Chapter VII.
On the Composition-Series of Group.

L et $G_1$ be a maximum self-conjugate sub-group (§ 27) of a given group $G$, $G_2$ a maximum self-conjugate sub-group of $G_1$, and so on. Since $G$ is a group of finite order, we must, after a finite number of sub-groups, arrive in this way at a sub-group $G_{n-1}$, whose only self-conjugate sub-group is that formed of the identical operation alone, so that $G_{n-1}$ is a simple group.

Definitions. The series of groups

The set of groups

Each of the set of factor-groups is necessarily (§ 30) a simple group.

The set of groups forming a composition-series of $G$ is not, in general, unique. Thus $G$ may have more than one maximum self-conjugate sub-group, in which case the second term in the series may be taken different from $G_1$. Moreover the groups succeeding $G_1$ are not all necessarily self-conjugate in $G$; and when some of them are not so, we obtain a new composition-series on transforming the whole set by a suitably chosen operation of $G$. That the new set thus obtained is again a composition-series is obvious; for if $G_{r+1}$ is a maximum self-conjugate sub-group of $G_r$, so also is $S^{-1}{G}_{r+1}S$ of $S^{-1}{G}_{r}S$. We proceed to prove that, if a group has two different composition-series, the number of terms in them is the same and the factor-groups derived from them are identical except as regards the sequence in which they occur.

This result, which is of great importance in the subsequent theory, is due to Herr Hölder³³; and the proof we here give does not differ materially from his.

The less general result, that, however the composition-series may be chosen, the composition-factors are always the same except as regards their sequence, had been proved by M. Jordan³⁴ some years before the date of Herr Hölder’s memoir.

90. THEOREM I. If $H$ is any self-conjugate sub-group of a group $G$; and if $K$, $K\prime$ are two self-conjugate sub-groups of $G$ contained in $H$, such that there is no self-conjugate sub-group of $G$ contained in $H$ and containing either $K$ or $K\prime$ except $H$, $K$ and $K\prime$ themselves; and if $L$ is the greatest common sub-group of $K$ and $K\prime$, so that $L$ is necessarily self conjugate in $G$; then the groups $\frac{H}{K}$ and $\frac{K\prime }{L}$ are simply isomorphic, as also are the groups $\frac{H}{K\prime }$ and $\frac{K}{L}$.

Since $K$ and $K\prime$ are self-conjugate sub-groups of $G$ contained in $H$,  must also be a self-conjugate sub-group of $G$ contained in $H$; and since, by supposition, there is in $H$ no self-conjugate sub-group of $G$ other than $H$ itself, which contains either $K$ or $K\prime$,  must coincide with $H$. Hence (§ 33) the product of the orders of $K$ and $K\prime$ is equal to the product of the orders of $H$ and $L$.

If the order of $\frac{K}{L}$ is $m$, the operations of $K$ may be divided into the $m$ sets

Consider now the $m$ sets of operations

No two operations of any one set can be identical. If operations from two different sets are the same, say

Now they all belong to the group $H$; and their number, being the order of $K\prime$ multiplied by the order of $\frac{K}{L}$, is equal to the order of $H$. Hence in respect of the self-conjugate sub-group $K\prime$, which $H$ contains, the operations of the group $H$ can be divided into the sets

Corollary. If $H$ coincides with $G$, $K$ and $K\prime$ are maximum self-conjugate sub-groups of $G$. Hence if $K$ and $K\prime$ are maximum self-conjugate sub-groups of $G$, and if $L$ is the greatest group common to $K$ and $K\prime$, then $\frac{G}{K}$ and $\frac{K\prime }{L}$ are simply isomorphic; as also are $\frac{G}{K\prime }$ and $\frac{K}{L}$.

Now $\frac{G}{K}$ and $\frac{G}{K\prime }$ are simple groups; and therefore, $\frac{K}{L}$ and $\frac{K\prime }{L}$ being simple groups, $L$ must be a maximum self-conjugate sub-group of both $K$ and $K\prime$.

91. We may now at once proceed to prove by a process of induction the properties of the composition-series of a group stated at the end of § 89. Let us suppose that, for groups whose orders do not exceed a given number $n$, it is already known that any two composition-series contain the same number of groups and that the factor-groups defined by them are the same except as regards their sequence. If $G$, a group whose order does not exceed $2n$, has more than one composition-series, let two such series be

If $H$ is the greatest common sub-group of $G_1$ and $G\prime$, and if

are two composition-series of $G$ which contain the same number of terms and give the same factor-groups. For it has there been shewn that $\frac{G}{G_1}$ and $\frac{G\prime }{H}$ are simply isomorphic; as also are $\frac{G}{G\prime }$ and $\frac{G_1}{H}$. Now the order of $G_1$, being a factor of the order of $G$, cannot exceed $n$. Hence the two composition-series

by supposition contain the same number of groups and give the same factor-groups; and the same is true of the two composition-series

Hence finally, the two original series are seen, by comparing them with the two new series that have been formed, to have the same number of groups and to lead to the same factor-groups. The property therefore, if true for groups whose order does not exceed $n$, is true also for groups whose order does not exceed $2n$. Now the simplest group, which has more than one composition-series, is that defined by

and for these the theorem is obviously true. It is therefore true generally. Hence:—

THEOREM II. Any two composition-series of a group consist of the same number of sub-groups, and lead to two sets of factor-groups which, except as regards the sequence in which they occur, are identical with each other.

The definite set of simple groups, which we thus arrive at from whatever composition-series we may start, are essential constituents of the group: the group is said to be compounded from them. The reader must not, however, conclude either that the group is defined by its set of factor-groups, or that it necessarily contains a sub-group simply isomorphic with any given one of them.

92. It has been already pointed out that the groups in a composition-series of $G$ are not necessarily, all of them, self-conjugate groups of $G$.

Suppose now that a series of groups

Definition. The series of groups, obtained in the manner just described, is called a chief composition-series, or a chief-series of $G$.

It should be noticed that such a series is not necessarily obtained by dropping out from a composition-series those of its groups which are not self-conjugate in the original group. Thus it follows immediately, from the results of § 54, that the composition-series of a group whose order is the power of a prime can be chosen, either (i) so that every group of the series is a self-conjugate sub-group, or (ii) so as to contain any given sub-group, self-conjugate or not.

A chief composition-series of a group is not necessarily unique; and when a group has more than one, the following theorem, exactly analogous to Theorem II, holds:—

THEOREM III. Any two chief-composition-series of a group consist of the same number of terms and lead to two sets of factor-groups, which, except as regards the sequence in which they occur, are identical with each other.

The formal proof of this theorem would be a mere repetition of the proof of § 91, Theorem I itself being used to start from instead of its Corollary; it is therefore omitted.

Although it is not always possible to pass from a composition-series to a chief-series, the process of forming a composition-series on the basis of a given chief-series can always be carried out. Thus if, in a chief-series, $H_{r+1}$ is not a maximum self-conjugate sub-group of $H_r$, the latter group must have a maximum self-conjugate sub-group $G_{r,1}$ which contains $H_{r+1}$. If $H_{r+1}$ is not a maximum self-conjugate sub-group of $G_{r,1}$, then such a group, $G_{r,2}$, may be found still containing $H_{r+1}$; and this process may be continued till we arrive at a group $G_{r,s-1}$, of which $H_{r+1}$ is a maximum self-conjugate sub-group. A similar process may be carried out for each pair of consecutive terms in the chief-series; the resulting series so obtained is a composition-series of the original group.

93. The factor-groups $\frac{H_r}{H_{r+1}}$ arising from a chief-series are not necessarily simple groups. If between $H_r$ and $H_{r+1}$ no groups of a corresponding composition-series occur, the group $\frac{H_r}{H_{r+1}}$ is simple; but when there are such intermediate groups, $\frac{H_r}{H_{r+1}}$ cannot be simple. We proceed to discuss the nature of this group in the latter case.

Let $G$ be multiply isomorphic with $G\prime$, so that the self-conjugate sub-group $H_{r+1}$ of $G$ corresponds to the identical operation of $G\prime$. Also let

Definition. If $\Gamma$ is a self-conjugate sub-group of $G$, and if $G$ has no self-conjugate sub-group, contained in $\Gamma$, whose order is less than that of $\Gamma$, then $\Gamma$ is called a minimum self-conjugate sub-group of $G$.

Making use of the phrase thus defined, the discussion of the factor-groups $\frac{H_r}{H_{r+1}}$ of a chief-series is the same as that of the minimum self-conjugate sub-groups of a given group.

94. To simplify the notation as much as possible, let $I$ be a minimum self-conjugate sub-group of $G$; and, if $I$ is not a simple group, suppose that in a composition-series of $G$ the term following $I$ is $g_1$. Since $g_1$ cannot be self-conjugate in $G$, it must be one of a set of conjugate sub-groups

Now, if $g_r=S^{-1}{g}_{1}S$, then as $g_1$ is a maximum self-conjugate sub-group of $I$, $S^{-1}{g}_{1}S$ or $g_r$ is a maximum self-conjugate sub-group of $S^{-1}IS$ or $I$; and hence every one of the above set of conjugate sub-groups is a maximum self-conjugate sub-group of $I$. If $g_{rs}$ represents the greatest sub-group common to $g_r$ and $g_s$, then, by Theorem I of the present chapter (§ 90),

are composition-series of $G$. Hence $\frac{g_1}{g_{1r}}$ is simply isomorphic with $\frac{I}{g_r}$. Now $\frac{I}{g_r}$ and $\frac{SI{S}^{-1}}{S{g}_r{S}^{-1}}$, or $\frac{I}{g_1}$ are simply isomorphic; therefore $\frac{g_1}{g_{1r}}$ and $\frac{I}{g_1}$ are simply isomorphic. Similarly we may shew that, whatever $r$ and $s$ may be, $\frac{g_r}{g_{rs}}$ and $\frac{I}{g_1}$ are simply isomorphic.

If $g_{1r}$ and $g_{1s}$ are the same group, whatever $r$ and $s$ may be, this group is common to the whole set of conjugate groups ($\alpha$). If these groups have any common sub-group, except identity, it is (Theorem V, § 27) a self-conjugate sub-group of $G$, and this self-conjugate sub-group would be contained in $I$, contrary to supposition. Hence if $g_{1r}$ and $g_{1s}$ are the same group, for all values of $r$ and $s$, this group consists of the identical operation alone, and a composition-series is

If $g_{1r}$ and $g_{1s}$ are distinct and if $g_{1rs}$ is their greatest common sub-group, then

are two composition-series, and $\frac{g_{1r}}{g_{1rs}}$ is simply isomorphic with with $\frac{I}{g_1}$ for all values of $r$ and $s$.

This reasoning may be repeated. If $g_{1rs}$ and $g_{1rt}$ are the same group whatever $s$ and $t$ may be, they must each be the identical operation. If not, $g_{1rst}$ is another term of the composition-series, and $\frac{g_{1rs}}{g_{1rst}}$ is simply isomorphic with $\frac{I}{g_1}$.

Hence however the composition-series from $I$ onwards be constructed, the corresponding factor-groups are all simply isomorphic with each other. Moreover every group after $I$ in the composition-series is self-conjugate in $I$. For $g_1$ and $g_r$ being self-conjugate sub-groups of $I$, so also is their greatest common sub-group $g_{1r}$; and $g_{1r}$, $g_{1s}$ being self-conjugate sub-groups of $I$, $g_{1rs}$ is also self-conjugate; and so on. Let the composition-series thus arrived at be now written

The final group ${\gamma }_{s-1}$ must be one of a conjugate set of, say, $\nu$ groups in $G$, no one of which has any operation except identity in common with any other. Since each of these $\nu$ groups is self-conjugate in $I$, and since no two of them have a common operation except identity, it follows by Theorem IX, § 34, that every operation of any one of them is permutable with every operation of the remaining $\nu -1$. The group generated by the $\nu$ groups conjugate to ${\gamma }_{s-1}$, being self-conjugate in $G$ and contained in $I$, must coincide with $I$. Now, if ${\gamma }_{s-1}^1$ and ${\gamma }_{s-1}^2$ are any two of this set of $\nu$ groups,  is their direct product, and it is a self-conjugate sub-group of $I$. If $s>2$, does not coincide with $I$; and therefore there must be another sub-group ${\gamma }_{s-1}^3$, of the set to which ${\gamma }_{s-1}$ belongs, which is not contained in . Since both the latter group and ${\gamma }_{s-1}^3$ are self-conjugate in $I$, while ${\gamma }_{s-1}^3$ is a simple group, no operation of ${\gamma }_{s-1}^3$ except identity can be contained in . Hence ${\gamma }_{s-1}^1$, ${\gamma }_{s-1}^2$ and ${\gamma }_{s-1}^3$ are independent, i.e. no operation of one of these groups can be expressed in terms of operations of the other two. The group is the direct product of ${\gamma }_{s-1}^1$, ${\gamma }_{s-1}^2$ and ${\gamma }_{s-1}^3$, and it is self-conjugate in $I$. If $s>3$, the same reasoning may be repeated. Finally, from the set of $\nu$ groups conjugate to ${\gamma }_{s-1}$, it must be possible to choose $s$ independent groups

THEOREM IV. If between two consecutive terms $H_r$ and $H_{r+1}$ in the chief-composition-series of a group there occur the groups $G_{r,1}$, $G_{r,2}$, …, $G_{r,s-1}$ of a composition-series; then (i) the factor groups

Corollary. If the order of $\frac{H_r}{H_{r+1}}$ is a power, $p^s$, of a prime, $\frac{H_r}{H_{r+1}}$ must be an Abelian group of type .

95. A chief-series of a group $G$ can always be constructed which shall contain among its terms any given self-conjugate sub-group of $G$. For if $\Gamma$ is a self-conjugate sub-group of $G$, and if $\frac{G}{\Gamma }$ is simple, we may take $\Gamma$ for the group which follows $G$ in the chief-series. If on the other hand $\frac{G}{\Gamma }$ is not simple, it must contain a minimum self-conjugate sub-group. Then ${\Gamma }_1$, the corresponding self-conjugate sub-group of $G$, contains $\Gamma$; and if there were a self-conjugate sub-group of $G$ contained in ${\Gamma }_1$, and containing $\Gamma$, the self-conjugate sub-group of $\frac{G}{\Gamma }$, which corresponds to ${\Gamma }_1$, would not be a minimum self-conjugate sub-group. We may now repeat the same process with ${\Gamma }_1$, and so on; the sub-groups thus introduced will, with $G$ and $\Gamma$, clearly form the part of a chief-series extending from $G$ to $\Gamma$. The series may be continued from $\Gamma$, till we arrive at the identical operation, in the usual way.

96. It will perhaps assist the reader if we illustrate the foregoing theory by one or two simple examples. We take first a group of order $12$, defined by the relations

From the last two equations, it follows that

Hence the only chief-series is

The orders of the factor-groups in the chief-series are $3$ and $2^2$, and the group of order $2^2$ is, as it should be, an Abelian group whose operations are all of order $2$. The composition-factors are $3$, $2$, $2$ in the order written.

97. As a rather less simple instance, we will now take a group generated by four permutable independent operations $A$, $B$, $P$, $Q$, of orders $2$, $2$, $3$, $3$ respectively and an operation $R$ of order $3$, for which

The sub-group , of order $36$, is clearly a maximum self-conjugate sub-group, and therefore the order of the group is $108$. Since $A$, $B$ and $AB$ are conjugate operations, every self-conjugate sub-group that contains $A$ must contain $B$; and since $Q$ and $QP$ are conjugate, every self-conjugate sub-group that contains $Q$ must contain $P$. Hence the only other possible maximum self-conjugate sub-groups are those of the form ; and since

Again, since $A$ and $B$ are conjugate in , the only maximum self-conjugate sub-groups of this group are those of the form . If $y$ is zero, this sub-group is Abelian; and we may take for the next term in the composition series any maximum sub-group $g_2$ of this Abelian sub-group, and for the last term but one any sub-group $g_3$ of $g_2$. But if $y$ is not zero, can only be followed by . Hence the remaining composition-series are of the forms:—

It should be noticed that if, in the last of these series, we drop out the terms which are not self-conjugate in the original group, here the third and fifth terms, we do not arrive at a chief-series. This illustrates a remark made in § 92.

98. THEOREM V. If $H$ is a sub-group of $G$, each composition-factor of $H$ must be equal to or be a factor of some composition-factor of $G$.

If $G$ is simple, its only composition-factor is equal to its order: the theorem in this case is obvious.

If $G$ is not simple, let $G_{r+1}$ be the first term in a composition-series of $G$ which does not contain $H$; and let $G_r$ be the term preceding $G_{r+1}$. If $H_1$ is the greatest common sub-group of $G_{r+1}$ and $H$, then $H_1$ is a self-conjugate sub-group of $H$. For every operation of $H$ transforms both $H$ and $G_{r+1}$ into themselves; and therefore every operation of $H$ transforms $H_1$, the greatest common sub-group of $H$ and $G_{r+1}$, into itself. Now the order of is equal to the product of the orders of $H$ and $G_{r+1}$ divided by the order of $H_1$; and is contained in $G_r$. Hence the order of $\frac{H}{H_1}$ is equal to or is a factor of the order of $\frac{G_r}{G_{r+1}}$. If then a composition-series of $H$ be taken, containing the term $H_1$, the orders of the factor-groups, formed by those terms of the series terminating with $H_1$, are equal to or are factors of the order of $\frac{G_r}{G_{r+1}}$. The same reasoning may now be used for $H_1$ that has been applied to $H$; and the theorem is therefore true.

Corollary. If all the composition-factors of $G$ are primes, so also are the composition-factors of every sub-group of $G$.

99. Definition. A group, all whose composition-factors are primes, is called a soluble group.

The Corollary of the last theorem may be stated in the form:—if a group is soluble, so also are all its sub-groups.

A soluble group of order $p^{\alpha }{q}^{\beta }\dots r^{\gamma }$, where $p$, $q$, …, $r$ are distinct primes, has $\alpha +\beta +\dots +\gamma$ composition-factors; these are capable of distinct arrangements.

For a specified group the composition-factors may, as we have already seen, occur in two or more distinct arrangements; but it is immediately obvious that two groups of the same order cannot be of the same type, i.e. simply isomorphic, unless the distinct arrangements, of which the composition-factors are capable, are the same for both. A first step therefore towards the enumeration of all distinct types of soluble groups of a given order, will be to classify them according to the distinct arrangements of which the composition-factors are capable; for no two groups belonging to different classes can be of the same type.

The case, in which the composition-factors are capable of all possible arrangements, is one which will always occur. Taking in this case $\beta$ $q$’s followed by $\alpha$ $p$’s for the last $\alpha +\beta$ composition-factors, the group contains a sub-group $G\prime$ of order $p^{\alpha }{q}^{\beta }$. In the composition-series of this group, with the composition-factors taken as proposed, there is a sub-group $H$ of order $p^{\alpha }$ contained self-conjugately in a sub-group $H_1$ of order $p^{\alpha }q$. This sub-group $H_1$ is contained self-conjugately in a group $H_2$ of order $p^{\alpha }{q}^2$. Hence (Theorem VII, Cor. IV, § 87) $H$ is contained self-conjugately in $H_2$. Again, $H_2$ is contained self-conjugately in a group $H_3$ of order $p^{\alpha }{q}^3$, and therefore again $H$ is self-conjugate in $H_3$. Proceeding thus, we shew that $H$ is self-conjugate in $G\prime$. It follows that $n$, the number of conjugate sub-groups of order $p^{\alpha }$ contained in the group, is not a multiple of $q$. Now $q$ may be any one of the distinct primes other than $p$ that divide the order of the group. Hence finally the group contains a self-conjugate sub-group of order $p^{\alpha }$. In the same way we shew that it contains self-conjugate sub-groups of orders $q^{\beta }$, …, $r^{\gamma }$. The group must therefore be the direct product of groups whose orders are $p^{\alpha }$, $q^{\beta }$, …, $r^{\gamma }$.

Hence:—

THEOREM VI. A soluble group, the composition-factors of which may be taken in any order, is the direct product of groups whose orders are powers of primes.

100. THEOREM VII. If $G$ is a soluble group of order $p^{\alpha }m$, where $p$ is prime and does not divide $m$, and if every operation of $G$ whose order is a power of $p$ is permutable with every operation whose order is relatively prime to $p$; then $G$ is the direct product of two groups of orders $p^{\alpha }$ and $m$.

Let $H$ be a sub-group of $G$ of order $p^{\alpha }$. This sub-group, from the conditions of the theorem, is necessarily self-conjugate. Let a chief-series of $G$ be formed which contains $H$, say

The order of $H\prime$ must be of the form $p^{\alpha }{q}^{\beta }$; and $H\prime$ must contain a single sub-group $h$ of order $q^{\beta }$. This sub-group $h$ is self-conjugate in $G$, since $H\prime$ is self-conjugate in $G$; and $h$ is the only sub-group of order $q^{\beta }$ contained in $H\prime$.

Consider now the group $\frac{G}{h}$. Every operation of this group whose order is a power of $p$ is permutable with every operation whose order is relatively prime to $p$. Hence we may repeat the same reasoning to shew that $\frac{G}{h}$ contains a self-conjugate sub-group of order $r^{\gamma }$ where $r$ is a prime distinct from $p$, but possibly the same as $q$. It follows that $G$ has a self-conjugate sub-group $h_1$ of order $q^{\beta }{r}^{\gamma }$. We may now repeat the same reasoning with $\frac{G}{h_1}$; and in this way we must at last reach a self-conjugate sub-group of $G$ of order $m$. Hence, since $G$ contains self-conjugate sub-groups of orders $p^{\alpha }$ and $m$, which are relatively prime, $G$ must be the direct product of these sub-groups.

It should be noticed that the above reasoning does not necessarily hold if $G$ is not soluble; for then the order of $H\prime$ may be of the form $p^{\alpha }{\mu }^{\beta }$, where $\mu$, the order of a simple group, contains more than one prime factor. In that case it would not be necessarily true that $H\prime$ contains a group of order ${\mu }^{\beta }$.

101. Though the actual determination of all types of soluble groups of a given order more properly forms part of the subject of Chapter XI, we will here, as a further illustration of the subject of the present Chapter, deal with the problem for groups whose orders are of the form $p^{2}q$, $p$ and $q$ being distinct primes.

Every such group must be soluble. In fact, if $p>q$, the group must contain a self-conjugate sub-group of order $p^2$; and if $p<q$, there must be a self-conjugate sub-group of order $q$ unless $p=2$, $q=3$; while in this case if there are $4$ sub-groups of order $3$, there must be a self-conjugate sub-group of order $4$. These statements all follow immediately from Sylow’s theorem.

The Abelian groups of order $p^{2}q$ may be specified immediately; and therefore in what follows we will assume that the group is not Abelian. There are $3$ possible arrangements for the composition-factors, viz. 

If the factors are capable of all three arrangements, the group is the direct product of groups of orders $p^2$ and $q$; it is therefore an Abelian group.

If the two arrangements $p$, $p$, $q$ and $q$, $p$, $p$ are possible, there are self-conjugate sub-groups of orders $p^2$ and $q$; the group again is Abelian, and all three arrangements are possible.

There are now five other possibilities.

I. $p$, $p$, $q$ and $p$, $q$, $p$, the only possible arrangements.

There must be here a sub-group of order $pq$, containing self-conjugate sub-groups of orders $p$ and $q$ and therefore Abelian. Let this be generated by operations $P$ and $Q$, of orders $p$ and $q$. Since the group has sub-groups of order $p^2$, there must be operations of orders $p$ or $p^2$, not contained in the sub-group of order $pq$, and permutable with $P$. Let $R$ be such an operation, so that $R^p$ belongs to the sub-group of order $pq$. $R$ cannot be permutable with $Q$, as the group would be then Abelian; hence

The two types are respectively defined by the relations

and

In each case, $\alpha$ is a primitive root of the congruence .

II. $p$, $q$, $p$ and $q$, $p$, $p$, the only possible arrangements.

There must be a self-conjugate sub-group of order $pq$, in which the sub-group of order $q$ is not self-conjugate, and a self-conjugate sub-group of order $p^2$. The sub-group of order $pq$ must be given by

so that in this case $q$ must be a factor of $p-1$. If the sub-group of order $p^2$ is not cyclical, there must be an operation $R$ of order $p$, not contained in the sub-group of order $pq$. Any such operation must be permutable with $P$. Moreover since the sub-group of order $pq$ is self-conjugate and contains only $p$ sub-groups of order $q$, the sub-group  must be permutable with some operation of order $p$. Hence we may assume that $R$ is permutable with , and, since $p>q$, with $Q$.

We thus obtain a single type defined by

If the sub-group of order $p^2$ is cyclical, all the operations, which have powers of $p$ for their orders and are not contained in the sub-group of order $pq$, must be of order $p^2$. There can therefore be no operation of order $p$, which is permutable with ; therefore there is no corresponding type.

III. $p$, $p$, $q$, the only possible arrangement.

There must be a self-conjugate sub-group of order $pq$, which has no self-conjugate sub-group of order $p$; it is therefore defined by

so that here $p$ must be a factor of $q-1$.

If the sub-groups of order $p^2$ are not cyclical, there must be an operation $R\prime$ of order $p$, not contained in this sub-group and permutable with $P$. Hence

If the sub-groups of order $p^2$ are cyclical, there must be an operation $R$ of order $p^2$, such that

IV. $p$, $q$, $p$, the only possible arrangement.

Here the self-conjugate sub-group of order $pq$ must be given by

and $q$ must be a factor of $p-1$. As in II, there must be an operation $R$ of order $p$, permutable with and therefore with $Q$; and since $R$ transforms into itself, it must be permutable with $P$. This however makes the sub-group self-conjugate, which requires $q$, $p$, $p$ to be a possible arrangement of the composition-factors. Hence there is no type corresponding to this case.

V. $q$, $p$, $p$, the only possible arrangement.

If the sub-group of order $p^2$ is cyclical, and is generated by $P$, while $Q$ is an operation of order $q$, we must have

If the sub-group of order $p^2$ is not cyclical, it can be generated by two permutable operations $P_1$ and $P_2$ of order $p$. Since a sub-group of order $q$ is not self-conjugate, either $p$ or $p^2$ must be congruent to unity ; and therefore $q$ must be a factor of either $p-1$ or $p+1$.

Suppose first, that $q$ is a factor of $p-1$.

There are $p+1$ sub-groups of order $p$. When these are transformed by any operation $Q$ of order $q$, those which are not permutable with $Q$ must be interchanged in sets of $q$. Hence at least two of these sub-groups must be permutable with $Q$, and we may take the generating operations of two such sub-groups for $P_1$ and $P_2$. Therefore

Now if either $\alpha$ or $\beta$, say $\beta$, were unity, then would be a self-conjugate sub-group and $p$, $q$, $p$ would be a possible arrangement of the composition-factors. Hence neither $\alpha$ nor $\beta$ can be unity, and we may take

It remains to determine how many distinct types these equations contain. When $q=2$, the only possible value of $x$ is unity; and there is a single type. When $q$ is an odd prime, and we take

Suppose next, that $q$ is a factor of $p+1$.

Any operation $Q$, of order $q$, will transform the sub-groups of order $p$, with which it is not permutable, so as to interchange them in sets of $q$; and hence if it is permutable with any sub-group, it must be permutable with $q$ at least. This, by the last case, is clearly impossible, and hence $Q$ is permutable with no sub-group of order $p$. We may therefore, by suitably choosing the generating operations of the group of order $p^2$, assume that

If now

Now since $Q^q$ is the lowest power of $Q$ that is permutable with $P_1$, ${\beta }_{q-1}$ must be the first term of the series ${\beta }_1$, ${\beta }_2$, … which vanishes. Hence $q$ is the least value of $z$ for which

From the quadratic congruence satisfied by $\iota$, it follows that

Finally, we may shew that, when $q$ is a factor of $p+1$, the equations

where $\iota$ is a primitive root of the congruence

Thus from the given equations it follows that

If then we take $P_1$, $P_3$ and $Q^x$ as generating operations in the place of $P_1$, $P_2$ and $Q$, the defining relations are reproduced with ${\iota }^x$ in the place of $\iota$. The relations therefore define a single type of group³⁷.

We have, for the sake of brevity, in each case omitted the verification that the defining relations actually give a group of order $p^{2}q$. This presents no difficulty, even for the last type; for the previous types it is immediately obvious.