where

m_{1} + m_{2} + \dots + m_{r} = m,

and

m_{1} \geq m_{2} \geq \dots \geq m_{r}

. Moreover if

μ

is any positive integer such that

m_{s} \geq μ \geq m_{s + 1}

, while

Q

is any operation of the group, then

Q^{p^{μ}} = P_{1}^{α_{1}} P_{2}^{α_{2}} \dots P_{s}^{α_{s}},

where

α_{1}

α_{2}

, …,

α_{s}

are divisible by

p^{μ}

The actual existence of the set of independent generating operations is demonstrated by the above inductive proof. The other inferences in the theorem may be established as follows. A set of independent permutable operations of orders $p^{m_{1}}$ , $p^{m_{2}}$ , …, $p^{m_{r}}$ generate a group of order $p^{m_{1} + m_{2} + \dots + m_{r}}$ , and therefore

m_{1} + m_{2} + \dots + m_{r} = m .

Since each number in this equation is a positive integer, the number of terms on the left-hand side cannot be greater than $m$ . Hence

r \leq m;

and, if

r = m

, every operation of the group except identity is of order

p

Finally, from the above inductive proof it follows that, if $Q$ is any operation of the group, then

Q^{p^{m_{s} + 1}} = P_{1}^{β_{1}} P_{2}^{β_{2}} \dots P_{s}^{β_{s}},

where

β_{1}

β_{2}

, …,

β_{s}

are all divisible by

p^{m_{s} + 1}

Hence

\begin{array}{l} Q^{p^{μ}} & = P_{1}^{β_{1} p^{μ - m_{s + 1}}} P_{2}^{β_{2} p^{μ - m_{s + 1}}} \dots P_{s}^{β_{s} p^{μ - m_{s + 1}}} \\ = P_{1}^{α_{1}} P_{2}^{α_{2}} \dots P_{s}^{α_{s}}, \end{array}

where $α_{1}$ , $α_{2}$ , …, $α_{s}$ are all divisible by $p^{μ}$ .

42. It is clear, from the synthetic process by which it has been proved that an Abelian group of order $p^{m}$ can be generated by a set of independent operations, that a considerable latitude exists in the choice of the actual generating operations; and the question arises as to the relations between the orders of the distinct sets of independent generating operations.

The discussion of this question is facilitated by a consideration of certain special sub-groups of $G$ . If $A$ and $B$ are two operations of $G$ , and if the order of $A$ is not less than that of $B$ , the order of $A B$ is equal to, or is a factor of, the order of $A$ . Hence the totality of those operations of $G$ whose orders do not exceed $p^{μ}$ , or in other words of those operations which satisfy the relation

S^{p^{μ}} = 1,

form a sub-group

G_{μ}

. The order of

G_{μ}

clearly depends on the orders of the various operations of

G

and in no way on a special choice of generating operations. Now if

P_{1}^{α_{1}} P_{2}^{α_{2}} \dots P_{r}^{α_{r}}

belongs to

G_{μ}

, then

P_{1}^{α_{1} p^{μ}} P_{2}^{α_{2} p^{μ}} \dots P_{r}^{α_{r} p^{μ}} = 1 .

Hence if $m_{s + 1}$ is the ﬁrst of the series

m_{1}, m_{2}, \dots, m_{r},

which is less than

μ

, then

α_{s + 1}

, …,

α_{r}

may have any values whatever; but

α_{t}

(

t = 1, 2, \dots, s

) must be a multiple of

p^{m_{t} - μ}

It follows from this that $G_{μ}$ is generated by the $r$ independent operations

P_{1}^{p^{m_{1} - μ}}, P_{2}^{p^{m_{2} - μ}}, \dots, P_{s}^{p^{m_{s} - μ}}, P_{s + 1}, \dots, P_{r} .

If then the order of

G_{μ}

p^{ν}

, we have

ν = μ s + \sum_{s + 1}^{r} m_{t} .

The order of $G_{1}$ , the sub-group formed of all operations of $G$ whose order is $p$ , is clearly $p^{r}$ .

43. Suppose now that by a fresh choice of independent generating operations, it were found that $G$ could be generated by the $r'$ independent operations

P', P', \dots, P'

of orders

p^{m'}, p^{m'}, \dots, p^{m'},

where

m' \geq m' \geq \dots \geq m' .

If $m'$ is the ﬁrst of this series which is less than $μ$ , the order of $G_{μ}$ will be $p^{ν'}$ , where

ν' = μ s' + \sum_{s' + 1}^{r'} m' .

The order of $G_{μ}$ is independent of the choice of generating operations; so that for all values of $μ$

ν = ν' .

Hence, by taking $μ = 1$ ,

r = r',

or the number of independent generating operations is independent of their choice.

If now

m_{t} = m' (t = s + 1, s + 2, \dots, r),

and

m_{s} > m',

and if we choose

μ

so that

m_{s} \geq μ > m';

then

ν = μ s + \sum_{s + 1}^{r} m_{t},

and

ν' = μ (s - 1) + m' + \sum_{s + 1}^{r} m_{t} .

The condition

ν = ν'

gives

μ = m',

in contradiction to the assumption just made.

Similarly we can prove that the assumption $m' > m_{s}$ cannot be maintained; hence

m_{s} = m';

and therefore, however the independent generating operations of

G

are chosen, their number is always

r

, and their orders are

p^{m_{1}}, p^{m_{2}}, \dots, p^{m_{r}} .

44. If $G'$ is a second Abelian group of order $p^{m}$ , simply isomorphic with $G$ , and if

P', P', \dots, P'

of orders

p^{m'}, p^{m'}, \dots, p^{m'},

where

m' \geq m' \geq \dots \geq m',

are a set of independent generating operations of

G'

, exactly the same process as that of the last paragraph may be used to shew that

r = r',

and

m_{s} = m' (s = 1, 2, \dots, r) .

In fact, since corresponding operations of two simply isomorphic groups have the same order, the order of $G_{μ}$ must be equal to the order of $G'$ ; and this is the condition that has been used to obtain the result of the last paragraph.

Two Abelian groups of order $p^{m}$ cannot therefore be simply isomorphic unless the series of integers $m_{1}$ , $m_{2}$ , …, $m_{r}$ is the same for each. On the other hand when this condition is satisﬁed, it is clear that the two groups are simply isomorphic, since by taking $P_{s}$ and $P'$ ( $s = 1, 2, \dots, r$ ) as corresponding operations, the isomorphism is actually established.

The number of distinct types of Abelian groups of order $p^{m}$ , where $p$ is a prime, i.e. the number of such groups no one of which is simply isomorphic with any other, is therefore equal to the number of partitions of $m$ . When the prime $p$ is given, each type of group may be conveniently, and without ambiguity, represented by the symbol of the corresponding partition. Thus the typical group $G$ that we have been dealing with would be represented by the symbol $(m_{1}, m_{2}, \dots, m_{r})$ .

45. Having thus determined all distinct types of Abelian groups of order $p^{m}$ , a second general problem in this connection is the determination of all possible types of sub-group when the group itself is given. This will be facilitated by the consideration of a second special class of sub-groups in addition to the sub-groups $G_{μ}$ already dealt with.

If $S$ and $S'$ are any two operations of $G$ , then

S^{p^{μ}} S'^{p^{μ}} = {(S S')}^{p^{μ}};

and therefore the totality of the distinct operations obtained by raising every operation of

G

to the power

p^{μ}

will form a sub-group

H_{μ}

m_{s} \geq μ > m_{s + 1},

then (Theorem II, § 41)

S^{p^{μ}} = P_{1}^{α_{1} p^{μ}} P_{2}^{α_{2} p^{μ}} \dots P_{s}^{α_{s} p^{μ}},

S

being any operation of

G

. Hence

H_{μ}

is generated by the

s

independent operations

P_{1}^{p^{μ}}, P_{2}^{p^{μ}}, \dots, P_{s}^{p^{μ}};

and the order of

H_{μ}

p^{\sum_{1}^{s} m_{t} - μ s}

46. Let now $Γ$ of type $(n_{1}, n_{2}, \dots, n_{s})$ be any sub-group of $G$ . The order of the group $Γ_{1}$ , formed of all the operations of $Γ$ which satisfy the equation

S^{p} = 1,

p^{s}

(§ 42). This group must be identical with or be a sub-group of

G_{1}

whose order is

p^{r}

. Hence

s \leq r,

i.e. the number of independent generating operations of any sub-group of

G

is equal to or is less than the number of independent generating operations of

G

itself.

If $p^{n}$ is the order of $Γ$ , there must be one or more sub-groups of $G$ of order $p^{n + 1}$ which contain $Γ$ ; for $\frac{G}{Γ}$ has operations of order $p$ . Hence $Γ$ must be contained in one or more sub-groups of $G$ of order $p^{m - 1}$ . We may begin then by considering all possible types of sub-groups of $G$ of order $p^{m - 1}$ . Let $g$ be such a sub-group and $(m', m', \dots, m')$ its type; and suppose that

m_{t} = m' (t = 1, 2, \dots, k - 1),

and

m_{k} \neq m' .

Then if

m_{k} < m',

the order of

h_{m_{k}}

, the sub-group of

g

which results by raising all its operations to the power

p^{m_{k}}

, is greater than the order of

H_{m_{k}}

. This is impossible, since

h_{m_{k}}

must coincide with

H_{m_{k}}

or with one of its sub-groups. Hence

m' < m_{k} .

Now

\sum_{1}^{r'} m' = \sum_{1}^{r} m_{t} - 1;

therefore

m' = m_{k} - 1,

and

m' = m_{t} (t = k + 1, \dots, r) .

In the particular case in which $m_{r}$ is unity if we take $k$ equal to $r$ , $m'$ is zero; and in this case $g$ would have $r - 1$ generating operations.

It is easy to see that sub-groups of order $p^{m - 1}$ and of all the types just determined actually exist. For

P_{1}, P_{2}, \dots, P_{k - 1}, P_{k^{p}}, P_{k + 1}, \dots, P_{r}

are a set of independent operations which generate a sub-group of the type

(m_{1}, m_{2}, \dots, m_{k - 1}, m_{k} - 1, m_{k + 1}, \dots, m_{r})

47. Returning to the general case, we will assume for all orders not less than $p^{n}$ the existence of a sub-group $Γ$ of type $(n_{1}, n_{2}, \dots, n_{s})$ , when the inequalities

\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u} (t = 1, 2, \dots, r)

are satisﬁed, where if

t > s

n_{t}

is zero. When

n

is equal to

m - 1

, the truth of this assumption has been established.

If $Γ'$ of type $(n', n', \dots, n')$ is a sub-group of order $p^{n - 1}$ , it must be contained in some sub-group of order $p^{n}$ , say $Γ$ of type $(n_{1}, n_{1}, \dots, n_{s})$ . Then, by the preceding discussion,

\sum_{1}^{t} n' \leq \sum_{1}^{t} n_{u} (t = 1, 2, \dots, s);

and therefore

\sum_{1}^{t} n' \leq \sum_{1}^{t} m_{u} (t = 1, 2, \dots, r),

so that the conditions assumed for sub-groups of order not less than

p^{n}

are necessary for sub-groups of order

p^{n - 1}

. Moreover if the conditions

\sum_{1}^{t} n' \leq \sum_{1}^{t} m_{u} (t = 1, 2, \dots, r)

are satisﬁed, and if

n'

is the ﬁrst of the series

n', n', \dots,

which is less than the corresponding term of the series

m_{1}, m_{2}, \dots,

we may take

\begin{array}{l} n_{u} & = n', u \neq k, \\ n_{k} & = n' + 1; \end{array}

and then the conditions

\sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u} (t = 1, 2, \dots, r);

are satisﬁed. But these conditions being satisﬁed it follows, from the assumption made, that

G

contains a sub-group of order

p^{n}

and type

(n_{1}, n_{2}, \dots, n_{s})

and therefore also a sub-group of order

p^{n - 1}

and type

(n', n', \dots, n')

. If the conditions assumed are suﬃcient for the existence of a sub-group of order

p^{n}

, they are thus proved to be suﬃcient for the existence of one of order

p^{n - 1}

. Hence since they are suﬃcient in the case of sub-groups of order

p^{m - 1}

, they are so generally.

We may summarize the results obtained in the last three paragraphs as follows:—

THEOREM III. The number of distinct types of Abelian groups of order $p^{m}$ , where $p$ is a prime, is equal to the number of partitions of $m$ ; and each type may be completely represented by the symbol $(m_{1}, m_{2}, \dots, m_{r})$ of the corresponding partition. If the numbers in the partition are written in descending order, a group of type $(m_{1}, m_{2}, \dots, m_{r})$ will have a sub-group of type $(n_{1}, n_{2}, \dots, n_{s})$ , when the conditions

\begin{matrix} s \leq r, \\ \sum_{1}^{t} n_{u} \leq \sum_{1}^{t} m_{u} (t = 1, 2, \dots, r), \end{matrix}

are satisﬁed; and the type of every sub-group must satisfy these conditions.

48. It will be seen later that the Abelian group of order $p^{m}$ and type $(1, 1, 1, \dots, with m units)$ is of special importance in the general theory, and we shall here discuss one or two of its simpler properties.

Since the generating operations of the group are all of order $p$ , every operation except identity is of order $p$ ; and therefore the type of any sub-group of order $p^{s}$ is $(1, 1, 1, \dots, with s units)$ . If the group be denoted by $G$ , every sub-group $G_{μ}$ coincides with $G$ ; while of the sub-groups $H_{μ}$ , the ﬁrst coincides with $G$ and all the rest consist of the identical operation only.

In choosing a set of independent generating operations, we may take for the ﬁrst, $P_{1}$ , any one of the $p^{m} - 1$ operations of the group, other than identity. The sub-group ${P_{1}}$ is of order $p$ ; and therefore $G$ has $p^{m} - p$ operations which are not contained in ${P_{1}}$ . If we choose any one of these, $P_{2}$ , it is necessarily independent of $P_{1}$ , and may be taken as a second generating operation. The sub-group ${P_{1}, P_{2}}$ is of order $p^{2}$ and type $(1, 1)$ ; and $G$ has $p^{m} - p^{2}$ operations which are not contained in this sub-group. If $P_{3}$ be any one of these, no power of $P_{3}$ other than identity is contained in ${P_{1}, P_{2}}$ ; and $P_{1}$ , $P_{2}$ , $P_{3}$ are therefore three independent operations which generate a sub-group of order $p^{3}$ . This process may clearly be continued till all $m$ generating operations have been chosen. If then the position which each generating operation occupies in the set of $m$ , when they are written in order, be taken into account, there are

(p^{m} - 1) (p^{m} - p) (p^{m} - p^{2}) \dots (p^{m} - p^{m - 1})

distinct ways in which a set may be chosen. If on the other hand the sets of generating operations which consist of the same operations written in diﬀerent orders be regarded as identical, the number of distinct sets is

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

49. No operation $P$ of the group can belong to two distinct sub-groups of order $p$ except the identical operation. Hence since every sub-group of order $p$ contains $p - 1$ operations besides identity, $G$ must contain $\frac{p^{m} - 1}{p - 1}$ sub-groups of order $p$ .

Let $N_{m, r}$ be the number of sub-groups of $G$ of order $p^{r}$ , so that

N_{m, 1} = \frac{p^{m} - 1}{p - 1} .

There are, in $G$ , $p^{m} - p^{r}$ operations not contained in any given sub-group of order $p^{r}$ . If $P$ occurs among these operations, so also do $P^{2}$ , $P^{3}$ , …, $P^{p - 1}$ . Hence there are $\frac{p^{m} - p^{r}}{p - 1}$ sub-groups of order $p$ in $G$ which are not contained in a given sub-group of order $p^{r}$ . Each of these may be combined with the given sub-group to give a sub-group of order $p^{r + 1}$ . When every sub-group of order $p^{r}$ is treated in this way, every sub-group of order $p^{r + 1}$ will be formed and each of them the same number, $x$ , of times. Hence

x N_{m, r + 1} = N_{m, r} \frac{p^{m} - p^{r}}{p - 1} .

Now a sub-group of order $p^{r + 1}$ contains $N_{r + 1, r}$ sub-groups of order $p^{r}$ , and $\frac{p^{r + 1} - p^{r}}{p - 1}$ sub-groups of order $p$ which are not contained in any given sub-group of order $p^{r}$ . Hence

x = \frac{p^{r + 1} - p^{r}}{p - 1} N_{r + 1, r},

and therefore

N_{m, r + 1} = \frac{N_{m, r}}{N_{r + 1, r}} \cdot \frac{p^{m - r} - 1}{p - 1} .

We will now assume that

N_{m, t} = \frac{(p^{m} - 1) (p^{m - 1} - 1) \dots (p^{m - t + 1} - 1)}{(p - 1) (p^{2} - 1) \dots (p^{t} - 1)},

for all values of

m

and for values of

t

not exceeding

r

. This has been proved for

r = 1

. Then it follows, from the above relation, that

N_{m, r + 1} = \frac{(p^{m} - 1) (p^{m - 1} - 1) \dots (p^{m - r} - 1)}{(p - 1) (p^{2} - 1) \dots (p^{r + 1} - 1)},

that is to say, if the result is true for values of

t

not exceeding

r

, it is also true when

t = r + 1

. Hence the formula is true generally.

It may be noticed that

N_{m, t} = N_{m, m - t} .

50. Ex. 1. Shew that a group whose operations except identity are all of order $2$ is necessarily an Abelian group.

Ex. 2. Prove that in a group of order $16$ , whose operations except identity are all of order $2$ , the $15$ operations of order $2$ may be divided into $5$ sets of $3$ each so that each set of $3$ with identity forms a sub-group of order $4$ ; and that this division into sets may be carried out in $56$ distinct ways.

Ex. 3. If $G$ is an Abelian group and $H$ a sub-group of $G$ , shew that $G$ contains one or more sub-groups simply isomorphic with $\frac{G}{H}$ .

Ex. 4. If the symbols in the successive rows of a determinant of $n$ rows are derived from those of the ﬁrst row by performing on them the substitutions of a regular Abelian group of order $n$ , prove that the determinant is the product of $n$ linear factors. (Messenger of Mathematics, Vol. XXIII p. 112.)

Ex. 5. Discuss the number of ways in which a set of independent generating operations of an Abelian group of order $p^{m}$ and given type may be chosen. Shew that, for a group of type $(m_{1}, m_{2}, \dots, m_{r})$ , where $m_{1} > m_{2} > \dots > m_{r}$ , the number of ways is of the form $p^{α} {(p - 1)}^{r}$ ; and in particular that for a group of order $p^{\frac{1}{2} n (n + 1)}$ and type $(n, n - 1, \dots, 2, 1)$ , the number of ways is $p^{ν} {(p - 1)}^{n}$ , where $ν = \frac{1}{6} n (n + 1) (2 n + 1) - n$ .

Ex. 6. Shew that for any Abelian group a set of independent generating operations

S_{1}, S_{2}, \dots, S_{r - 1}, S_{r}, \dots, S_{n}

can be chosen such that, for each value of

r

, the order of

S_{r}

is equal to, or is a factor, of the order of

S_{r - 1}

ptail

top

Chapter IV.On Abelian Groups.

Chapter IV.
On Abelian Groups.