V. On Groups Whose Orders Are the Powers of Primes.

Chapter V.
On Groups Whose Orders Are the Powers of Primes.

H aving in the last chapter dealt in some detail with Abelian groups of order $p^{m}$ , where $p$ is a prime, we shall now investigate some of the more important properties of groups, which have the power of a prime for their order but are not necessarily Abelian. Besides illustrating and leading to many interesting applications of the general theorems of Chapter III, the discussion of groups, whose order is the power of a prime, will be found in many ways to facilitate the subsequent discussion of other groups, whose order is not thus limited.

52. If $G$ is a group whose order is $p^{m}$ , where $p$ is a prime, the order of every sub-group of $G$ must also be a power of $p$ ; and therefore (§ 25) the total number of operations of $G$ which are conjugate with any given operation must be a power of $p$ . The identical operation of $G$ is self-conjugate. Hence if the operations of $G$ , other than the identical operation, are distributed in conjugate sets containing $p^{α}$ , $p^{β}$ , … operations, the total number of operations of $G$ is $1 + p^{α} + p^{β} + \dots$ ; and therefore

p^{m} = 1 + p^{α} + p^{β} + \dots .

This equation can only be true if $p^{r_{1}} - 1$ of the indices $α$ , $β$ , … are zero, $r_{1}$ being some integer not less than unity. Hence $G$ must contain $p^{r_{1}}$ self-conjugate operations, which form (§ 27) a self-conjugate sub-group¹⁴. Hence:—

THEOREM I. Every group whose order is the power of a prime contains self-conjugate operations, other than the identical operation; and no such group can be simple.

53. Let $H_{1}$ be the sub-group of $G$ of order $p^{r_{1}}$ , formed of its self-conjugate operations. Then $\frac{G}{H_{1}}$ is a group of order $p^{m - r_{1}}$ , and it must contain self-conjugate operations other than identity, forming a self-conjugate sub-group of order $p^{r_{2}}$ . Corresponding to this self-conjugate sub-group of $\frac{G}{H_{1}}$ , $G$ must have a self-conjugate sub-group $H_{2}$ of order $p^{r_{1} + r_{2}}$ , which contains $H_{1}$ . If $r_{1} + r_{2} < m$ , $\frac{G}{H_{2}}$ is a group of order $p^{m - r_{1} - r_{2}}$ ; and it again will have self-conjugate operations forming a sub-group of order $p^{r_{3}}$ . Corresponding to this, $G$ will have a self-conjugate sub-group $H_{3}$ of order $p^{r_{1} + r_{2} + r_{3}}$ , which contains $H_{2}$ . If the process thus indicated is continued, till $G$ itself is arrived at, a series of groups

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

is formed, each of which is self-conjugate in

G

and contains all that precede it¹⁵. Moreover

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

is the group formed of the self-conjugate operations of

\frac{G}{H_{r}}

, and it is therefore an Abelian group.

The series of sub-groups thus arrived at is of considerable importance. From the mode of formation, it is clear that a second group $G'$ of order $p^{m}$ cannot be simply isomorphic with $G$ , unless $\frac{H_{r + 1}}{H_{r}}$ and $\frac{H'}{H'}$ are simply isomorphic for each value of $r$ . It by no means however follows that, when this latter condition is satisﬁed, $G$ and $G'$ are always simply isomorphic. This is indeed not necessarily the case; instances to the contrary will be found among the groups of order $p^{3}$ and $p^{4}$ in §§ 69–72. While in general there is no limitation on the type of any Abelian factor-group $\frac{H_{r + 1}}{H_{r}}$ formed from two consecutive terms of the series, it is to be noted that the last factor-group $\frac{G}{H_{n}}$ cannot be cyclical¹⁶, a restriction that can be established as follows. If $\frac{G}{H_{n}}$ were cyclical, then, because $\frac{H_{n}}{H_{n - 1}}$ is the group formed of the self-conjugate operations of $\frac{G}{H_{n - 1}}$ , the operations of $\frac{G}{H_{n - 1}}$ could be arranged in the sets

\frac{H_{n}}{H_{n - 1}}, P \frac{H_{n}}{H_{n - 1}}, P^{2} \frac{H_{n}}{H_{n - 1}}, \dots, P^{p^{α - 1}} \frac{H_{n}}{H_{n - 1}},

p^{α}

being the order of

\frac{G}{H_{n}}

and

P^{p^{α}}

being an operation of

\frac{H_{n}}{H_{n - 1}}

. But since

P

is permutable with its own powers and with every operation of

\frac{H_{n}}{H_{n - 1}}

, these operations would form an Abelian group. Now the self-conjugate operations of

\frac{G}{H_{n - 1}}

form the group

\frac{H_{n}}{H_{n - 1}}

, and therefore

\frac{G}{H_{n - 1}}

cannot be Abelian. Hence the assumption, that

\frac{G}{H_{n}}

is cyclical, is incorrect.

One immediate consequence of this result is that a group $G$ of order $p^{2}$ is necessarily Abelian. For if it were not, its self-conjugate operations would form a sub-group $H$ of order $p$ , and $\frac{G}{H}$ being of order $p$ would be cyclical.

A second consequence is that if $G$ , of order $p^{m}$ , is not Abelian, the order of the sub-group $H_{1}$ , formed of the self-conjugate operations of $G$ , cannot exceed $p^{m - 2}$ .

54. It was proved in the last chapter that an Abelian group has one or more sub-groups the order of which is any given factor of the order of the group itself. Hence, since $\frac{H_{r + 1}}{H_{r}}$ is Abelian, $H_{r + 1}$ must have one or more sub-groups which contain $H_{r}$ , of any given order greater than that of $H_{r}$ and less than its own.

In particular, if $p^{s}$ and $p^{t}$ are the orders of $H_{r}$ and $H_{r + 1}$ , then $\frac{H_{r + 1}}{H_{r}}$ must have a series of sub-groups of orders

p^{t - s - 1}, p^{t - s - 2}, \dots, p, 1,

each of which contains those that follow it, while each one is of necessity self-conjugate in

\frac{G}{H_{r}}

. Hence between

H_{r}

and

H_{r + 1}

, a series of groups

h_{s + 1}

h_{s + 2}

, …,

h_{t - 1}

of orders

p^{s + 1}

p^{s + 2}

, …,

p^{t - 1}

can be chosen, so that each of the series

H_{r}, h_{s + 1}, h_{s + 2}, \dots, h_{t - 1}, H_{r + 1}

is self-conjugate in

G

and contains all that precede it. We can therefore always form, in at least one way, a series of sub-groups of

G

of orders

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

such that each is self-conjugate in

G

and contains all that precede it.

It will be shewn in the next paragraph that every sub-group of order $p^{s}$ of $G$ is contained self-conjugately in some sub-group of order $p^{s + 1}$ , and that every sub-group of order $p^{s}$ of a group of order $p^{s + 1}$ is self-conjugate. Assuming this result, it follows at once that a series of sub-groups of orders

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

can always be formed, which shall contain any one given sub-group and which shall be such that each group of the series contains the previous one self-conjugately. It will not however, in general, be the case, as in the previous series, that each group of the series is self-conjugate in

G

55. Let $G_{s}$ be any sub-group of $G$ of order $p^{s}$ , and let $H_{r + 1}$ be the ﬁrst of the series of sub-groups

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

which is not contained in

G_{s}

. Then

\frac{G_{s}}{H_{r}}

is a sub-group of

\frac{G}{H_{r}}

which does not contain all the operations of

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

. Every operation of

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

is self-conjugate in

\frac{G}{H_{r}}

; and therefore

\frac{G_{s}}{H_{r}}

is self-conjugate in

\{\frac{G_{s}}{H_{r}}, \frac{H_{r + 1}}{H_{r}}\}

, a group of order greater than its own. Hence

G_{s}

must be contained self-conjugately in some group

G_{s + t}

of order

p^{s + t}

, where

t

is not less than unity. Moreover since

\frac{G_{s + t}}{G_{s}}

must contain operations of order

p

, there must be one or more groups of order

p^{s + 1}

which contain

G_{s}

self-conjugately. Hence:—

THEOREM II. If $G_{s}$ of order $p^{s}$ is a sub-group of $G$ , which is of order $p^{m}$ , then $G$ must contain a sub-group of order $p^{s + t}$ , $t ≮ 1$ , within which $G_{s}$ is self-conjugate. In particular, every sub-group of order $p^{m - 1}$ of $G$ is a self-conjugate sub-group¹⁷.

Suppose now that $G_{s + t}$ , of order $p^{s + t}$ , is the greatest sub-group of $G$ which contains a given sub-group $G_{s}$ , of order $p^{s}$ , self-conjugately; so that $G_{s}$ is one of $p^{m - s - t}$ conjugate sub-groups. Suppose also that $G_{s + t + u}$ of order $p^{s + t + u}$ , ( $u ≮ 1$ ), is the greatest sub-group of $G$ that contains $G_{s + t}$ self-conjugately. Every operation of $G_{s + t + u}$ transforms $G_{s + t}$ into itself; and no operation of $G_{s + t + u}$ that is not contained in $G_{s + t}$ transforms $G_{s}$ into itself. Hence, in $G_{s + t + u}$ , $G_{s}$ is one of $p^{u}$ conjugate sub-groups and each of these is self-conjugate in $G_{s + t}$ .

The $p^{m - s - t}$ sub-groups conjugate with $G_{s}$ may therefore be divided into $p^{m - s - t - u}$ sets of $p^{u}$ each, ( $u ≮ 1$ ), such that any operation of one of the sets transforms each sub-group of that set into itself.

Similarly if $G_{r}$ , of order $p^{r}$ , is the greatest sub-group of $G$ that contains a given operation $P$ self-conjugately, and if $G_{r + s}$ , $s ≮ 1$ , is the greatest sub-group that contains $G_{r}$ self-conjugately, then $G_{r}$ must contain self-conjugately $p^{s}$ operations of the conjugate set to which $P$ belongs, and therefore any two of these $p^{s}$ operations are permutable. Hence the $p^{m - r}$ conjugate operations of the set to which $P$ belongs can be divided into $p^{m - r - s}$ sets of $p^{s}$ each, ( $s ≮ 1$ ), such that all the operations of any one set are permutable with each other. In particular, if $P$ is one of a set of $p$ conjugate operations, all the operations of the set are permutable.

56. If $P$ is an operation of $G$ , of order $p^{n}$ , which is conjugate to one of its own powers $P^{α}$ , there must be some other operation $Q$ such that

Q^{- 1} P Q = P^{α} .

From this equation it follows that

Q^{- 1} P^{x} Q = {(Q^{- 1} P Q)}^{x} = P^{x α} .

Again

\begin{array}{l} Q^{- 2} P Q^{2} & = Q^{- 1} P^{α} Q & = P^{α^{2}}, \\ Q^{- 3} P Q^{3} & = Q^{- 1} P^{α^{2}} Q & = P^{α^{3}}, \end{array}

and so on. Hence

Q^{- y} P^{x} Q^{y} = P^{x α^{y}} .

If $Q^{β}$ is the lowest power of $Q$ which is permutable with $P$ , then in ${P, Q}$ $P$ will be one of $β$ conjugate operations, and therefore $β$ must be a power of $p$ , say $p^{r}$ , less than $p^{n}$ . Now

Q^{- p^{r}} P Q^{p^{r}} = P^{α^{p^{r}}};

and therefore

α^{p^{r}} \equiv 1 (mod p^{n}),

while

α^{p^{r - 1}} ≢ 1 (mod p^{n}) .

First, we will suppose that $p$ is an odd prime. Then since

x^{p^{r}} = x (mod p),

whatever integer

x

may be, we may assume that

α = 1 + k p^{s}

, where

k

is not a multiple of

p

; and then

\begin{matrix} α^{p^{r}} = 1 + k p^{r + s} + \dots \\ α^{p^{r - 1}} = 1 + k p^{r + s - 1} + \dots \end{matrix}

Hence

r + s = n,

and

α \equiv 1 (mod p^{n - r}) .

In particular, we see that no operation of order $p$ can be conjugate to one of its powers. Hence if $P$ and $P'$ are two conjugate operations of order $p$ , ${P}$ and ${P'}$ have no operation in common except identity. Also, if ${P}$ be a self-conjugate sub-group of order $p$ , each of its operations is self-conjugate.

If $p$ is $2$ , we must take

α = \pm 1 + k 2^{s},

where

k

is odd, and we are led by the same process to the result

α \equiv \pm 1 (mod 2^{n - r}) .

57. If $P$ is an operation of $G$ which belongs to the sub-group $H_{r + 1}$ (§ 53) and to no previous sub-group in the set

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

then

{P, H_{r}}

is a sub-group of

H_{r + 1}

; and therefore every operation of

\frac{{P, H_{r}}}{H_{r}}

is self-conjugate in

\frac{G}{H_{r}}

. The set of operations

P H_{r}

is therefore transformed into itself by every operation of

G

, and hence every operation that is conjugate to

P

is contained in the set

P H_{r}

. Suppose now, if possible, that all the operations conjugate with

P

are contained in the set

P H_{r - 1}

. Then every operation of

G

transforms this set into itself, and therefore every operation of

\frac{{P, H_{r - 1}}}{H_{r - 1}}

is self-conjugate in

\frac{G}{H_{r - 1}}

. This however cannot be the case, since by supposition

P

does not belong to

H_{r}

. Hence among the operations conjugate to

P

, there must be some which belong to

P H_{r}

and not to

P H_{r - 1}

. In particular, if

P

is one of

p

conjugate operations, no operation conjugate to

P

can belong to the set

P H_{r - 1}

. For there must be an operation

Q

such that

Q^{- 1} P Q

belongs to

P H_{r}

and not to

P H_{r - 1}

; and this being the case, no one of the operations

Q^{- x} P Q^{x}

, (

x = 1, 2, \dots, p - 1

), can belong to

P H_{r - 1}

. But these operations with

P

constitute the conjugate set.

It follows from the above reasoning that, if $P$ belongs to $H_{r + 1}$ , then each operation of the set $P^{- α} Q^{- 1} P^{α} Q$ , where $Q$ is any operation of $G$ , belongs to $H_{r}$ . If $Q$ belongs to $H_{s + 1}$ , ( $s < r$ ), the operation $P^{- α} Q^{- 1} P^{α} Q$ , regarded as the product of $P^{- α} Q^{- 1} P^{α}$ and $Q$ , must similarly belong to $H_{s}$ . Hence unless every operation of $H_{s + 1}$ is permutable with $P$ , there must be operations conjugate to $P$ , which belong to the set $P H_{s}$ . Moreover those operations of $G$ which transform $P$ into operations of the set $P H_{s}$ form a sub-group. For if

S^{- 1} P S = P h_{s},

and

S'^{- 1} P S' = P h',

then

S'^{- 1} S^{- 1} P S S' = P h' \cdot S'^{- 1} h_{s} S,

and when both

h_{s}

and

h'

belong to

H_{s}

, so also does

h' S'^{- 1} h_{s} S'

58. In illustration of the foregoing paragraph, we will consider a group $G$ of order $p^{r + s}$ where $p$ is an odd prime in which the group $H_{1}$ , is of type $(r)$ , while $\frac{G}{H_{1}}$ is an Abelian group of type $(1, 1, \dots with s units)$ .

Let $Q$ be an operation of order $p^{r}$ that generates $H_{1}$ ; and let $P_{1}$ , $P_{2}$ , …, $P_{s}$ be $s$ operations no one of which is a power of any other, such that, with $Q$ , they generate $G$ .

If $P_{1}$ and $P_{2}$ are not permutable, then

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

and

P_{2}^{- p} P_{1} P_{2}^{p} = P_{1} Q^{p α} .

Now $P_{2}^{p}$ belongs to $H_{1}$ , and therefore

Q^{p α} = 1,

so that

α = k p^{r - 1} .

Hence the only operations which are conjugate to $P_{1}$ are

P_{1}, P_{1} Q^{p^{r - 1}}, P_{1} Q^{2 p^{r - 1}}, \dots, P_{1} Q^{(p - 1) p^{r - 1}};

and

P_{1}

is one of a set of

p

conjugate operations, which are transformed among themselves by

P_{2}

. Similarly

P_{2}

is one of a set of

p

conjugate operations which are transformed among themselves by

P_{1}

. Hence both

P_{1}

and

P_{2}

are permutable with each of the operations

P_{3}

P_{4}

, …,

P_{s}

. Again,

P_{3}

is not self-conjugate, and there must therefore be another, say

P_{4}

, of the set

P_{3}

P_{4}

, …,

P_{s}

which, with its powers, transforms

P_{3}

into a set of

p

conjugate operations. Then

P_{3}

will similarly transform

P_{4}

; and

P_{3}

P_{4}

will be permutable with each of the set

P_{5}

, …,

P_{s}

P_{1}

P_{2}

. Hence ﬁnally the set of

s

operations

P_{1}

P_{2}

, …,

P_{s}

must be divisible into sets of two, such that each pair are permutable with all the remaining operations, but are not permutable with each other. The group can therefore only exist if

s

is even¹⁸.

The sub-group ${P_{1}, P_{2}, Q}$ is a group which satisﬁes the same conditions as $G$ , when $s = 2$ . Its order is $p^{r + 2}$ , and it contains a self-conjugate operation of order $p^{r}$ . Now we shall see in §§ 65, 66 that such a group is necessarily of one of two types¹⁹, the generating operations of which satisfy one or the other of the sets of equations

Q^{p^{r}} = 1, P_{1}^{p} = 1, P_{2}^{p} = 1, P_{2}^{- 1} P_{1} P_{2} = P_{1} Q^{p^{r - 1}} :

Q^{p^{r}} = 1, P_{1}^{p} = Q, P_{2}^{p} = 1, P_{2}^{- 1} P_{1} P_{2} = P_{1} Q^{p^{r - 1}} .

The same is true for the sub-groups ${P_{3}, P_{4}, Q}$ , etc.; and all the operations of any one of these sub-groups are permutable with each operation of ${P_{1}, P_{2}, Q}$ .

Hence ﬁnally, since the equations to be satisﬁed by each pair of operations, such as $P_{1}$ and $P_{2}$ , may be chosen in two distinct ways, there are in all $2^{\frac{1}{2} s}$ distinct types of group of order $p^{r + s}$ , for which $H_{1}$ is a cyclical group of order $p^{r}$ , and $\frac{G}{H_{1}}$ is an Abelian group of type $(1, 1, \dots, to s units)$ .

59. It has been seen in § 55 that every sub-group $G'$ of order $p^{m - 1}$ of a group $G$ of order $p^{m}$ is self-conjugate. Suppose now that $G$ contains two such sub-groups $G'$ and $G''$ . Then since $G'$ and $G''$ are permutable with each other, while the order of ${G', G''}$ is $p^{m}$ , the order of the greatest group $g'$ common to them must (§ 33) be $p^{m - 2}$ ; and since $g'$ is the greatest common sub-group of two self-conjugate sub-groups of $G$ , it must itself be a self-conjugate sub-group of $G$ . The factor group $\frac{G}{g'}$ of order $p^{2}$ contains the two distinct sub-groups $\frac{G'}{g'}$ and $\frac{G''}{g'}$ , which are of order $p$ and permutable with each other. Hence $\frac{G}{g'}$ must be an Abelian group of type $(1, 1)$ , and it therefore contains (§ 49) $p + 1$ sub-groups of order $p$ . Hence, besides $G'$ , $G$ must contain $p$ other sub-groups of order $p^{m - 1}$ which have in common with $G'$ the sub-group $g'$ . If the $p + 1$ sub-groups thus obtained do not exhaust the sub-groups of $G$ of order $p^{m - 1}$ , let $G'''$ be a new one. Then, as before, $G'$ and $G'''$ must have a common sub-group $g''$ , of order $p^{m - 2}$ , which is self-conjugate in $G$ . If $g''$ were the same as $g'$ , $\frac{G'''}{g'}$ would be contained in $\frac{G}{g'}$ , which by supposition is not the case. It may now be shewn as above that there are, in addition to $G'$ , $p$ sub-groups of order $p^{m - 1}$ which have in common with $G'$ the group $g''$ . These are therefore necessarily distinct from those before obtained. This process may clearly be repeated till all the sub-groups of order $p^{m - 1}$ are exhausted. Hence ﬁnally, if the number of sub-groups of $G$ , of order $p^{m - 1}$ be $r_{m - 1}$ , we have $r_{m - 1} \equiv 1 (mod p)$ .

60. The self-conjugate operations of a group $G$ of order $p^{m}$ , whose orders are $p$ , form with identity a self-conjugate sub-group whose order is some power of $p$ ; and therefore their number must be congruent to $- 1, (mod p)$ . On the other hand, if $P$ is any operation of $G$ of order $p$ which is not self-conjugate, the number of operations in the conjugate set to which $P$ belongs is a power of $p$ . Hence the total number of operations of $G$ , of order $p$ , is congruent to $- 1, (mod p)$ . Now if $r_{1}$ is the total number of sub-groups of $G$ of order $p$ , the number of operations of order $p$ is $r_{1} (p - 1)$ , since no two of these sub-groups can have a common operation, except identity. It follows that

r_{1} (p - 1) \equiv - 1, (mod p),

and therefore

r_{1} \equiv 1, (mod p) .

If now $G_{s}$ is any sub-group of $G$ of order $p^{s}$ , and if $G_{s + t}$ is the greatest sub-group of $G$ in which $G_{s}$ is contained self-conjugately, then every sub-group of $G$ which contains $G_{s}$ self-conjugately is contained in $G_{s + t}$ . But every sub-group of order $p^{s + 1}$ , which contains $G_{s}$ , contains $G_{s}$ self-conjugately; and therefore every sub-group of order $p^{s + 1}$ , which contains $G_{s}$ , is itself contained in $G_{s + t}$ . By the preceding result, the number of sub-groups of $\frac{G_{s + t}}{G_{s}}$ of order $p$ is congruent to unity, $(mod p)$ . Hence the number of sub-groups of $G$ of order $p^{s + 1}$ , which contain $G_{s}$ of order $p^{s}$ , is congruent to unity, $(mod p)$ .

61. Let now $r_{s}$ represent the total number of sub-groups of order $p^{s}$ contained in a group $G$ of order $p^{m}$ . If any one of them is contained in $a_{x}$ sub-groups of order $p^{s + 1}$ , and if any one of the sub-groups of order $p^{s + 1}$ contains $b_{y}$ sub-groups of order $p^{s}$ , then

\sum_{x = 1}^{x = r} a_{x} = \sum_{y = 1}^{y = r_{s + 1}} b_{y};

for the numbers on either side of this equation are both equal to the number of sub-groups of order

p^{s + 1}

, when each of the latter is reckoned once for every sub-group of order

p^{s}

that it contains. It has however been shewn, in the two preceding paragraphs, that for all values of

x

and

y

a_{x} \equiv 1, b_{y} \equiv 1 (mod p) .

Hence

r_{s} \equiv r_{s + 1} (mod p) .

Now it has just been proved that

r_{1} \equiv 1 and r_{m - 1} \equiv 1 (mod p);

and therefore ﬁnally, for all values of

s

r_{s} \equiv 1 (mod p) .

We may state the results thus obtained as follows:—

THEOREM III. The number of sub-groups of any given order $p^{s}$ of a group of order $p^{m}$ is congruent to unity, $(mod p)$ ²⁰.

Corollary. The number of self-conjugate sub-groups of order $p^{s}$ of a group of order $p^{m}$ is congruent to unity, $(mod p)$ .

This is an immediate consequence of the theorem, since the number of sub-groups in any conjugate set is a power of $p$ .

62. Having shewn that the number of sub-groups of $G$ of order $p^{s}$ is of the form $1 + k p$ , we may now discuss under what circumstances it is possible for $k$ to be zero, so that $G$ then contains only one sub-group $G_{s}$ of order $p^{s}$ .

If this is the case, and if $P$ is any operation of $G$ not contained in $G_{s}$ , the order of $P$ must be not less than $p^{s + 1}$ ; for if it were less, $G$ would have some sub-group of order $p^{s}$ containing $P$ and this would necessarily be diﬀerent from $G_{s}$ . If the order of $P$ is $p^{s + t}$ , then ${P^{p^{t}}}$ is a cyclical sub-group of order $p^{s}$ ; and it must coincide with $G_{s}$ . Hence, if $G_{s}$ is the only sub-group of order $p^{s}$ , it must be cyclical.

Suppose now that $G$ contains operations of order $p^{r}$ ( $r > s$ ), but no operations of order $p^{r + 1}$ ; and let $P$ be an operation of $G$ of order $p^{r}$ . Then ${P}$ must be contained self-conjugately in a non-cyclical sub-group of order $p^{r + 1}$ .

We will take ﬁrst the case in which $p$ is an odd prime. Then (§ 55) $G$ must contain an operation $P'$ which does not belong to ${P}$ , such that

P'^{- 1} P P' = P^{α}, P'^{p} = P^{β} .

α

were unity,

{P, P'}

would be an Abelian group of order

p^{r + 1}

containing no operation of order

p^{r + 1}

. Its type would therefore be

(r, 1)

, and it would necessarily contain an operation of order

p

not occurring in

{P}

. It has been shewn that this is impossible if

G_{s}

is the only sub-group of order

p^{s}

, and therefore

α

cannot be unity.

We may then without loss of generality (§ 56) assume that

α = 1 + p^{r - 1} .

Moreover if

β

were not divisible by

p

, the order of

P'

would be

p^{r + 1}

, contrary to supposition. Hence we must have the relations

P'^{- 1} P P' = P^{1 + p^{r - 1}}, P'^{p} = P^{γ p} .

By successive applications of the ﬁrst of these equations, we get

P'^{- y} P^{x} P'^{y} = P^{x (1 + y p^{r - 1})},

for all values of

x

and

y

; and from this it immediately follows that

\begin{array}{l} {(P^{x} P')}^{p} & = {P'}^{p} P^{x {p + \frac{1}{2} p (p + 1) p^{r - 1}}} \\ = P^{(x + γ) p} . \end{array}

Hence if

x = p^{r - 1} - γ,

the order of

P^{x} P'

, an operation not contained in

{P}

, is

p

. This is impossible if

G_{s}

is the only sub-group of order

p^{s}

. If then

r < m

G

must contain operations of order greater than

p^{r}

; and

G

is therefore a cyclical group. Hence:—

THEOREM IV. If $G$ , of order $p^{m}$ , where $p$ is an odd prime, contains only one sub-group of order $p^{s}$ , $G$ must be cyclical.

63. When $p = 2$ , the result is not so simple.

Let $Q$ be an operation that transforms ${P}$ into itself, and suppose that the lowest power of $Q$ that occurs in ${P}$ is $Q^{4}$ . Then ${P, Q}$ must be deﬁned by

Q^{- 1} P Q = P^{α}, Q^{4} = P^{β},

where

α = \pm 1 + k 2^{r - 2} .

Moreover

β

must be a multiple of

4

, as otherwise the order of

Q

would be greater than

2^{r}

. A simple calculation now gives

{(P^{x} Q^{2})}^{2} = P^{β + x (1 + α^{2})};

and since

β \equiv 0 (mod 4),

and

α^{2} \equiv 1 (mod 2^{r - 1}),

x

can always be chosen so that

β + x (1 + α^{2}) \equiv 0 (mod 2^{r}) .

When

x

is thus chosen,

P^{x} Q^{2}

is an operation of order

2

which is not contained in

{P}

. But this is inconsistent with

G_{s}

being the only sub-group of order

2^{s}

; hence

{P}

must contain

Q^{2}

If now ${P}$ is not a self-conjugate sub-group, there must (§ 55) be some operation $P'$ of order $p^{r}$ , conjugate with $P$ and not contained in ${P}$ , which transforms ${P}$ into itself; and then ${P, P'}$ is deﬁned by

P'^{- 1} P P' = P^{α}, P'^{2} = P^{2 β},

where

α

- 1

\pm 1 + 2^{r - 1}

, and

β

is odd.

α = 1 + 2^{r - 1},

then

{(P^{x} P')}^{2} = P^{2 {β + x (1 + 2^{r - 2})}};

and

x

can be chosen so that

P^{x} P'

is of order

2

. Hence this case cannot occur.

If $α$ is either $- 1$ or $- 1 + 2^{r - 1}$ , the deﬁning relations of ${P, P'}$ are not self-consistent unless $r$ is $2$ ; for they lead to

P'^{- 1} P^{2 β} P' = P^{- 2 β},

and

P'^{- 1} P^{2 β} P' = P^{2 β} .

If $r$ is $2$ , and if $P''$ is an operation that transforms $P$ into $P'$ , so that

P {''}^{- 1} P P'' = P',

then

{(P P'')}^{2} = P P {''}^{2} P' = P^{- 1} P',

since

P^{2} = P'^{2} = P {''}^{2} .

Hence $P P''$ would be of order $8$ ; and therefore this case cannot occur.

Hence, ﬁnally, ${P}$ must be self-conjugate.

If $r + 1 < m$ , ${P}$ must be transformed into itself by some operation $Q'$ not contained in ${P, Q}$ , so that

Q^{- 1} P Q = P^{- 1 + k 2^{r - 1}},

and

Q'^{- 1} P Q' = P^{- 1 + k' 2^{r - 1}},

where

k

and

k'

are each either zero or unity. If both are zero or both unity,

Q Q'

is permutable with

P

; and if one is zero and the other unity,

{(Q Q')}^{- 1} P (Q Q') = P^{1 + 2^{r - 1}} .

In either case, the group contains an operation of order

2

that does not belong to

{P}

, and this is inadmissible. Hence, lastly,

r

must not be less than

m - 1

Suppose now that $r$ is equal to $m - 1$ , and that of the operations of $G$ , not contained in ${P}$ , $Q$ has as small an order as possible, say $2^{t}$ ( $t ≮ s + 1$ ). Then $Q^{2}$ of order $2^{t - 1}$ is contained in ${P}$ ; without loss of generality we may assume

Q^{2} = P^{2^{m - t}} .

Moreover

Q^{- 1} P Q = P^{- 1} or P^{- 1 + 2^{m - 2}},

and hence

Q^{- 1} P^{2^{m - t}} Q = P^{- 2^{m - t}} .

Now

Q

is permutable with

P^{2^{m - t}}

, one of its own powers, and therefore

P^{2^{m - t}}

is an operation of order

2

. Hence

t

2

and

s

is unity. If then

s

is greater than unity,

G

must contain operations of order

2^{m}

and must therefore be cyclical. Moreover if

s

is unity, the relations

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

lead to

{(P Q)}^{2} = 1,

and they are therefore inadmissible.

On the other hand, the relations

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

give

{(P^{x} Q)}^{2} = P^{2^{m - 2}};

and every operation of

G

, not contained in

{P}

, is of order

4

. Also these relations are clearly self-consistent, and they deﬁne a group of order

2^{m}

. Hence ﬁnally:—

THEOREM V. If a group $G$ , of order $2^{m}$ , has a single sub-group of order $2^{s}$ , $s > 1$

, it must be cyclical; if it has a single sub-group of order $2$ , it is either cyclical or of the type deﬁned by

P^{2^{m - 1}} = 1, Q^{2} = P^{2^{m - 2}}, Q^{- 1} P Q = P^{- 1} .

64. We shall now proceed to discuss, in application of the foregoing theorems and for the importance of the results themselves, the various types of groups of order $p^{m}$ which contain self-conjugate cyclical sub-groups of orders $p^{m - 1}$ and $p^{m - 2}$ respectively. It is clear from Theorem V that the case $p = 2$ requires independent investigation; we shall only deal in detail with the case in which $p$ is an odd prime, and shall state the results for the case when $p = 2$ .

The types of Abelian groups of order $p^{m}$ which contain operations of order $p^{m - 2}$ are those corresponding to the symbols $(m)$ , $(m - 1, 1)$ , $(m - 2, 2)$ , and $(m - 2, 1, 1)$ . We will assume that the groups which we consider in the following paragraphs are not Abelian.

65. We will ﬁrst consider a group $G$ , of order $p^{m}$ , which contains an operation $P$ of order $p^{m - 1}$ . The cyclical sub-group ${P}$ is self-conjugate and contains a single sub-group ${P^{p^{m - 2}}}$ of order $p$ . By Theorem IV, since $G$ is not cyclical, it must contain an operation $Q'$ , of order $p$ , which does not occur in ${P}$ . Since ${P}$ is self-conjugate and the group is not Abelian, $Q'$ must transform $P$ into one of its own powers. Hence

Q'^{- 1} P Q' = P^{α},

and since

Q'^{p}

is permutable with

P

it follows, from § 56, that

α = 1 + k p^{m - 2} .

Since the group is not Abelian, $k$ cannot be zero; but it may have any value from $1$ to $p - 1$ . If now

k x \equiv 1 (mod p),

then

Q'^{- x} P Q'^{x} = P^{1 + p^{m - 2}};

and therefore, writing

Q

for

Q'^{x}

, the group is deﬁned by

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

These relations are clearly self-consistent, and they deﬁne a group of order $p^{m}$ .

There is therefore a single type of non-Abelian group of order $p^{m}$ which contains operations of order $p^{m - 1}$ , because, for any such group, a pair of generating operations may be chosen which satisfy the above relations.

From the relation

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

it follows by repetition and multiplication that

Q^{- y} P^{x} Q^{y} = P^{x (1 + y p^{m - 2})},

and therefore that

{(P^{x} Q^{y})}^{z} = P^{x z {1 + \frac{1}{2} (z + 1) y p^{m - 2}} (1 - y z p^{m - 2})} Q^{y z},

and

{(P^{x} Q^{y})}^{p} = P^{x p} .

Hence $G$ contains $p$ cyclical sub-groups of order $p^{m - 1}$ of which $P$ and $P Q^{y}$ ( $y = 1, 2, \dots, p - 1$ ) may be taken as the generating operations. Since $Q$ and $P^{p}$ are permutable, $G$ also contains an Abelian non-cyclical sub-group ${Q, P^{p}}$ of order $p^{m - 1}$ It is easy to verify that the $1 + p$ sub-groups thus obtained exhaust the sub-groups of order $p^{m - 1}$ ; and that, for any other order $p^{s}$ , there are also exactly $p + 1$ sub-groups of which $p$ are cyclical and one is Abelian of type $(s - 1, 1)$ .

The reader will ﬁnd it an instructive exercise to verify the results of the corresponding case where $p$ is $2$ ; they may be stated thus. There are four distinct types of non-Abelian group of order $2^{m}$ , which contain operations of order $2^{m - 1}$ , when $m > 3$ . Of these, one is the type given in Theorem V, and the remaining three are deﬁned by

\begin{array}{l} P^{2^{m - 1}} & = 1, & Q^{2} & = 1, & Q P Q & = P^{1 + 2^{m - 2}}; \\ P^{2^{m - 1}} & = 1, & Q^{2} & = 1, & Q P Q & = P^{- 1 + 2^{m - 2}}; \\ P^{2^{m - 1}} & = 1, & Q^{2} & = 1, & Q P Q & = P^{- 1} . \end{array}

When $m = 3$ , there are only two distinct types. In this case, the second and the fourth of the above groups are identical, and the third is Abelian.

66. Suppose next that $G$ , a group of order $p^{m}$ , has a self-conjugate cyclical sub-group ${P}$ of order $p^{m - 2}$ , and that no operation of $G$ is of higher order than $p^{m - 2}$ . We may at once distinguish two cases for separate discussion; viz. (i) that in which $P$ is a self-conjugate operation, (ii) that in which $P$ is not self-conjugate.

Taking the ﬁrst case, there can be no operation $Q'$ in $G$ such that $Q'^{p^{2}}$ is the lowest power of $Q'$ contained in ${P}$ , for if there were, ${Q', P}$ would be Abelian and, its order being $p^{m}$ , it would necessarily coincide with $G$ . Hence any operation $Q'$ , not contained in ${P}$ , generates with $P$ an Abelian group of type $(m - 2, 1)$ , and we may choose $P$ and $Q$ as independent generators of this sub-group, the order of $Q$ being $p$ . If now $R'$ is any operation of $G$ not contained in ${Q, P}$ , $R'^{p}$ must occur in this sub-group, and therefore

R'^{p} = P^{α} Q^{β} .

If $β$ were not zero, $R^{p^{2}}$ would be the lowest power of $R$ that occurs in ${P}$ , and we have just seen that this cannot be the case. Hence

R'^{p} = P^{α},

and

{R', P}

is again an Abelian group of type

(m - 2, 1)

. If

P

and

R

are independent generators of this group, the latter cannot occur in

{Q, P}

. Now since

Q

is not self-conjugate,

R^{- 1} Q R = Q P^{β};

and since

R^{p}

is permutable with

Q

P^{p β} = 1,

so that

β \equiv 0 (mod p^{m - 3}) .

Hence

R^{- 1} Q R = Q P^{k p^{m - 3}},

where

k

is not a multiple of

p

. If ﬁnally,

P^{k}

be taken as a generating operation in the place of

P

, the group is deﬁned by

\begin{matrix} P^{p^{m - 2}} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} Q R = Q P^{p^{m - 3}}, \\ P Q = Q P, P R = R P . \end{matrix}

There is therefore a single type of group of order $p^{m}$ , which contains a self-conjugate operation of order $p^{m - 2}$ and no operation of order $p^{m - 1}$ .

67. Taking next the second case, we suppose that $G$ contains a cyclical self-conjugate sub-group ${P}$ of order $p^{m - 2}$ , but no self-conjugate operation of this order.

If $G$ contains an operation $Q'$ such that $Q'^{p^{2}}$ is the lowest power of $Q'$ occurring in ${P}$ , it follows (§ 56), since ${P}$ is self-conjugate, that

Q'^{- 1} P Q' = P^{1 + k p^{m - 4}} .

Moreover, since the order of $Q'$ cannot exceed $p^{m - 2}$ , we have

Q'^{p^{2}} = P^{α p^{2}} .

A simple calculation, similar to that of § 65, will now shew that

{(P^{x} Q')}^{p^{2}} = P^{(x + α) p^{2}},

while

{(P^{x} Q')}^{p} = Q'^{p} P^{x {p + \frac{1}{2} k (p + 1) p^{m - 3}}} .

Hence $P^{- α} Q'$ is an operation of order $p^{2}$ , none of whose powers, except identity, is contained in ${P}$ . If this operation be represented by $Q$ , $G$ is deﬁned by

P^{p^{m - 2}} = 1, Q^{p^{2}} = 1, Q^{- 1} P Q = P^{1 + k p^{m - 4}} .

If $k$ is a multiple of $p^{2}$ , the group is Abelian.

If $k$ is $k' p$ , and if $Q^{α}$ is taken for a new generating operation, where $α$ is chosen so that

α k' \equiv 1 (mod p),

we have a single type deﬁned by

P^{p^{m - 2}} = 1, Q^{p^{2}} = 1, Q^{- 1} P Q = P^{1 + p^{m - 3}} .

If $k$ is not a multiple of $p$ , we may take it equal to $k_{1} + k_{2} p$ , where $k_{1}$ and $k_{2}$ are both less than $p$ , and $k_{1}$ is not zero. (This case cannot occur if $m < 5$ .) Then

Q^{- x_{1} - x_{2} p} P Q^{x_{1} + x_{2} p} = P^{1 + k_{1} x_{1} p^{m - 4} + (k_{1} x_{2} + k_{2} x_{1}) p^{m - 3}},

when

m > 5

; while if

m = 5

, the coeﬃcient of

p^{m - 3}

, viz. of

p^{2}

, in the index of

P

must be increased by

\frac{1}{2} x_{1} (x_{1} - 1) k_{1}^{2}

If now we choose $x_{1}$ so that

k_{1} x_{1} = 1 + y_{1} p,

and then

x_{2}

so that

y_{1} + k_{1} x_{2} + k_{2} x_{1} \equiv 0 (mod p),

(with the suitable modiﬁcation when

m = 5

), then

Q^{x_{1} + x_{2} p}

transforms

P

into

P^{1 + p^{m - 4}}

. We therefore again in this case get a single type of group, which, if

Q^{x_{1} + x_{2} p}

is represented by

R

, is deﬁned by

P^{p^{m - 2}} = 1, R^{p^{2}} = 1, R^{- 1} P R = P^{1 + p^{m - 4}} .

As already stated, this type exists only when

m > 4

If every operation of $G$ is such that its $p$ th power occurs in ${P}$ , $P$ must be one of a set of $p$ conjugate operations; for if $P$ is one of $p^{2}$ conjugate operations, there must be an operation $Q$ which transforms $P$ into $P^{1 + k p^{m - 4}}$ , where $k$ is not a multiple of $p$ , and then $Q^{p^{2}}$ is the lowest power of $Q$ which is permutable with $P$ . Since $P$ is one of $p$ conjugate operations, it must be self-conjugate in an Abelian group ${Q, P}$ of type $(m - 2, 1)$ . Again, since $P$ and $P^{1 + p^{m - 3}}$ are conjugate operations, $P$ must be contained in a group of order $p^{m - 1}$ , deﬁned by

P^{p^{m - 2}} = 1, R^{p} = 1, R^{- 1} P R = P^{1 + p^{m - 3}} .

If $Q$ is not a self-conjugate operation, it must be one of $p$ conjugate operations and (§ 57) these must be

Q, Q P^{p^{m - 3}}, Q P^{2 p^{m - 3}}, \dots, Q P^{(p - 1) p^{m - 3}} .

Now if

R^{- 1} Q R = Q P^{α p^{m - 3}},

then

R^{- 1} P^{- α} Q R = P^{- α} Q .

Hence, if

Q

is not self-conjugate,

P^{- α} Q

is a self-conjugate operation of order

p^{m - 2}

: and the group is that determined in the last paragraph. Therefore

Q

must be self-conjugate, and the group is deﬁned by

\begin{matrix} P^{p^{m - 2}} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} P R = P^{1 + p^{m - 3}}, \\ P Q = Q P, R Q = Q R . \end{matrix}

It is clear from these relations that the group thus arrived at is the direct product of the groups ${P, R}$ and ${Q}$ ²¹.

It is to be expected, from the result of the corresponding case at the end of § 65, that the number of distinct types when $p = 2$ is much greater than when $p$ is an odd prime. There are, in fact, when $m > 5$ , fourteen distinct types of groups of order $2^{m}$ , which contain a self-conjugate cyclical sub-group of order $2^{m - 2}$ and no operation of order $2^{m - 1}$ . They may be classiﬁed as follows.

Suppose ﬁrst that the group has a self-conjugate operation of order $2^{m - 2}$ . There is then a single type deﬁned by the relations

(i) A^{2^{m - 2}} = 1, B^{2} = 1, C^{2} = 1, C B C = B A^{2^{m - 3}} .

Suppose next that the group $G$ has no self-conjugate operation of order $2^{m - 2}$ , and let ${A}$ be a self-conjugate cyclical sub-group of order $2^{m - 2}$ . If $\frac{G}{{A}}$ is cyclical, there are, when $m > 5$ , ﬁve distinct types. The common deﬁning relations of these are

A^{2^{m - 2}} = 1, B^{4} = 1, B^{- 1} A B = A^{α};

and the ﬁve distinct types are

\begin{matrix} (ii) α = - 1, (iii) α = 1 + 2^{m - 3}, (iv) α = - 1 + 2^{m - 3}, \\ (v) α = 1 + 2^{m - 4}, (vi) α = - 1 + 2^{m - 4} . \end{matrix}

If $m = 5$ , then (iv) and (v) are identical, and (vi) is Abelian; so that there are only three distinct types. If $m = 4$ , there is a single type; it is given by (ii).

When $\frac{G}{{A}}$ is not cyclical, the square of every operation of $G$ is contained in ${A}$ . If all the self-conjugate operations of $G$ are not contained in ${A}$ , there must be a self-conjugate operation $B$ , of order $2$ , which does not occur in ${A}$ . If $C$ is any operation of $G$ , not contained in ${A, B}$ , then ${A, C}$ is a self-conjugate sub-group of order $2^{m - 1}$ , which has no operation except identity in common with ${B}$ . Hence $G$ is a direct product of a group of order $2$ and a group of order $2^{m - 1}$ . There are therefore, for this case, four types (vii), (viii), (ix), (x), when $m > 4$ , corresponding to the four groups of order $2^{m - 1}$ of § 65. If $m = 4$ , there are two types.

Next, let all the self-conjugate operations of $G$ be contained in ${A}$ ; and suppose that $A$ is one of two conjugate operations. Then $G$ must contain an Abelian sub-group of type $(m - 2, 1)$ , in which $A$ occurs; and it may be shewn that, when $m > 4$ , there are two types deﬁned by the relations

(x i) a n d (x i i) \begin{gathered} A^{2^{m - 2}} = 1, B^{2} = 1, B A B = A, C^{2} = 1, \\ C B C = B A^{2^{m - 3}}, C A C = A^{- 1} or A^{- 1 + 2^{m - 3}} . \end{gathered}

When

m = 4

, there is, for this case, no type.

Lastly, suppose that $A$ is one of four conjugate operations. Then $G$ must contain sub-groups of order $2^{m - 1}$ , of the second and the third types of § 65, and a sub-group of order $2^{m - 1}$ of either the ﬁrst or fourth type (l.c.). In this last case, there are two distinct types deﬁned by

(x i i i) a n d (x i v) \begin{aligned} A^{2^{m - 2}} = 1, B^{2} = 1, B A B = A^{1 + 2^{m - 3}}, \\ C^{2} = 1, C A C = A^{- 1 + 2^{m - 3}}, C B C = B or B A^{2^{m - 3}} . \end{aligned}

These two types exist only when

m > 4

68. We shall now, as a ﬁnal illustration, determine and tabulate all types of groups of orders $p^{2}$ , $p^{3}$ and $p^{4}$ . It has been already seen that when $p = 2$ the discussion must, in part at least, be distinct from that for an odd prime; for the sake of brevity we shall not deal in detail with this case, but shall state the results only and leave their veriﬁcation as an exercise to the reader.

It has been shewn (§ 53) that all groups of order $p^{2}$ are Abelian; and hence the only distinct types are those represented by $(2)$ and $(1, 1)$ .

For Abelian groups of order $p^{3}$ , the distinct types are $(3)$ , $(2, 1)$ and $(1, 1, 1)$ .

If a non-Abelian group of order $p^{3}$ contains an operation of order $p^{2}$ , the sub-group it generates is self-conjugate; hence (§ 65) in this case there is a single type of group deﬁned by

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

If there is no operation of order $p^{2}$ , then since there must be a self-conjugate operation of order $p$ , the group comes under the head discussed in § 66; there is again a single type of group deﬁned by

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

These two types exhaust all the possibilities for non-Abelian groups of order $p^{3}$ .

69. For Abelian groups of order $p^{4}$ , the possible distinct types are $(4)$ , $(3, 1)$ , $(2, 2)$ , $(2, 1, 1)$ and $(1, 1, 1, 1)$ .

For non-Abelian groups of order $p^{4}$ which contain operations of order $p^{3}$ there is a single type, namely that given in § 65 when $m$ is put equal to $4$ .

For non-Abelian groups which contain a self-conjugate cyclical sub-group of order $p^{2}$ and no operation of order $p^{3}$ , there are three distinct types, obtained by writing $4$ for $m$ in the group of § 66 and in the ﬁrst and the last groups of § 67. The deﬁning relations of these need not be here repeated, as they will be given in the summarizing table (§ 73).

It remains now to determine all distinct types of groups of order $p^{4}$ , which contain no operation of order $p^{3}$ and no self-conjugate cyclical sub-group of order $p^{2}$ . We shall ﬁrst deal with groups which contain operations of order $p^{2}$ .

Let $S$ be an operation of order $p^{2}$ in a group $G$ of order $p^{4}$ . The cyclical sub-group ${S}$ must be self-conjugate in a non-cyclical sub-group ${S, T}$ of order $p^{3}$ , deﬁned by

S^{p^{2}} = 1, T^{p} = 1, T^{- 1} S T = S^{1 + k p} .

If $R$ is any operation of $G$ , not contained in ${S, T}$ , then since ${S}$ is not self-conjugate, we must have (§ 57)

R^{- 1} S R = S^{1 + α p} T^{β},

and therefore

R^{- 1} S^{p} R = S^{p} .

Now

T^{- 1} S^{p} T = S^{p},

and therefore the

p

th power of every operation of order

p^{2}

G

is a self-conjugate operation.

First let us suppose that $G$ contains other self-conjugate operations besides those of ${S^{p}}$ . Every such operation must occur in the group that contains ${S}$ self-conjugately; hence in this case $T$ must be self-conjugate.

We now therefore have

\begin{matrix} S^{p^{2}} = 1, T^{p} = 1, T^{- 1} S T = S, \\ R^{- 1} S R = S^{1 + α p} T^{β}, R^{- 1} T R = T, \\ R^{p} = S^{γ p} T^{δ} . \end{matrix}

These equations give

{(S^{x} R)}^{p} = R^{p} S^{x p} = S^{(x + γ) p} T^{δ} .

Hence, if $δ = 0$ , $S^{- γ} R$ is an operation of order $p$ . Denoting this by $R'$ and $S^{α p} T^{β}$ by $T'$ , the group is deﬁned by

\begin{matrix} S^{p^{2}} = 1, T'^{p} = 1, R'^{p} = 1, R'^{- 1} S R' = S T', \\ T'^{- 1} S T' = S, R'^{- 1} T' R' = T' . \end{matrix}

On the other hand, if $δ$ is not zero, $S^{\frac{δ α}{β} - γ} R$ is an operation of order $p^{2}$ such that $R$ transforms it into a power of itself. This is contrary to the supposition that the group contains no cyclical self-conjugate sub-group of order $p^{2}$ . Hence $δ$ cannot be diﬀerent from zero: we therefore have only one type of group.

70. Next, let ${S^{p}}$ contain all the self-conjugate operations of $G$ ; and as before, let ${S, T}$ be the group that contains ${S}$ self-conjugately. If $G$ contains an operation $S'$ of order $p^{2}$ which does not occur in ${S, T}$ , there must also be a non-cyclical sub-group ${S', T'}$ of order $p^{3}$ which contains ${S'}$ self-conjugately. Now ${S, T}$ and ${S', T'}$ must have a common sub-group of order $p^{2}$ ; since this is self-conjugate in $G$ , it cannot be cyclical. The only non-cyclical sub-groups of orders $p^{2}$ that ${S, T}$ and ${S', T'}$ contain are ${S^{p}, T}$ and ${S'^{p}, T'}$ . Hence these must be identical, and therefore $T$ must occur in ${S', T'}$ . If now ${S, T}$ and ${S', T'}$ were both Abelian, $T$ would be permutable both with $S$ and with $S'$ , and would therefore, contrary to supposition, be a self-conjugate operation. Hence either (i) $G$ must contain a non-Abelian group ${S, T}$ of order $p^{3}$ , in which $S$ is an operation of order $p^{2}$ ; or (ii) the Abelian group ${S, T}$ , in which $S$ is an operation of order $p^{2}$ , must contain all the operations of $G$ of order $p^{2}$ .

In the case (i), the group is deﬁned by

\begin{matrix} S^{p^{2}} = 1, T^{p} = 1, T^{- 1} S T = S^{1 + p}, \\ R^{- 1} S R = S^{1 + α p} T^{β}, R^{- 1} T R = S^{γ p} T, \\ R^{p} = S^{δ p} . \end{matrix}

These equations give

{(S^{x} R)}^{p} = R^{p} S^{x p {1 + \frac{1}{6} (p - 1) p (p + 1) β (γ - 2)}},

and therefore, except when

p = 3

S^{- δ} R

is an operation of order

p

. Hence, if

p > 3

, we may, without loss of generality, put zero for

δ

. In this case a simple calculation gives

R^{- x} S R^{x} = S^{1 + α_{x} p} T^{β_{x}}, R^{- x} T R^{x} = S^{γ_{x} p} T,

where

α_{x} = \frac{1}{2} β γ x (x - 1) + x α, β_{x} = x β, γ_{x} = x γ .

Hence if $R^{x} = R'$ , $S^{y} = S'$ , $S^{z p} T = T'$ , where $x$ , $y$ and $z$ are not multiples of $p$ , then

\begin{matrix} S'^{p^{2}} = 1, T'^{p} = 1, T'^{- 1} S' T' = S'^{1 + p}, \\ R'^{- 1} S' R' = S'^{1 + (α_{x} - z β_{x}) p} T'^{y β}, \\ R'^{- 1} T' R' = S'^{\frac{γ_{x} p}{y}} T', R'^{p} = 1 . \end{matrix}

If now we take

z \equiv \frac{α_{x}}{β_{x}}, y \equiv \frac{1}{β_{x}} (mod p),

then

R'^{- 1} S' R' = S' T', R'^{- 1} T' R' = S'^{x^{2} β γ p} T' .

Dropping the accents, the deﬁning relations now become

\begin{matrix} S^{p^{2}} = 1, T^{p} = 1, R^{p} = 1, T^{- 1} S T = S^{1 + p}, \\ R^{- 1} S R = S T, R^{- 1} T R = S^{α p} T; \end{matrix}

where $α$ is either zero, unity, or any given non-residue. Since the sub-group ${R, T, S^{p}}$ contains all the operations of order $p$ of the group, it follows at once that these three cases give three distinct types.

When $p = 3$ , it will be found that the deﬁning relations again give three distinct types. The operation $R$ may always be chosen so that $α$ and $γ$ are zero. If it is so chosen, the three types correspond to the values $1$ , $0$ ; $- 1$ , $1$ ; and $- 1$ , $- 1$ ; of $β$ and $δ$ .

In case (ii), the group is deﬁned by

\begin{matrix} S^{p^{2}} = 1, T^{p} = 1, T^{- 1} S T = S, \\ R^{- 1} S R = S^{1 + α p} T^{β}, R^{- 1} T R = S^{γ p} T, \\ R^{p} = 1; \end{matrix}

with the condition that all operations of $G$ , not contained in ${S, T}$ , are of order $p$ .

The formulæ for $R^{- x} S R^{x}$ and $R^{- x} T R^{x}$ enable us to calculate directly the power of any given operation of $G$ . Thus they give

{(S^{x} R)}^{p} = S^{p x {1 + \frac{1}{6} β γ (p + 1) p (p - 1) + \frac{1}{2} α p (p + 1)}} .

If $p > 3$ , this gives

{(S^{x} R)}^{p} = S^{p x},

so that, if

x

is not a multiple of

p

, the order of

S^{x} R

p^{2}

. Hence the type of group under consideration can only occur when

p = 3

In this case

{(S^{x} R)}^{3} = S^{3 x (1 + β γ)} .

Hence if $p = 3$ and $β γ \equiv - 1 (mod 3)$ , we obtain a new type. A reduction similar to that in the previous case may now be eﬀected; and taking unity for $x$ , the group is deﬁned by

\begin{matrix} S^{9} = 1, T^{3} = 1, R^{3} = 1, T^{- 1} S T = S, \\ R^{- 1} S R = S T, R^{- 1} T R = S^{- 3} T . \end{matrix}

71. It only remains to determine the distinct types which contain no operation of order $p^{2}$ .

Suppose ﬁrst that the self-conjugate operations of $G$ form a group of order $p^{2}$ . This must be generated by two independent operations $P$ and $Q$ of order $p$ .

If now $R$ is any other operation of the group, ${P, Q, R}$ must be an Abelian group of type $(1, 1, 1)$ . If again $S$ is any operation not contained in ${P, Q, R}$ , it cannot be permutable with $R$ ; for if it were, $R$ would be self-conjugate. There must therefore be a relation of the form (§ 57)

S^{- 1} R S = R P^{α} Q^{β} .

Since any operation of ${P, Q}$ may be taken for one of its generating operations, we may take $P^{α} Q^{β}$ or $P'$ for one. If then $Q'$ is an independent operation of ${P, Q}$ , $G$ will be deﬁned by

P'^{p} = 1, Q'^{p} = 1, R^{p} = 1, S^{p} = 1, S^{- 1} R S = R P',

in addition to the relations expressing that

P'

and

Q'

are self-conjugate. There is then in this case a single type. That all the operations of

G

in this case are actually of order

p

follows from the obvious fact that every sub-group of order

p^{3}

is either Abelian or of the second type of non-Abelian groups of order

p^{3}

72. Suppose, secondly, that the self-conjugate operations of $G$ form a sub-group of order $p$ , generated by $P$ . There must then be some operation $Q$ which belongs to a set of $p$ conjugate operations; for if every operation of $G$ which is not self-conjugate were one of a set of $p^{2}$ conjugate operations, the total number of operations in the group would be congruent to $p (mod p^{2})$ . It follows that $Q$ must be self-conjugate in a group of order $p^{3}$ ; and since $P$ is also self-conjugate in this group, it must be Abelian. Let $P$ , $Q$ , $R$ be generators of this group and $S$ any operation of $G$ not contained in it. We may now assume that $Q$ belongs to the sub-group $H_{2}$ (§ 53), and therefore that

S^{- 1} Q S = Q P^{α},

while

S^{- 1} R S = R Q^{β} P^{γ} .

If $β$ were zero, $Q^{\frac{1}{α}} R^{- \frac{1}{γ}}$ would be a self-conjugate operation not contained in ${P}$ ; and therefore $β$ must be diﬀerent from zero. We may now put

Q^{β} P^{γ} = Q', P^{α β} = P';

and the group is then deﬁned by the relations

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

together with the relations expressing that $P$ is self-conjugate. There is thus again in this case, at most, a single type. It remains to determine whether the operations are all actually of order $p$ .

The deﬁning relations give

S^{- 1} P^{α_{x}} Q^{β_{x}} R^{γ_{x}} S = P^{α_{x + 1}} Q^{β_{x + 1}} R^{γ_{x + 1}},

where

α_{x + 1} = α_{x} + β_{x}, β_{x + 1} = β_{x} + γ_{x}, γ_{x + 1} = γ_{x};

and therefore

S^{- x} P^{α} Q^{β} R^{γ} S^{x} = P^{α_{x}} Q^{β_{x}} R^{γ_{x}},

where

α_{x} = α + x β + \frac{1}{2} x (x - 1) γ, β_{x} = β + x γ, γ_{x} = γ .

Hence

\begin{array}{l} {(P^{α} Q^{β} R^{γ} S)}^{p} & = P^{\sum_{1}^{p} α_{x}} Q^{\sum_{1}^{p} β_{x}} R^{\sum_{1}^{p} γ_{x}} \\ = P^{p α + \frac{1}{2} p (p - 1) β + \frac{1}{6} (p + 1) p (p - 1) γ} Q^{p β + \frac{1}{2} p (p - 1) γ} R^{p γ} . \end{array}

If $p$ is a greater prime than $3$ , the indices of $P$ , $Q$ , $R$ are all multiples of $p$ ; hence $P^{α} Q^{β} R^{γ} S$ is of order $p$ , $S$ being any operation not contained in $G$ . If however $p = 3$ , then

{(P^{α} Q^{β} R^{γ} S)}^{3} = P^{γ},

so that, if

γ

is not a multiple of

3

P^{α} Q^{β} R^{γ} S

is an operation of order

9

. Hence this last type of group exists as a distinct type for all primes greater than

3

; but for

p = 3

, it is not distinct from one of the previous types containing operations of order

9

73. In tabulating, as follows, the types of group thus obtained, we give with each group $G$ a symbol of the form

(a, b, \dots) (a', b', \dots) (a'', b'', \dots) \dots,

indicating the types of

H_{1}

\frac{H_{2}}{H_{1}}

\frac{H_{3}}{H_{2}}

, …, where

H_{1}, H_{2}, H_{3}, \dots, H_{n}, G

is the series of self-conjugate sub-groups deﬁned in § 53. This symbol is to be read from the left so that

(a, b, \dots)

is the type of

H_{1}

Moreover in each group there is no operation of higher order than that denoted by $P$ .

Table of groups of order $p^{n}$ , $p$ an odd prime²².

I. $n = 2$ , two types.

(i) $(2)$ ; (ii) $(1, 1)$ .

II. $n = 3$ , ﬁve types.

(i) $(3)$ ; (ii) $(2, 1)$ ; (iii) $(1, 1, 1)$ ;

(iv) $P^{p^{2}} = 1$ , $Q^{p} = 1$ , $Q^{- 1} P Q = P^{1 + p}$ , $(1) (11)$ ;

(v) $\begin{gathered} P^{p} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} Q R = Q P, \\ R^{- 1} P R = P, Q^{- 1} P Q = P, (1) (11) . \end{gathered}$

III. $n = 4$ , ﬁfteen types.

(i) $(4)$ ; (ii) $(3, 1)$ ; (iii) $(2, 2)$ ;

(iv) $(2, 1, 1)$ ; (v) $(1, 1, 1, 1)$ ;

(vi) $P^{p^{3}} = 1$ , $Q^{p} = 1$ , $Q^{- 1} P Q = P^{1 + p^{2}}$ , $(2) (11)$ ;

(vii) $\begin{gathered} P^{p^{2}} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} Q R = Q P^{p}, \\ Q^{- 1} P Q = P, R^{- 1} P R = P, (2) (11); \end{gathered}$

(viii) $P^{p^{2}} = 1$ , $Q^{p^{2}} = 1$ , $Q^{- 1} P Q = P^{1 + p}$ , $(11) (11)$ ;

(ix) $\begin{gathered} P^{p^{2}} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} P R = P^{1 + p}, \\ P^{- 1} Q P = Q, R^{- 1} Q R = Q, (11) (11), \end{gathered}$
this group (ix) being the direct product of ${Q}$ and ${P, R}$ ;

(x) $\begin{gathered} P^{p^{2}} = 1, Q^{p} = 1, R^{p} = 1, R^{- 1} P R = P Q, \\ Q^{- 1} P Q = P, R^{- 1} Q R = Q, (11) (11); \end{gathered}$

(xi), (xii), and (xiii) $p > 3$ ,

\begin{matrix} P^{p^{2}} = 1, Q^{p} = 1, R^{p} = 1, Q^{- 1} P Q = P^{1 + p}, \\ R^{- 1} P R = P Q, R^{- 1} Q R = P^{α p} Q, (1) (1) (11), \end{matrix}

where for (xi) $α = 0$ , for (xii) $α = 1$ , for (xiii) $α =$ any non-residue, $(mod p)$ ;

(xi), (xii), and (xiii) $p = 3$ ,

\begin{matrix} P^{9} = 1, Q^{3} = 1, R^{3} = P^{3 α}, Q^{- 1} P Q = P^{4}, \\ R^{- 1} P R = P Q^{β}, R^{- 1} Q R = Q, (1) (1) (11), \end{matrix}

where for (xi) $α = 0$ , $β = 1$ , for (xii) $α = 1$ , $β = - 1$ , for (xiii) $α = - 1$ , $β = - 1$ .

(xiv) $\begin{gathered} P^{p} = 1, Q^{p} = 1, R^{p} = 1, S^{p} = 1, S^{- 1} R S = R P, \\ S^{- 1} Q S = Q, S^{- 1} P S = P, R^{- 1} Q R = Q, \\ R^{- 1} P R = P, Q^{- 1} P Q = P, (11) (11), \end{gathered}$
this group (xiv) being the direct product of ${Q}$ and ${P, R, S}$ ;

(xv) $p > 3$ ,

\begin{matrix} P^{p} = 1, Q^{p} = 1, R^{p} = 1, S^{p} = 1, S^{- 1} R S = R Q, S^{- 1} Q S = Q P, \\ S^{- 1} P S = P, R^{- 1} Q R = Q, R^{- 1} P R = P, \\ Q^{- 1} P Q = P, (1) (1) (11); \end{matrix}

(xv) $p = 3$ ,

\begin{matrix} P^{9} = 1, Q^{3} = 1, R^{3} = 1, Q^{- 1} P Q = P, R^{- 1} P R = P Q, \\ R^{- 1} Q R = P^{- 3} Q, (1) (1) (11) . \end{matrix}

74. To complete the list, we add, as was promised in § 68, the types of non-Abelian groups of orders $2^{3}$ and $2^{4}$ ; the possible types of Abelian groups being the same as for an odd prime.

Non-Abelian groups of order $2^{3}$ ; two types.

(i) identical with II (iv), writing $2$ for $p$ ;

(ii) $P^{4} = 1$ , $Q^{4} = 1$ , $Q^{- 1} P Q = P^{- 1}$ , $Q^{2} = P^{2}$ , $(1) (11)$ .

Non-Abelian groups of order $2^{4}$ ; nine types.

(i), (ii), (iii), (iv) and (v) identical with III (vi), (vii), (viii), (ix) and (x), writing $2$ for $p$ ;

(vi) $\begin{gathered} P^{4} = 1, Q^{4} = 1, R^{2} = 1, Q^{- 1} P Q = P^{- 1}, \\ Q^{2} = P^{2}, R^{- 1} Q R = Q, R^{- 1} P R = P, (11) (11), \end{gathered}$
this group (vi) being the direct product of ${R}$ and ${P, Q}$ ;

(vii) $P^{8} = 1$ , $Q^{2} = 1$ , $Q^{- 1} P Q = P^{- 1}$ , $(1) (1) (11)$ ;

(viii) $P^{8} = 1$ , $Q^{2} = 1$ , $Q^{- 1} P Q = P^{3}$ , $(1) (1) (11)$ ;

(ix) $P^{8} = 1$ , $Q^{4} = 1$ , $Q^{- 1} P Q = P^{- 1}$ , $Q^{2} = P^{4}$ , $(1) (1) (11)$ .

75. In the following examples, $G$ is a non-Abelian group whose order is the power of a prime $p$ ; and the sub-groups $H_{1}$ , $H_{2}$ , …, $H_{n}$ referred to are the series of self-conjugate sub-groups of § 53.

Ex. 1. If $P$ and $Q$ are any two operations of $G$ , the totality of operations of the form $P^{- 1} Q^{- 1} P Q$ generate a self-conjugate sub-group identical with or contained in $H_{n}$ .

Ex. 2. Shew that, if every sub-group of $G$ is Abelian, $\frac{G}{H_{1}}$ is an Abelian group of type $(1, 1)$ ; and any two operations of $G$ , neither of which belongs to $H_{1}$ and neither of which is a power of the other, generate $G$ .

Ex. 3. If $\frac{G}{H_{n}}$ , $\frac{H_{n}}{H n - 1}$ , $\frac{H_{n - 1}}{H_{n - 2}}$ , … are of types $(1, 1)$ , $(1)$ , $(1)$ , …, shew that $G$ can be generated by two operations.

Ex. 4. If $G$ , of order $p^{m}$ , is not Abelian, and if every sub-group of $G$ is self-conjugate, shew that $p$ must be $2$ . (Dedekind.)

Ex. 5. If $G$ is of order $p^{m}$ , and $H_{1}$ of type $(1, 1, \dots, with m - 2 units)$ , and if $m > 5$ , $G$ is the direct product of two groups. If $m = 5$ , there is one type for which $G$ is not a direct product, viz.

\begin{matrix} Q_{1}^{p^{2}} = 1, Q_{2}^{p^{2}} = 1, P^{p} = 1, Q_{2}^{- 1} Q_{1} Q_{2} = Q_{1} P, \\ Q_{1}^{- 1} P Q_{1} = P, Q_{2}^{- 1} P Q_{2} = P . \end{matrix}

Ex. 6. A group $G$ , of order $p^{m}$ , ( $m > 4$ ), where $p$ is an odd prime, contains an operation $P$ of order $p^{m - 2}$ , and no cyclical self-conjugate sub-group of order $p^{m - 2}$ . Shew that, if $P$ is one of a set of $p^{2}$ conjugate operations, the deﬁning relations of the group are of the form

\begin{matrix} P^{p^{m - 2}} = 1, Q^{p} = 1, Q^{- 1} P Q = P^{1 + p^{m - 3}}, R^{p} = P^{α p} Q^{β}, \\ R^{- 1} Q R = Q, R^{- 1} P R = P^{1 + β p^{m - 4}} Q; \end{matrix}

and that, if $P$ is one of a set of $p$ conjugate operations, the deﬁning relations are of the form

\begin{matrix} P^{p^{m - 2}} = 1, Q^{p} = 1, Q^{- 1} P Q = P, R^{p} = Q^{α}, \\ R^{- 1} Q R = P^{β p^{m - 3}} Q, R^{- 1} P R = P Q . \end{matrix}

Determine in each case the number of distinct types.

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Chapter V.On Groups Whose Orders Are the Powers of Primes.

Table of groups of order pn, p an odd prime22.

Chapter V.
On Groups Whose Orders Are the Powers of Primes.

Table of groups of order $p^{n}$ , $p$ an odd prime²².