Chapter XV.
On Soluble and Composite Groups.

T he most general problem of pure group-theory, so far as it is concerned with groups of finite order, is the determination and analysis of all distinct types of group whose order is a given integer. The solution of this problem clearly involves the previous determination of all types of simple groups whose orders are factors of the given integer.

Now there is no known criterion by means of which we can say whether, corresponding to an arbitrarily given composite integer $N$ as order, there exists a simple group or not. For certain particular forms of $N$, this question can be answered in the affirmative. For instance, when $N=\frac{1}{2}n!$, previous investigation enables us to state that there is a simple group of order $N$; but even in these cases we cannot, in general, say how many distinct types of simple group of order $N$ there are.

On the other hand, we have seen that, for certain forms of $N$, it is possible to state that there is no simple group of order $N$. Thus, when $N$ is the power of a prime, or is the product of two distinct primes, there is no simple group whose order is $N$; and further, every group of order $N$ is soluble. Again, if $N$ is divisible by a prime $p$, and contains no factor of the form $1+kp$, a group of order $N$ cannot be simple; for, by Sylow’s theorem, it must contain a self-conjugate sub-group having a power of $p$ for its order.

We propose, in the present Chapter, to prove a series of theorems which enable us to state, in a considerable variety of cases, that a group, which has a number of given form for its order, is either soluble or composite. If the results appear fragmentary, it must be remembered that this branch of the subject has only recently received attention: it should be regarded rather as a promising field of investigation than as one which is thoroughly explored.

The symbols $p_1$, $p_2$, $p_3$, … will be used throughout the Chapter to denote distinct primes in ascending order of magnitude; while distinct primes, without regard to their magnitude, will be denoted by $p$, $q$, $r$, ….

242. THEOREM I. If $H$ is one of $n$ conjugate sub-groups of a group $G$ of order $N$, and if $N$ is not a factor of $n!$, $G$ cannot be simple.

The group $G$ is isomorphic with the group of substitutions given on transforming the set of $n$ conjugate sub-groups by each of the operations of $G$. This is a transitive group of degree $n$, and its order therefore is equal to or is a factor of $n!$. Hence, if $N$ is not a factor of $n!$, the isomorphism cannot be simple; $G$ is therefore not simple.

Corollary⁹⁴. If a prime $p$ divides $N$, and if $m$ is the greatest factor of $N$ which is congruent to unity , $G$ will be composite unless $N$ is a factor of $m!$.

There cannot, in fact, be more than $m$ conjugate sub-groups whose order is the highest power of $p$ that divides $N$; so that this result follows from the theorem.

In dealing with a given integer as order, it may happen that, though no single application of this theorem will prove the corresponding group to be composite, a repeated application of Sylow’s theorem, on which the theorem depends, will lead to that result. As an example, let us consider groups whose order is $3^2\cdot 5\cdot 11$. Any such group must contain either $1$ or $45$ sub-groups of order $11$, and either $1$ or $11$ sub-groups of order $5$. Hence, if the group is simple, it must contain $45$ sub-groups of order $11$, and $11$ sub-groups of order $5$; and each one of the latter must be contained self-conjugately in a sub-group of order $45$. Now the $45$ sub-groups of order $11$ would contain $450$ distinct operations of order $11$, leaving only $45$ others; and therefore a sub-group of order $45$, if the group contain such a sub-group, must in this case be self-conjugate. This is, however, in direct contradiction to the assumption that the group contains $11$ sub-groups of order $5$; therefore the group cannot be simple.

243. THEOREM II. Every group whose order is the power of a prime is soluble.

This has already been proved in § 54.

THEOREM III⁹⁵. A group $G$ whose order is $p^{\alpha }{q}^{\beta }$, where $\alpha$ is less than $2m$, $m$ being the index to which $p$ belongs , is soluble.

If $\alpha <m$, $G$ contains a self-conjugate sub-group $H$ of order $q^{\beta }$.

If $\alpha =m$, $G$ contains either $1$ or $p^m$ sub-groups of order $q^{\beta }$. In the latter case, if $q^r$ is the order of the greatest sub-group common to two groups of order $q^{\beta }$, and if $h$ is such a sub-group, $h$ must (§ 80) be contained self-conjugately in a sub-group $k$ of order $p^x{q}^{r+s}$, $x>0$, $s>0$; and this sub-group must contain more than one sub-group of order $q^{r+s}$. Hence $x$ must be $m$; and therefore $h$ must be common to all the $p^m$ sub-groups of order $q^{\beta }$, so that $h$ is self-conjugate, and $s$ is $\beta -r$. If $r$ is zero, no two sub-groups of order $q^{\beta }$ have a common operation except identity, and the $p^m$ sub-groups contain distinct operations, so that a sub-group of order $p^m$ is self-conjugate. Hence when $\alpha$ is equal to $m$, $G$ contains either a self-conjugate sub-group of order $q^r$ ($r>0$) or else one of order $p^m$.

If $m<\alpha <2m$, $G$ contains $1$ or $p^m$ sub-groups of order $q^{\beta }$. If there is only one, it is self-conjugate.

If there are $p^m$ sub-groups of order $q^{\beta }$, and if $q^r$ ($r>0$) is the order of the greatest group common to any two of them, it may be shewn, exactly as in the preceding case, that $G$ contains a self-conjugate sub-group of order $q^r$.

If $r=0$, the $q^{\beta }-1$ operations, other than identity, of each of the $p^m$ sub-groups of order $q^{\beta }$ are all distinct. Now $G$ is isomorphic with a group of degree $p^m$. If the isomorphism is multiple, $G$ must have a self-conjugate sub-group whose order is a power of $p$. If the isomorphism is simple, we may regard $G$ as a transitive group of degree $p^m$; and every operation, except identity, of a sub-group of order $q^{\beta }$ will leave one symbol only unchanged. Hence $q^{\beta }$ is equal to or is a factor of $p^m-1$. Moreover, when $G$ is thus regarded, the sub-groups of order $p^{\alpha }$ must be transitive; for all the sub-groups of order $q^{\beta }$ will be given on transforming any one of them by the operations of a sub-group of order $p^{\alpha }$.

Consider now a sub-group of order $p^{\alpha -m}{q}^{\beta }$, which keeps one symbol fixed and contains a sub-group of order $q^{\beta }$ self-conjugately. It will contain $q^{\gamma }$ ($\gamma \le \beta$) sub-groups of order $p^{\alpha -m}$. Any one of these must keep $p^x$ symbols fixed; for it is the sub-group of a transitive group of order $p^{\alpha }$ and degree $p^m$, which keeps one symbol fixed. If $\gamma <\beta$, there are operations of order $q$ which transform one of these sub-groups into itself, and therefore interchange among themselves the $p^x$ symbols unchanged by the sub-group. Hence $p^x-1$ must be divisible by $q$. This requires that $x=m$, which is impossible. Hence $\gamma =\beta$, and the sub-group of order $p^{\alpha -m}{q}^{\beta }$ contains $q^{\beta }$ sub-groups of order $p^{\alpha -m}$. It is obvious that no two of these can enter in the same sub-group of order $p^{\alpha }$; for if they did, they would generate a group of order $p^{\alpha -m+x}$ ($x>0$), a sub-group of this order cannot enter in a group of order $p^{\alpha -m}{q}^{\beta }$. Therefore $G$ contains $q^{\beta }$ sub-groups of order $p^{\alpha }$.

Since the sub-groups of $G$, of order $p^{\alpha }$, are transitive, their self-conjugate operations must (§ 106) displace all the symbols, and they cannot therefore be permutable with any operation whose order is a power of $q$. Suppose now that every operation of a sub-group $Q$, of order $q^x$, is permutable with an operation $P$ of order $p^y$, and that there is no sub-group of order $q^{x+1}$ of which this is true. If $S$ is a self-conjugate operation of a sub-group of order $p^{\alpha }$ to which $P$ belongs, $P$ is transformed into itself and $Q$ into a new sub-group $Q\prime$ by $S$. Hence the sub-group which contains $P$ self-conjugately has two, and therefore $p^m$, sub-groups of order $q^x$. No two of these can belong to the same sub-group of order $q^{\beta }$, for if they did they would generate a group of order $q^{x+x\prime }$ ($x\prime >0$). Now since $P$ is permutable with $Q$, it must leave unchanged the symbol which $Q$ leaves unchanged. Hence $P$ must leave every symbol unchanged; i.e. it must be the identical operation. The order of every operation of $G$ is therefore either a power of $p$ or a power of $q$.

Suppose now that $p^r$ is the order of the greatest sub-group that is common to any two sub-groups of order $p^{\alpha }$, and that $h$ is such a sub-group. Then (§ 80) $h$ must be permutable with an operation of order $q$. Since no operation of $h$ is permutable with any operation of order $q$, it follows that $p^r-1$ is divisible by $q$, and therefore that $r=m$.

Let $k$, of order $p^{m+t}{q}^{\gamma }$, be the greatest sub-group that contains $h$ self-conjugately. If $\gamma =\beta$, $h$ is common to each of the $q^{\beta }$ sub-groups of order $p^{\alpha }$, and it is therefore a self-conjugate sub-group of $G$.

If $\gamma <\beta$, let $Q$ be a sub-group, of order $q^{\gamma }$, of $k$. In $G$ there must be a group, whose order is divisible by $q^{\gamma +1}$, containing $Q$ self-conjugately. Hence there must be operations in $G$, which transform $Q$ into itself and $h$ into a conjugate sub-group $h\prime$. If $h$ and $h\prime$ have a common sub-group, it must be transformed into itself by every operation of $Q$. This is impossible, since $p^s-1$ is not divisible by $q$ when $s<m$.

In the group , $Q$ is one of $p^m$ conjugate sub-groups, and therefore no operation of $h\prime$ can transform $Q$ into itself. Hence if an operation $P\prime$ of $h\prime$ transforms $h$ into itself, it must transform $Q$ into a group $Q\prime$, which is conjugate to $Q$ in $k$. Hence, $k\prime$ being the greatest sub-group that contains $h\prime$ self-conjugately,  must be common to $k$ and $k\prime$. But containing two sub-groups of order $q^{\gamma }$, must contain $p^m$ such sub-groups; and therefore the $p^m$ sub-groups of order $q^{\gamma }$ that enter in $k$ are identical with those that enter in $k\prime$. This is impossible if $H$ and $H\prime$ are distinct; for the $p^m$ sub-groups of $k$, of order $q^{\gamma }$, generate . Hence no operation of $h\prime$, except identity, transforms $h$ into itself, and the number of sub-groups in the conjugate set to which $h$ belongs must not be less than $p^m$; so that

Now we have seen that

from these congruences, and the inequality $\alpha -m<m$, it follows that

This inequality is inconsistent with the previous one, which follows directly from the supposition that $\gamma$ is less than $\beta$. Hence $\gamma =\beta$, and $h$ is a self-conjugate sub-group of $G$.

It follows therefore that, on every possible supposition, $G$ must have a self-conjugate sub-group. If this sub-group be represented by $G_1$, the same reasoning may be repeated with respect to the groups $\frac{G}{G_1}$ and $G_1$. Hence $G$ is soluble.

Corollary. All groups whose orders are $p_1{p}_2^{\alpha }$, $p_1^2{p}_2^{\alpha }$, $p_1^3{p}_2^{\alpha }$, $p_1^4{p}_2^{\alpha }$, $p_1^5{p}_2^{\alpha }$, $p_1^{\alpha }{p}_2$ are soluble.

Since the congruence

If in these cases there are $16$ sub-groups of order $3^{\alpha }$ ($\alpha >1$), there must (§ 78) be sub-groups of order $3^{\alpha -1}$ common to two sub-groups of order $3^{\alpha }$; and, in the groups of order $2^4{3}^{\alpha }$ or $2^5{3}^{\alpha }$, such a sub-group, if not self-conjugate, must be one of either $4$ or $8$ conjugate sub-groups. The groups must therefore be isomorphic with groups of degree $4$ or $8$; from this it follows immediately that they must be soluble (§ 146).

244. THEOREM IV. Groups of order $p_1^{\alpha }{p}_2^2$ are soluble.

If a group $G$, of order $p_1^{\alpha }{p}_2^2$, contains only $p_2$ sub-groups of order $p_1^{\alpha }$, it cannot be simple, for a group of order $p_2^2$ cannot be expressed as a substitution-group of $p_2$ symbols. Similarly, if $G$ contains a single group of order $p_1^{\alpha }$, it is not simple.

If $G$ contains $p_2^2$ sub-groups of order $p_1^{\alpha }$, and if these sub-groups have no common operations, except identity, there are operations in $G$ whose orders are powers of $p_1$; and therefore a group of order $p_2^2$ is self-conjugate. If the operations of the $p_2^2$ sub-groups of order $p_1^{\alpha }$ are not all distinct, let $p_1^r$ be the order of the greatest sub-group common to any two of them; and let $h$ be such a sub-group. Then (§ 80) $h$ must be self-conjugate in a group $k$ of order $p_1^{r+s}{p}_2^{\beta }$ ($s>0$, $\beta =1$ or $2$). If $\beta =2$, $h$ is self-conjugate in $G$; and if $\beta =1$, $k$ must contain $p_2$ sub-groups of order $p_1^{r+s}$, and therefore

Now (§ 78)

Again, if $s=\alpha -r$, $h$ is one of $p_2$ conjugate sub-groups; as before, $G$ cannot then be simple.

Suppose now that $H$ is a sub-group of order $p_1^{\alpha }$, and that the self-conjugate operations of $H$ form a sub-group $h$. This must be one of $1$, $p_2$ or $p_2^2$ conjugate sub-groups; in the first two cases, $G$ cannot be simple. Moreover, if any two sub-groups of the conjugate set to which $h$ belongs have a common sub-group, it must be self-conjugate in a group containing more than one sub-group of order $p_1^{\alpha }$; again, $G$ cannot be simple.

Let

Finally, suppose that $h$ contains the $p_2$ sub-groups

If $h$ contains further sub-groups of the conjugate set to which $h$ belongs, they must be permuted among themselves in sets of $p_2$ when $h$ is transformed by an operation $Q$, of order $p_2$, belonging to $k$. Hence we may assume that

Now, since by supposition $h$ is not a self-conjugate sub-group of $H$, there must (§ 55) be in $H$ a sub-group $h\prime$ conjugate to and permutable with $h$. Let this be the sub-group of order $p_1^r$, common to

In other words, $k$ must contain a self-conjugate sub-group of $G$.

On every supposition that can be made, we have shewn that $G$ must have a self-conjugate sub-group. If this sub-group is $G_1$, and if the order of either $G_1$ or $\frac{G}{G_1}$ contains $p_2^2$, we may repeat the same reasoning; while if the orders of $\frac{G}{G_1}$ and $G_1$ are both divisible by $p_2$, we have already seen that they must be soluble. Finally, then, $G$ itself must be soluble.

245. THEOREM V. A group of order $p_1^{\alpha }{p}_2^{\beta }$, in which the sub-groups of orders $p_1^{\alpha }$ and $p_2^{\beta }$ are Abelian, is soluble.

Suppose, first, that there are $p_2^{\gamma }$ ($\gamma <\beta$) sub-groups of order $p_1^{\alpha }$. The group is then isomorphic with a substitution group of degree $p_2^{\gamma }$. Now the operations of a sub-group of order $p_2^{\beta }$ transform the $p_2^{\gamma }$ sub-groups of order $p_1^{\alpha }$ transitively among themselves; and therefore, in the isomorphism between the given group and the substitution group of degree $p_2^{\gamma }$, there corresponds to an Abelian sub-group of order $p_2^{\beta }$ a transitive substitution group of degree $p_2^{\gamma }$. But (§ 124) an Abelian group can only be represented as a transitive substitution group of a number of symbols equal to its order. Hence the isomorphism between the given group and the substitution group is multiple; and the given group is not simple.

Suppose, next, that there are $p_2^{\beta }$ sub-groups of order $p_1^{\alpha }$, and therefore (§ 81) $p_1^{\alpha }-1$ distinct sets of conjugate operations whose orders are powers of $p_1$. Let $P$ be an operation whose order is a power of $p_1$; and let $H$, of order $p_1^{\alpha }{p}_2^{\delta }$, be the greatest group which contains $P$ self-conjugately. Then $H$ contains at least $p_2^{\delta }$ operations whose orders are powers of $p_2$, and therefore at least $p_2^{\delta }$ distinct operations of the form $PQ$, where $P$ and $Q$ are permutable and

The group therefore always contains a self-conjugate sub-group. A repetition of the same reasoning shews that it is soluble.

246. THEOREM VI. A group of order $p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_n^{{\alpha }_n}$, in which the sub-groups of order $p_1^{{\alpha }_1}$, $p_2^{{\alpha }_2}$, …, $p_{n-1}^{{\alpha }_{n-1}}$ are cyclical, is soluble⁹⁷.

If the number of operations of the group, whose orders divide $p_r^{{\alpha }_r}{p}_{r+1}^{{\alpha }_{r+1}}\dots p_n^{{\alpha }_n}$, is exactly equal to this number, it follows from Theorem VII, Cor. I, § 87, that the same is true for the number of operations of the group whose orders divide $p_{r+1}^{{\alpha }_{r+1}}\dots p_n^{{\alpha }_n}$. Now, when $r=1$, this relation obviously holds; and therefore it is true for all values of $r$. The group therefore contains just $p_n^{{\alpha }_n}$ operations whose orders divide $p_n^{{\alpha }_n}$; hence a sub-group of order $p_n^{{\alpha }_n}$ is self-conjugate. Any sub-group of order $p_{n-1}^{{\alpha }_{n-1}}$ must transform this self-conjugate sub-group of order $p_n^{{\alpha }_n}$ into itself, so that the group must contain a sub-group of order $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$. If there were more than one sub-group of this order, the group would contain more than $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$ operations whose orders divide this number. Hence the group contains a single sub-group of order $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$, and this sub-group must be self-conjugate. By continuing this reasoning it may be shewn that, for each value of $r$, the group contains a single sub-group of order $p_r^{{\alpha }_r}{p}_{r+1}^{{\alpha }_{r+1}}\dots p_n^{{\alpha }_n}$, such a sub-group being necessarily self-conjugate. The group is therefore soluble; for we have already seen that a group whose order is the power of a prime is always soluble.

The self-conjugate sub-group of order $p_r^{{\alpha }_r}\dots p_n^{{\alpha }_n}$ must contain the complete set of sub-groups of order $p_r^{{\alpha }_r}$. If then there are $m$ of these sub-groups, $m$ must be a factor of $p_{r+1}^{{\alpha }_{r+1}}\dots p_n^{{\alpha }_n}$; and a sub-group of order $p_r^{{\alpha }_r}$ must be contained self-conjugately in a sub-group of order

Ex. Shew that, when the groups of order $p_n^{{\alpha }_n}$ are also cyclical, there are sub-groups whose orders are any factors whatever of the order of the group.

247. A special case of the class of groups under consideration is that in which the order contains no repeated prime factor. These groups have formed the subject of a memoir by Herr O. Hölder⁹⁸. He shews that they are capable of a specially simple form of representation, which we shall now consider.

Let

248. The theorem of § 246 is a particular case of the following more general one, due to Herr Frobenius⁹⁹, from which a number of interesting and important results may be deduced.

THEOREM VII. If $G$ is a group, of order $N=mn$, where $m$ and $n$ are relatively prime, and where

The theorem will be proved inductively, by shewing that, if it is true for any similar group of order $m\prime n\prime$, where $m\prime$ is a factor of $m$, it is also true for $G$.

Suppose that the greatest group, which contains $R$ self-conjugately, is a group $H\prime$ of order $p^{\alpha \prime }{q}^{\beta \prime }\dots r^{\gamma }n\prime$, where $n\prime$ is a factor of $n$. Let $N\prime$ be any operation of this sub-group, whose order $\nu \prime$ is a factor of $n\prime$. Then (§ 177) since and $\nu \prime$ are relatively prime, $N\prime$ is permutable with every operation of $R$. Hence every operation of $H\prime$, whose order divides $n\prime$, is permutable with every operation of $R$.

Let now $S$ be any operation of $R$; and suppose that the order of $K$, the greatest sub-group which contains $S$ self-conjugately, is $p^{{\alpha }_1}{q}^{{\beta }_1}\dots r^{\gamma }{n}_1$. If the order of $S$ is $r^{\delta }$, the order of  is $p^{{\alpha }_1}{q}^{{\beta }_1}\dots r^{\gamma -\delta }{n}_1$, say $m_1{n}_1$. We have seen in § 176 that, if $P_1$ is a sub-group of $P$, then  is equal to or is a factor of . Hence if $P_1$ is a sub-group of order $p^{{\alpha }_1}$ of ,  and $\frac{N}{p^{\alpha }}$ are relatively prime; therefore, a fortiori, and $\frac{m_1{n}_1}{p^{{\alpha }_1}}$ are relatively prime. Similarly and $\frac{m_1{n}_1}{p^{{\alpha }_1}{q}^{{\beta }_1}}$ are relatively prime, and so on. Moreover $m_1$ is a factor of $m$. We may therefore assume that the theorem holds for . Hence this group contains exactly $n_1$ operations whose orders divide $n_1$, and it also contains a sub-group of order $m_1$ in which a sub-group of order $r^{\gamma -\delta }$ is self-conjugate. Therefore $K$ contains a sub-group of order $p^{{\alpha }_1}{q}^{{\beta }_1}\dots r^{\gamma }$ in which a sub-group of order $r^{\gamma }$ is self-conjugate, and exactly $n_1$ operations of the form $SN$, where $S$ and $N$ are permutable and the order of $N$ divides $n$.

Now $S$ is one of $p^{\alpha -{\alpha }_1}{q}^{\beta -{\beta }_1}\dots \frac{n}{n_1}$ conjugate operations in $G$; corresponding to each of these, there is such a set of $n_1$ operations of the form $SN$, while (§ 16) no two of these operations can be identical. Hence $G$ contains $p^{\alpha -{\alpha }_1}{q}^{\beta -{\beta }_1}\dots n$ operations of the form $SN$, arising from the conjugate set to which $S$ belongs. If then we sum for the distinct conjugate sets of operations of $G$ whose orders are powers of $r$, the number of operations of $G$, whose orders divide $r^{\gamma }n$ and do not divide $n$, is $n\sum p^{\alpha -{\alpha }_1}{q}^{\beta -{\beta }_1}\dots$.

Now since $H\prime$ is the greatest group that contains $R$ self-conjugately, the greatest common sub-group of $K$ and $H\prime$ is the greatest sub-group of $K$ that contains $R$ self-conjugately. The order of this group certainly contains $p^{{\alpha }_1}{q}^{{\beta }_1}\dots r^{\gamma }$ as a factor; and if the order is $p^{{\alpha }_1}{q}^{{\beta }_1}\dots r^{\gamma }\nu$, then $S$ is one of $p^{\alpha \prime -{\alpha }_1}{q}^{\beta \prime -{\beta }_1}\dots \frac{n\prime }{\nu }$ conjugate operations in $H\prime$. Every operation of $H\prime$ however, whose order divides $n\prime$, is permutable with every operation of $R$; and therefore $\nu =n\prime$. Hence the number of operations of $H\prime$, excluding identity, whose orders are powers of $r$, is $\sum \prime p^{\alpha \prime -{\alpha }_1}{q}^{\beta \prime -{\beta }_1}\dots$, where the summation is extended to all the distinct conjugate sets of operations of $H\prime$ whose orders are powers of $r$. It has been shewn (Theorem IV, Corollary, § 81) that the number of distinct conjugate sets of operations in $G$, whose orders are powers of $r$, is the same as the number in $H\prime$. Hence the symbols $\sum$ and $\sum \prime$ represent summations of the same number of terms.

Finally, in accordance with the induction we are using, we may assume that the number of operations of $G$ whose orders divide $r^{\gamma }n$ is exactly equal to $r^{\gamma }n$. For the order of $G$ may be separated into the factors

Hence the number of operations of $G$ whose orders divide $n$, is equal to

The free use of the inductive process that has been made in the preceding proof may possibly lead the reader to doubt its validity. He will find it instructive to verify the truth of the theorem directly in the simpler cases. In fact, the direct verification for the case in which $m$ is a power of a prime is essential to the formal proof; it has been omitted for the sake of brevity.

Corollary I. If $m=m_1{m}_2$, where $m_1$ and $m_2$ are relatively prime, $G$ has a single conjugate set of sub-groups of order $m_1$.

That $G$ contains groups of order $m_1$ may be shewn inductively at once. For since it is clearly true when $m$ contains only one, or two, distinct prime factors, we may assume it true when $m_1$ does not contain $r^{\gamma }$ as a factor. Now $G$ has a sub-group $g$ of order $p^{\alpha }{q}^{\beta }\dots r^{\gamma }$, which contains a sub-group $R$ of order $r^{\gamma }$ self-conjugately. The factor-group $\frac{g}{R}$ contains then a sub-group of order $m\prime$ where $m\prime$ is any factor of $\frac{m}{r^{\gamma }}$ which is relatively prime to $\frac{m}{m\prime r^{\gamma }}$; and therefore $g$ contains a sub-group of order $m\prime r^{\gamma }$.

If now $I$ and $I\prime$ are two sub-groups of $G$ of order $m\prime r^{\gamma }$, they may be assumed to contain the same sub-group $R$ of order $r^{\gamma }$. For the sub-groups of $G$ of order $r^{\gamma }$ form a single conjugate set; and therefore, if $I\prime$ contained a sub-group $R\prime$ of this set, $S^{-1}I\prime S$ would contain $S^{-1}R\prime S$, which may be taken to be $R$. Now $R$ is a self-conjugate sub-group of both $I$ and $I\prime$, and therefore the factor-groups $\frac{I}{R}$ and $\frac{I\prime }{R}$ of order $m\prime$ are sub-groups of the factor-group $\frac{H\prime }{R}$. Hence if the result is true when $m_1$ does not contain the factor $r^{\gamma }$, it is also true when $m_1$ does contain $r^{\gamma }$. But when $m_1$ is equal to $p^{\alpha }$, the result is obviously true; and therefore it is true generally.

Corollary II. If the sub-groups $P$, $Q$, …, $R$ of $G$ contain characteristic sub-groups $P_0$, $Q_0$, …, $R_0$ of orders $p^{{\alpha }_0}$, $q^{{\beta }_0}$, …, $r^{{\gamma }_0}$; then $G$ contains a sub-group of order $m_0=p^{{\alpha }_0}{q}^{{\beta }_0}\dots r^{{\gamma }_0}$, which has a self-conjugate sub-group of order $r^{{\gamma }_0}$.

Assuming the truth of this statement when $m_0$ does not contain the factor $r$, the factor-group $\frac{H\prime }{R}$ must contain a sub-group of order $\frac{m_0}{r^{{\gamma }_0}}$; and therefore $G$ contains a sub-group $J$ of order $p^{{\alpha }_0}{q}^{{\beta }_0}\dots r^{\gamma }$ which has $R$ for a self-conjugate sub-group. Hence, since $R_0$ is a characteristic sub-group of $R$, $J$ contains a sub-group of order $p^{{\alpha }_0}{q}^{{\beta }_0}\dots$ and a self-conjugate sub-group of order $r^{{\gamma }_0}$. It therefore contains a sub-group of order $p^{{\alpha }_0}{q}^{{\beta }_0}\dots r^{{\gamma }_0}$, which has a self-conjugate sub-group of order $r^{{\gamma }_0}$.

249. The reader will have no difficulty in seeing, as has already been stated, that Theorem VI is a direct result of the theorem proved in the last paragraph. We shall proceed at once to further applications of it.

THEOREM VIII. A group $G$ of order

are Abelian, with either one or two generating operations, is generally soluble; the special case $p_1=2$, $p_2=3$, may constitute an exception¹⁰⁰.

If an Abelian sub-group $P_r$ of order $p_r^{{\alpha }_r}$ is generated by a single operation, it is cyclical and  is (§ 176) equal to $p_r-1$. If $P_r$ is generated by two independent and permutable operations,  is equal to . Now no prime greater than $p_r$ can divide unless $p_r+1$ be a prime; and this is only possible when $p_r$ is equal to $2$. Hence unless $P_1=2$, $p_2=3$,  and $\frac{N}{p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_r^{{\alpha }_r}}$ are relatively prime for all values of $r$ from $1$ to $n-1$. Omitting for the present this exceptional case, the conditions of Theorem VII are satisfied by $G$; it therefore contains exactly $p_n^{a_n}$ operations whose orders are divisible by $p_n$. Therefore $G$ has a self-conjugate sub-group $P_n$ of order $p_n^{{\alpha }_n}$, and hence also one or more sub-groups of order $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$. There are however only $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$ operations in $G$ whose orders are divisible by no prime smaller than $p_{n-1}$, and therefore the sub-group of order $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$ must be self-conjugate. This process clearly may be continued to shew that $G$ has a self-conjugate sub-group of every order $p_r^{{\alpha }_r}{p}_{r+1}^{{\alpha }_{r+1}}\dots p_n^{{\alpha }_n}$ ($r=2,3,\dots ,n$); from which it follows immediately that $G$ is soluble.

Returning now to the exceptional case, suppose that a group of order $2^{{\alpha }_1}$ is contained self-conjugately in a maximum group of order $2^{{\alpha }_1}{3}^{\beta }m$. If every operation of the group of order $2^{{\alpha }_1}$ is self-conjugate within this sub-group, $G$ contains $2^{{\alpha }_1}-1$ distinct conjugate sets of operations whose orders are powers of $2$; hence it follows that there are just $\frac{N}{2^{{\alpha }_1}}$ operations in $G$ whose orders are not divisible by $2$. In this case, the conditions of Theorem VII are satisfied by $G$; and it is still soluble.

If, lastly, the operations of the sub-groups of order $2^{{\alpha }_1}{3}^{\beta }m$, whose orders are powers of $2$, are not all self-conjugate, there must be an operation $B$, whose order is a power of $3$, in this sub-group which is not permutable with every operation of the sub-group of order $2^{{\alpha }_1}$. Let now

Corollary. A group of order $p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_n^{{\alpha }_n}$ in which no one of the indices ${\alpha }_1$, ${\alpha }_2$, …, ${\alpha }_{n-1}$ is greater than $2$, is soluble unless it has a sub-group which is isomorphic with a tetrahedral group. For a group of order $p_r^2$ is necessarily an Abelian group which is generated by either one or two independent operations.

250. We shall next consider certain groups of even order in which the operations of odd order form a self-conjugate sub-group. If $G$ is a group of order $N$, where

251. Suppose, secondly, that, under the same conditions, $G$ contains a self-conjugate sub-group $𝒬$ of order $2^{\alpha }$. Since and $n$ are relatively prime, every operation of $𝒬$ is permutable with every operation of $G$ (§ 177). If $𝒬\prime$ is a sub-group of $𝒬$ of order $2^{\alpha -1}$, $𝒬\prime$ must therefore be self-conjugate in $G$. Now the order of the factor group $\frac{G}{𝒬\prime }$ is $2n$; hence, by the preceding result, it has a self-conjugate sub-group of order $n$. It follows that $G$ has a self-conjugate sub-group of order $2^{\alpha -1}n$. This group, again, has a self-conjugate sub-group of order $2^{\alpha -2}n$, and so on. Hence $G$ has a sub-group of order $n$. Now by Theorem VII (§ 248), $G$ has just $n$ operations whose divide $n$. Hence the sub-group $H$, of order $n$, must be self-conjugate; and $G$ is the direct product of the two groups $𝒬$ and $H$.

252. Suppose next that $𝒬$ is not self-conjugate in $G$; and let $I$, of order $2^{\alpha }n\prime$, be the greatest group that contains $𝒬$ self-conjugately. Every operation of $I$ is permutable with every operation of $𝒬$; and therefore (§ 81) $G$ contains $2^{\alpha -1}$ distinct sets of conjugate operations whose orders are powers of $2$. The case, in which $𝒬$ is cyclical, has been already dealt with and will now be excluded; we may thus assume that $𝒬$ contains $2^{\beta }-1$ ($\beta \nless 2$) operations of order $2$, and that $G$ contains an equal number of sets of conjugate operations of order $2$.

If possible, let no two sub-groups of order $2^{\alpha }$ have a common operation except identity. Then if two operations $S$ and $T$, of order $2$, are chosen, belonging to distinct conjugate sets and to different sub-groups of order $2^{\alpha }$, they cannot be permutable with each other. They will therefore generate a dihedral sub-group of order $2m$. If $m$ were odd, $S$ and $T$ would be conjugate operations. Hence $m$ must be even; and the dihedral sub-group must contain a self-conjugate operation $U$ of order $2$. Since $U$ is permutable with both $S$ and $T$, it must occur in at least two different sub-groups of order $2^{\alpha }$; and therefore the supposition, that no two sub-groups of order $2^{\alpha }$ have a common operation except identity, is impossible.

Let now $𝒬\prime$ of order $2^{\alpha \prime }$ be a sub-group common to $𝒬$ and ${𝒬}_1$, two sub-groups of order $2^{\alpha }$; and suppose that no two sub-groups of order $2^{\alpha }$ have a common sub-group which contains $𝒬\prime$ and is of greater order. Since every operation of $𝒬$ is self-conjugate in $I$, $𝒬\prime$ must be self-conjugate in $I$. If then $J$, of order $2^{\alpha }n\prime r$, is the greatest sub-group of $G$ that contains $𝒬\prime$ self-conjugately, $J$ contains $r$ sub-groups of order $2^{\alpha }$. Hence the factor-group $\frac{J}{𝒬\prime }$ contains $r$ sub-groups of order $2^{\alpha -\alpha \prime }$, and no two of these have a common sub-group; for if they had, some two sub-groups of order $2^{\alpha }$ contained in $J$ would have a common sub-group greater than and containing $𝒬\prime$, contrary to supposition. The $r$ sub-groups of order $2^{\alpha -\alpha \prime }$ of $\frac{J}{𝒬\prime }$ must therefore be cyclical.

253. We will apply the results of the last paragraph to the case in which all the operations of $𝒬$ are of order $2$. Then $𝒬\prime$ must be of order $2^{\alpha -1}$, since $\frac{𝒬}{𝒬\prime }$ is cyclical. Suppose now that $R$ is an operation of $J$ of odd order, which is permutable with $𝒬\prime$ but not with $𝒬$. If $R$ were self-conjugate in a sub-group whose order is divisible by $2^{\alpha }$, this sub-group would contain $𝒬\prime$ and therefore one or more sub-groups of order $2^{\alpha }$ containing $𝒬\prime$. But $R$ is not permutable with any sub-group of order $2^{\alpha }$ that contains $𝒬\prime$; and therefore the highest power of $2$ that divides the order of the group in which $R$ is self-conjugate is $2^{\alpha -1}$, so that $R$ is one of $2\mu$ ($\mu$ odd) conjugate operations. If now $A$ is an operation of order $2$ that belongs to $𝒬$ and not to $𝒬\prime$, no operation conjugate to $R$ can be permutable with $A$. Hence in the substitution group of degree $2\mu$, that results on transforming the set of operations conjugate to $R$ among themselves by all the operations of $G$, the substitution corresponding to $A$ is an odd substitution. This substitution group has therefore a self-conjugate sub-group whose order is half its own, and therefore $G$ has a self-conjugate sub-group of order $2^{\alpha -1}n$. In the same way it may be shewn that this sub-group has a self-conjugate sub-group of order $2^{\alpha -2}n$; and so on. Hence $G$ has a sub-group $H$ of order $n$. But it follows from Theorem VII, § 248, that $G$ has exactly $n$ operations whose orders divide $n$; and therefore $H$ is a self-conjugate sub-group. Moreover, since $\frac{G}{H}$ is an Abelian group of order $2^{\alpha }$, it must contain self-conjugate sub-groups of every order $2^r$ ($r=1,2,\dots ,\alpha -1$); and $G$ therefore has self-conjugate sub-groups of every order $2^{r}n$. In general however these sub-groups are not characteristic, as is the case when $𝒬$ is cyclical.

254. We will consider next the case where $𝒬$ is generated by two operations $A$ and $B$, of orders $2^{\alpha -1}$ and $2$. There are in $𝒬$ three operations of order $2$, namely, $A^{2^{\alpha -2}}$, $A^{2^{\alpha -2}}B$, and $B$; and in $𝒬$ there are no operations, of order greater than $2$, of which the two latter are powers. Suppose that $G$ contains an operation $B\prime$, conjugate to $B$, with which $A^{2^{\alpha -2}}$ is not permutable. Then is a dihedral group, and if $R$ is an operation of odd order of this group, no power of $A$ is permutable with $R$. Since $A^{2^{\alpha -2}}$ and $B\prime$ are not conjugate operations in $G$, there must be an operation of order $2$, conjugate to $A^{2^{\alpha -2}}B$, and permutable with $R$. The sub-group, in which is self-conjugate, therefore contains representatives of each of the three sets of conjugate operations of order $2$ that belong to $G$; and in this sub-group these representatives form three distinct conjugate sets. Hence no operation conjugate to $A^{2^{\alpha -2}}$ can be permutable with $R$; and therefore no power, except identity, of an operation conjugate to $A$ is permutable with $R$. Hence $R$ is one of $2^{\alpha -1}\mu$ ($\mu$ odd) conjugate operations, with none of which is $A$ permutable. It follows that, in the substitution group of degree $2^{\alpha -1}\mu$, with which $G$ is isomorphic, $A$ is an odd substitution. Hence $G$ contains a self-conjugate sub-group of order $2^{\alpha -1}n$; and since this is of the same type as $G$, the same reasoning may be applied to it. Exactly as before, $G$ will contain self-conjugate sub-groups of every order $2^{r}n$ ($r=0,1,\dots ,\alpha -1$).

We have assumed that there is an operation $B\prime$, conjugate to $B$, with which $A^{2^{\alpha -2}}$ is not permutable. If $A^{2^{\alpha -2}}$ were permutable with every operation that is conjugate to $B$, the sub-group in which $A^{2^{\alpha -2}}$ is permutable would contain a self-conjugate sub-group $G\prime$ of $G$ whose order is divisible by $2$. In this case, we may deal with $\frac{G}{G\prime }$ exactly as we have been dealing with $G$.

The results of §§ 250–254 may be summed up in the following form:—

THEOREM IX. If in a group $G$ of order $2^{\alpha }n$, where $n$ is odd, a sub-group $𝒬$ of order $2^{\alpha }$ is an Abelian group of type , , or , and if and $n$ are relatively prime, then $G$ contains self-conjugate sub-groups of each order $2^{\beta }n$ $$

.

255. Still representing the order of $G$ by $2^{\alpha }n$, where $n$ is odd, there are two cases, in which the sub-groups of order $2^{\alpha }$ are not Abelian, where it may be shewn without difficulty that $G$ contains a self-conjugate sub-group of order $n$.

THEOREM X. If the order of $G$ is $2^{\alpha }n$, where $n$ is odd and not divisible by $3$, and if a sub-group $𝒬$ of order $2^{\alpha }$ is of the type

$G$ contains a self-conjugate sub-group of each order $2^{\beta }n$ $$

.

For each of these types,  is $3$ or $1$; and if $𝒬\prime$ is any sub-group of $𝒬$,  is either $3$ or $1$. Hence if any operation of $G$ of odd order is permutable with $𝒬$ or $𝒬\prime$, it must be permutable with every operation of $𝒬$ or $𝒬\prime$. In each type, the self-conjugate operations form a cyclical sub-group of order $2$, namely .

We will consider first the case where $𝒬$ is of type (i). In this case,  and are sub-groups of $G$ of order $4$, the former being self-conjugate in $𝒬$ while the latter is not. If they are conjugate sub-groups in $G$, we have seen (§ 82) that $G$ must contain an operation of odd order $S$, such that the sub-groups ($n=0,1,2,\dots$) are permutable with each other. This is impossible, since every operation that is permutable with a sub-group $𝒬\prime$ is permutable with all its operations. Hence  and are not conjugate sub-groups; and $A^{2^{\alpha -3}}$ and $B$ are not conjugate operations.

Now $A^{2^{\alpha -3}}$ is self-conjugate in  and is one of two conjugate operations in $𝒬$; therefore in $G$ it must be one of $2\mu$ conjugate operations, where $\mu$ is odd. When these $2\mu$ conjugate operations are transformed by $A^{2^{\alpha -3}}$, two only, namely, $A^{2^{\alpha -3}}$ and $A^{-2^{\alpha -3}}$, remain unchanged. Let us suppose that, in the resulting substitution, there are $x$ cycles of $2$ symbols and $y$ cycles of $4$ symbols.

When the $2\mu$ conjugate operations are transformed by $B$, none can remain unchanged; and we may suppose that the resulting substitution has $x\prime$ cycles of $2$ symbols and $y\prime$ cycles of $4$ symbols.

Now since

and therefore

Hence one of the two substitutions $A^{2^{\alpha -3}}$ and $B$ must be odd; $G$ has therefore a self-conjugate sub-group of order $2^{\alpha -1}n$. The groups of order $2^{\alpha -1}$ contained in this self-conjugate sub-group are either of type (i), with $\alpha -1$ written for $\alpha$: or they are cyclical; for, like $𝒬$, they can only contain a single operation of order $2$. If they are cyclical, the reasoning of § 250 may be applied; and if they are of type (i) the same reasoning will apply to the self-conjugate sub-group of order $2^{\alpha -1}n$ that has been used for $G$. Finally, then, $G$ must contain sub-groups of orders $n$, $2n$, $2^{2}n$, …, $2^{\alpha -1}n$, each of which is contained self-conjugately in the next; and therefore the sub-group $H$ of order $n$ must be self-conjugate.

If $𝒬$ is of type (ii), $A$ is one of $2\mu$ conjugate operations in $G$, where $\mu$ is odd. It may be shewn, as in the previous case, that $B$ and $A^{2^{\alpha -2}}$ cannot be conjugate in $G$, so that no one of the operations conjugate to $A$ has $B$ for one of its powers. If now $B$ were permutable with any one of these $2\mu$ conjugate operations, the group would contain an Abelian sub-group of order $2^{\alpha }$, which is not the fact. Hence the substitution, given on transforming the $2\mu$ operations by $B$, consists of $\mu$ transpositions and is an odd substitution. Therefore $G$ contains a self-conjugate sub-group of order $2^{\alpha -1}n$, in which the sub-groups of order $2^{\alpha -1}$ are cyclical. Hence, again, there is a self-conjugate sub-group of order $n$.

256. The only non-Abelian groups of order $2^3$ are those of types (i) and (ii) of § 255, when $\alpha =3$. The Abelian groups of order $2^2$ and $2^3$ are all included in the types considered in Theorem IX, § 254. For an Abelian group $𝒬$, of order $2^3$ and type ,  is $3$; and for one of order $2^3$ and type ,  is $21$. Hence Theorems IX and X shew that, if $2$, $2^2$ or $2^3$ divide the order of a group, but not $2^4$, then the operations of odd order form a self-conjugate sub-group, with possible exceptions when $12$ or $56$ is a factor of the order. Hence:—