Chapter VI.
On Sylow’s Theorem.

I t has been proved (§ 22) that the order of any sub-group of a group $G$ is a factor of the order of $G$; and it results at once from the investigation of §§ 45–47 that, in an Abelian group, there is always at least one sub-group whose order is any given factor of the order of the group. The latter result is not however generally true for groups which are not Abelian; and for factors of the order of a group which contain more than one distinct prime, no general law is known as to the existence or non-existence of corresponding sub-groups. If however $p^m$, where $p$ is a prime, divides the order of the group, it may be shewn that the group will always contain a sub-group of order $p^m$. The special form of this theorem, that a group whose order is divisible by a prime $p$ contains operations of order $p$, is due originally to Cauchy²³. The more general result was first established by Sylow. He has shewn²⁴ that, if $p^{\alpha }$ is the highest power of a prime $p$ which divides the order of a group, the group contains a single conjugate set of $kp+1$ sub-groups of order $p^{\alpha }$.

For the special case $p=2$, the following proof of Cauchy’s theorem is perhaps worth giving for its simplicity. Arrange the operations of a group of even order in pairs $S$ and $S^{-1}$ of inverse operations. If at any stage no further pairs can be formed, the remaining operations must all, except the identical operation, be of order $2$. The number of these remaining operations, which include the identical operation, is even. Hence there must be at least one operation of order $2$, and the total number of such operations is odd.

We shall devote the present chapter to the proof of Sylow’s theorem; and a consideration of some of its more immediate consequences. These constitute, as will be seen later on, a most important set of results.

77. We shall divide the proof of Sylow’s theorem into two parts. First we shew that, if $p^{\alpha }$ is the highest power of a prime $p$ which divides the order of a group, the group must have a sub-group of order $p^{\alpha }$; and secondly that the sub-groups of order $p^{\alpha }$ form a single conjugate set and that their number is congruent to unity, .

Lemma. If $p^{\alpha }$ is the highest power of a prime $p$ by which the order of a group $G$ is divisible, $G$ must contain a sub-group whose order is divisible by $p^{\alpha }$.

If the group $G$ is Abelian, this has been already proved in § 37. Suppose then that $G$ of order $N$ ($=p^{\alpha }m$, where $m$ is prime to $p$) is not Abelian; and let $G\prime$ of order $N\prime$ be the sub-group of $G$ formed of its self-conjugate operations. If $S$ is any operation of $G$ which is not self-conjugate, and if $n$ is the order of the greatest sub-group within which $S$ is permutable, $S$ forms one of a set of $\frac{N}{n}$ conjugate operations; hence, by equating the order of the group to the sum of the numbers of operations contained in the different conjugate sets, we obtain the equation

If $N\prime$ is not divisible by $p$, or in other words if $G$ contains no self-conjugate operations of order $p$, this equation requires that at least one of the symbols $n$ should be divisible by $p^{\alpha }$; therefore in this case the lemma is true.

If $N\prime$ is divisible by $p^{\alpha \prime }$, the group $G\prime$, being Abelian, must contain a sub-group $g\prime$ of order $p^{\alpha \prime }$; and this sub-group is self-conjugate in $G$. Consider now the group $\frac{G}{g\prime }$ of order $p^{\alpha -\alpha \prime }m$. If it has no self-conjugate operations of order $p$, it must (by the result just obtained) contain a sub-group whose order is divisible by $p^{\alpha -\alpha \prime }$; and hence $G$ contains a sub-group whose order is divisible by $p^{\alpha }$. If, on the other hand, $\frac{G}{g\prime }$ has a sub-group of self-conjugate operations of order $p^{\alpha \prime \prime }$, then $G$ must contain a self-conjugate sub-group of order $p^{\alpha \prime +\alpha \prime \prime }$, say $g\prime \prime$. We may now repeat the same reasoning with the group $\frac{G}{g\prime \prime }$ of order $p^{\alpha -\alpha \prime -\alpha \prime \prime }m$. In this way we must in any case be ultimately led to a sub-group of $G$ whose order is divisible by $p^{\alpha }$. The lemma is therefore established.

We have as an immediate inference:—

THEOREM I. If $p^{\alpha }$ is the highest power of a prime $p$ which divides the order of a group, the group contains a sub-group of order $p^{\alpha }$.

For the group contains a sub-group whose order is divisible by $p^{\alpha }$; this sub-group must itself contain a sub-group whose order is divisible by $p^{\alpha }$, and so on. Hence at last a sub-group must be arrived at whose order is $p^{\alpha }$.

Corollary. Since it has been seen in § 53 that a group of order $p^{\alpha }$ contains sub-groups of every order $p^{\beta }$ ($\beta <\alpha$), it follows that, if the order of a group is divisible by $p^{\beta }$, the group must contain a sub-group of order $p^{\beta }$. In particular, a group whose order is divisible by a prime $p$ has operations of order $p$.

78. THEOREM II. If $p^{\alpha }$ is the highest power of a prime $p$ which divides the order of a group $G$, the sub-groups of $G$ of order $p^{\alpha }$ form a single conjugate set, and their number is congruent to unity, .

If $H$ is a sub-group of $G$ of order $p^{\alpha }$, the only operations of $G$, which are permutable with $H$ and have powers of $p$ for their orders, are the operations of $H$ itself. For if $P$ is an operation of order $p^r$, not contained in $H$ and permutable with it, and if $p^s$ is the order of the greatest group common to  and $H$, the order of is $p^{\alpha +r+s}$. But $G$ can have no such sub-group, since $p^{\alpha +r-s}$ is not a factor of the order of $G$.

Suppose now that $H\prime$ is any sub-group conjugate to $H$; and let $p^{\beta }$ be the order of the group $h$ common to $H$ and $H\prime$. When $H\prime$ is transformed by all the operations of $H$, the operations of $h$ are the only ones which transform $H\prime$ into itself. Hence the operations of $H$ can be divided into $p^{\alpha -\beta }$ sets of $p^{\beta }$ each, such that the operations of each set transform $H\prime$ into a distinct sub-group. In this way, $p^{\alpha -\beta }$ sub-groups are obtained distinct from each other and from $H$ and conjugate to $H$. If these sub-groups do not exhaust the set of sub-groups conjugate to $H$, let $H\prime \prime$ be a new one. From $H\prime \prime$ another set of $p^{\alpha -\beta \prime }$ sub-groups can be formed, distinct from each other and from $H$ and conjugate to $H$. Moreover no sub-group of this latter set can coincide with one of the previous set. For if

If now $G$ contains another sub-group $H_1$, of order $p^{\alpha }$, it must belong to a different conjugate set. The number of sub-groups in the new set may be shewn, as above, to be congruent to unity, . But on transforming $H_1$ by the operations of $H$, a set of $p^{\alpha -\gamma }$ conjugate sub-groups is obtained, where $p^{\gamma }$ is the order of the sub-group common to $H$ and $H_1$. A further sub-group of the set, if it exists, gives rise to $p^{\alpha -\gamma \prime }$ additional conjugate sub-groups, distinct from each other and from the previous $p^{\alpha -\gamma }$. Proceeding thus we shew that the number of sub-groups in the conjugate set is a multiple of $p$; and as it cannot be at once a multiple of $p$ and congruent to unity, , the set does not exist. The sub-groups of order $p^{\alpha }$ therefore form a single conjugate set and their number is congruent to unity, .

Corollary I. If $p^{\alpha }m$ is the order of the greatest group $I$, within which the group $H$ of order $p^{\alpha }$ is contained self-conjugately, the order of the group $G$ must be of the form

Corollary II. The number of groups of order $p^{\alpha }$ contained in $G$, i.e. the factor $1+kp$ in the preceding expression for the order of the group, can be expressed in the form

This follows immediately from the arrangement of the set of groups given in the proof of the theorem. Thus each of the $p^{\alpha -\beta }$ groups, obtained on transforming $H\prime$ by the operations of $H$, has in common with $H$ a greatest common sub-group of order $p^{\beta }$. It may of course happen that any one or more of the numbers $k_1$, $k_2$, …, $k_{\alpha }$ is zero. If no two sub-groups of the set have a common sub-group whose order is greater than $p^r$, then $k_1$, $k_2$, …, $k_{\alpha -r-1}$ all vanish; and the number of sub-groups in the set is congruent to unity, . Conversely, if $p^s$ is the highest power of $p$ that divides $kp$, some two sub-groups of the set must have a common sub-group whose order is not less than $p^{\alpha -s}$; for if there were no such common sub-groups, the number of sub-groups in the set would be congruent to unity, .

Corollary III. Every sub-group of $G$ whose order is $p^{\beta }$, ($\beta <\alpha$), must be contained in one or more sub-groups of order $p^{\alpha }$.

For if the sub-group of order $p^{\beta }$ is contained in no sub-group of order $p^{\beta +1}$, the only operations whose orders are powers of $p$ that transform it into itself are its own. In this case, the preceding method may be used to shew that the number of sub-groups in the conjugate set to which the given sub-group belongs must be congruent to unity, . But this is impossible, as the number of such sub-groups must, on the assumption made, be a multiple of $p^{\alpha -\beta }$. The sub-group of order $p^{\beta }$ is therefore contained in one of order $p^{\beta +1}$, and hence repeating the same reasoning in one of order $p^{\alpha }$.

79. We shall refer to Theorems I and II together as Sylow’s theorem. In discussing in this and the following paragraphs some of the results that follow from Sylow’s theorem, we shall adhere to the notation that has been used in establishing the theorem itself. Thus $p^{\alpha }$ will always denote the highest power of a prime $p$ which divides the order of $G$; the sub-groups of $G$ of order $p^{\alpha }$ will be denoted by $H$, $H_1$, …, and the greatest sub-groups of $G$ that contain these self-conjugately by $I$, $I_1$, …. These latter form a single conjugate set of sub-groups of $G$, whose orders are $p^{\alpha }m$, the order of $G$ itself being . Moreover the number of groups in this conjugate set is $1+kp$.

Suppose now that $S$ is any operation of $G$ whose order is a power of $p$. When the $1+kp$ sub-groups

A sub-group (or operation), having a power of $p$ for its order and occurring in $1+lp$ sub-groups of order $p^{\alpha }$, will not necessarily be one of the same number of conjugate sub-groups (or operations) in each of these $1+lp$ sub-groups. If then $h$ is a sub-group, of order $p^r$, that occurs in $1+lp$ sub-groups of order $p^{\alpha }$, we may choose one of these, say $H$, in which $h$ is one of as small a number as possible of conjugate sub-groups. Let this number be $p^{\alpha -r-s}$, so that, in $H$, $h$ is self-conjugate in a group of order $p^{r+s}$. Then in $G$, the order of the greatest group $i$, in which $h$ is self-conjugate, must be $p^{r+s}n$, where $n$ is relatively prime to $p$. The number of sub-groups of $i$ of order $p^{r+s}$ must be congruent to unity, . No two of these groups of order $p^{r+s}$ can occur in the same group of order $p^{\alpha }$; for if they did, they would generate a group of order $p^{r+s+t}$, ($t\nless 1$), and this is impossible, $p^{r+s}$ being the highest power of $p$ that divides the order of $i$. Moreover the number of groups of order $p^{\alpha }$ in which any one of these groups of order $p^{r+s}$ enters is congruent to unity, . Hence the number of groups of order $p^{\alpha }$, in which $h$ is one of $p^{\alpha -r-s}$ (the least possible number) conjugate sub-groups, is congruent to unity, . Suppose now that, in $H\prime$, $h$ is one of $p^{\alpha -r-s\prime }$ ($s\prime <s$) conjugate sub-groups; so that in $H\prime$ it is self-conjugate in a sub-group of order $p^{r+s\prime }$ and in no greater sub-group. Then the highest power of $p$ that divides the order of the group common to $I\prime$ and $i$ is $p^{r+s\prime }$; and therefore, when $H\prime$ is transformed by all the operations of $i$, the number of groups of order $p^{\alpha }$ formed is a multiple of $p^{s-s\prime }$. In each of these groups, $h$ is one of $p^{\alpha -r-s\prime }$ conjugate sub-groups. If this does not exhaust all the groups of order $p^{\alpha }$ in which $h$ is one of $p^{\alpha -r-s\prime }$ conjugate sub-groups, let $H\prime \prime$ be another. Then from this another set of groups, whose number is a multiple of $p^{s-s\prime }$, may be formed, which are distinct from each other and from the previous set, such that in each of them $h$ is one of $p^{\alpha -r-s\prime }$ conjugate sub-groups. This process may clearly be continued till all such groups are exhausted. Hence the number of groups of order $p^{\alpha }$, in which $h$ is one of $p^{\alpha -r-s\prime }$ ($s\prime <s$) conjugate sub-groups, is a multiple of $p^{s-s\prime }$. If $h$ enters as one of $p^{\alpha -r-s\prime }$ conjugate sub-groups in $H\prime$, a sub-group conjugate to $h$ must enter as one of $p^{\alpha -r-s\prime }$ conjugate sub-groups in $H$. Hence if $p^{r+s}$, $p^{r+s\prime }$, $p^{r+s\prime \prime }$, … are the orders of the greatest sub-groups that contain $h$ self-conjugately in the various sub-groups of order $p^{\alpha }$ in which it appears, then $H$ must contain sub-groups of the conjugate set (in $G$) to which $h$ belongs in conjugate sets of $p^{r+s}$, $p^{r+s\prime }$, $p^{r+s\prime \prime }$, … only. The total number $x$ of such sub-groups contained in $H$ is clearly connected with the total number $1+lp$ of sub-groups of order $p^{\alpha }$, in which $h$ appears, by the relation

It is to be noticed that, with the above notation,

80. THEOREM III. Let $p^{\alpha }$ be the highest power of a prime $p$ which divides the order of a group $G$, and let $H$ be a sub-group of $G$ of order $p^{\alpha }$. Let $h$ be a sub-group common to $H$ and some other sub-group of order $p^{\alpha }$, such that no sub-group, which contains $h$ and is of greater order, is common to any two sub-groups of order $p^{\alpha }$. Then there must be some operation of $G$, of order prime to $p$, which is permutable with $h$ and not with $H$²⁵.

Suppose that $H$ and $H\prime$ are two groups of order $p^{\alpha }$ to which $h$ is common; and let $h_1$ and $h\prime$ be sub-groups of $H$ and $H\prime$, of greater order than $h$, in which $h$ is self-conjugate. If $h_1$ and $h\prime$ generate a group whose order is a power of $p$, it must occur in some group $H\prime \prime$ of order $p^{\alpha }$; and then $H$ and $H\prime \prime$ have a common group $h_1$, which contains $h$ and is of greater order. This is contrary to supposition, and therefore the order of the group generated by $h_1$ and $h\prime$ is not a power of $p$. Hence $h$ is permutable with some operation whose order is prime to $p$.

Let $p^r$ be the order of $h$, and $p^{r+s}n$ be the order of the greatest sub-group $i$ of $G$ that contains $h$ self-conjugately. If $i$ contained a self-conjugate sub-group of order $p^{r+s}$, $h_1$ and $h\prime$ would be sub-groups of it and they would generate a group whose order is a power of $p$. This is not the case, and therefore $i$ must contain $1+k\prime p$ sub-groups of order $p^{r+s}$, so that we may write for $n$; and then, in $i$, a sub-group of order $p^{r+s}$ is self-conjugate in a sub-group of order $p^{r+s}m\prime$. No sub-group of $i$ of order $p^{r+t}$ ($t>0$) can occur in more than one sub-group of order $p^{\alpha }$; and the $1+k\prime p$ sub-groups of $i$ of order $p^{r+s}$ belong therefore to $1+k\prime p$ distinct sub-groups of order $p^{\alpha }$. Moreover $h$ occurs in no sub-groups of order $p^{\alpha }$ other than these $1+k\prime p$. For if $h$ occurred in another sub-group $H_1$, it would in this sub-group be self-conjugate in a group of order $p^{r+s\prime }$ ($s\prime >0$); and this group would occur in $i$. This group would then be common to two sub-groups of order $p^{\alpha }$, contrary to supposition.

An operation of $i$, which transforms one of its sub-groups of order $p^{r+s}$ into another, must transform the sub-group of order $p^{\alpha }$ containing the one into that containing the other. Hence $i$ must contain operations which are not permutable with $H$. The greatest common sub-group of $i$ and $I$ is that sub-group of $i$ of order $p^{r+s}m\prime$ which contains the sub-group of order $p^{r+s}$ belonging to $H$ self-conjugately. For every operation, that transforms this sub-group of order $p^{r+s}$ into itself, must transform $H$ into itself; and no operation can transform $H$ into itself which transforms this sub-group of order $p^{r+s}$ into another.

81. Let $P$ be an operation, or sub-group, which is self-conjugate in $H$; and let $Q$ be another operation, or sub-group, of $H$, which is conjugate to $P$ in $G$, but not conjugate to $P$ in $I$. Suppose first that, if possible, $Q$ is self-conjugate in $H$. There must be an operation $S$ which transforms $P$ into $Q$ and $H$ into some other sub-group $H\prime$, so that

Now in the sub-group which contains $Q$ self-conjugately, the sub-groups of order $p^{\alpha }$ form a single conjugate set, and $H$ must occur among them. Hence this sub-group must contain an operation $T$ such that

or that, contrary to supposition, $P$ and $Q$ are conjugate in $I$. Hence:—

THEOREM IV. Let $G$ and $H$ be defined as in the previous theorem, and let $I$ be the greatest sub-group of $G$ which contains $H$ self-conjugately. Then if $P$ and $Q$ are two self-conjugate operations or sub-groups of $H$, which are not conjugate in $I$, they are not conjugate in $G$.

Corollary. If $H$ is Abelian, no two operations of $H$ which are not conjugate in $I$ can be conjugate in $G$. Hence the number of distinct sets of conjugate operations in $G$, which have powers of $p$ for their orders, is the same as the number of such sets in $I$.

82. Suppose next that $Q$ is not self-conjugate in $H$. Then every operation that transforms $Q$ into $P$ must transform $H$ into a sub-group of order $p^{\alpha }$ in which $P$ is not self-conjugate. Of the sub-groups of order $p^{\alpha }$, to which $P$ belongs and in which $P$ is not self-conjugate, choose $H\prime$ so that, in $H\prime$, $P$ forms one of as small a number of conjugate operations or sub-groups as possible. Let $g$ be the greatest sub-group of $H\prime$ that contains $P$ self-conjugately. Among the sub-groups of order $p^{\alpha }$ that contain $P$ self-conjugately, there must be one or more to which $g$ belongs. Let $H$ be one of these; and suppose that $h$ and $h\prime$ are the greatest sub-groups of $H$ and $H\prime$ respectively that contain $g$ self-conjugately. The orders of both $h$ and $h\prime$ must (Theorem II, § 55) be greater than the order of $g$; and in consequence of the assumption made with respect to $H\prime$, every sub-group, having a power of $p$ for its order and containing $h$, must contain $P$ self-conjugately.

Now consider the sub-group . Since it does not contain $P$ self-conjugately, its order cannot be a power of $p$. Also if $p^{\beta }$ is the highest power of $p$ that divides its order, it must contain more than one sub-group of order $p^{\beta }$. For any sub-group of order $p^{\beta }$, to which $h$ belongs, contains $P$ self-conjugately; and any sub-group of order $p^{\beta }$, to which $h\prime$ belongs, does not. Suppose now that $S$ is an operation of , having its order prime to $p$ and transforming a sub-group of  of order $p^{\beta }$, to which $h$ belongs, into one to which $h\prime$ belongs. Then $S$ cannot be permutable with $P$; for if it were, $P$ would be self-conjugate in each of these sub-groups of order $p^{\beta }$.

When $P$ is an operation, we may reason in the same way with respect to .

Since $g$ is self-conjugate in both $h$ and $h\prime$, $S$ must transform $g$ into itself. Now $P$ is self-conjugate in $g$, and therefore $S^{-r}P{S}^r$ is also self-conjugate in $g$ for all values of $r$. If then $S^t$ is the first power of $S$ which is permutable with $P$, the series of groups $P$, $S^{-1}PS$, …, $S^{-t+1}P{S}^{t-1}$ are all distinct and each is a self-conjugate sub-group of $g$. Every group in this series is therefore permutable with every other. Hence:—

THEOREM V. If $G$ and $H$ are defined as in the two preceding theorems, and if $P$ is a self-conjugate sub-group or operation of $H$, then either (i) $P$ must be self-conjugate in every sub-group of $G$, of order $p^{\alpha }$, in which it enters, or (ii) there must be an operation $S$, of order $q$ prime to $p$, such that the set of sub-groups $$

are all distinct and permutable with each other.

83. In illustration of Sylow’s theorem and its consequences, we will now consider certain special types of group; and we will deal first with a group whose order is the product of two different primes.

If $p_1$ and $p_2$ ($p_1<p_2$) are distinct primes, a group of order $p_1{p}_2$ must, by Sylow’s theorem, have a self-conjugate sub-group of order $p_2$. If $p_2$ is not congruent to unity, , the group must also have a self-conjugate sub-group of order $p_1$. The two self-conjugate sub-groups of orders $p_1$ and $p_2$ can have no common operation except the identical operation; and therefore (Theorem IX, § 34) every operation of one must be permutable with every operation of the other. Hence if $p_2$ is not congruent to unity, , a group of order $p_1{p}_2$ must be Abelian, and therefore also cyclical.

If $p_2$ is congruent to unity, , there may be either $1$ or $p_2$ conjugate sub-groups of order $p_1$.

If there is one, it is self-conjugate and the group is cyclical.

If there are $p_2$ sub-groups of order $p_1$, let $P_1$ denote an operation which generates one of them. Then if $P_2$ is an operation of order $p_2$,

A group of order $p_1{p}_2$, which has $p_2$ conjugate sub-groups of order $p_1$, must therefore, when it exists, be defined by the relations

It follows from § 33 that the order of the group defined by these relations cannot exceed $p_1{p}_2$. But also from the given relations it is clearly impossible to deduce new relations of the form

Again, let $\beta$ be a primitive root of the congruence, distinct from $\alpha$, so that

Then if, in the group defined by

Hence finally, when $p_2$ is congruent to unity, , there is a single type of group of order $p_1{p}_2$, which contains $p_2$ conjugate sub-groups of order $p_1$.

84. We will next deal with the problem of determining all distinct types of group of order $24$.

A group of order $24$ must contain either $1$ or $3$ sub-groups of order $8$, and either $1$ or $4$ sub-groups of order $3$. If it has one sub-group of order $8$ and one sub-group of order $3$, the group must, since each of these sub-groups is self-conjugate, be their direct product. We have seen (§§ 68, 74) that there are five distinct types of group of order $8$; there are therefore five distinct types of group of order $24$, which are obtained by taking the direct product of any group of order $8$ and a group of order $3$.

If there are $3$ groups of order $8$, some two of them must (Theorem II, Cor. II, § 78) have a common sub-group of order $4$; and (Theorem III, § 80) this common sub-group must be a self-conjugate sub-group of the group of order $24$. Moreover if, in this case, a sub-group of order $8$ is Abelian, each operation of the self-conjugate sub-group of order $4$ must (Theorem IV, Cor. § 81) be a self-conjugate operation of the group of order $24$.

With the aid of these general considerations, it now is easy to determine for each type of group of order $8$, the possible types of group of order $24$, in addition to the five types already obtained.

(i) Suppose a group of order $8$ to be cyclical, and let $A$ be an operation that generates it. If  is self-conjugate and $B$ is an operation of order $3$, then

(ii) Next, let a group of order $8$ be an Abelian group defined by

If this is self-conjugate, then, by considerations similar to those of the preceding case, we infer that the group is the direct product of groups of orders $8$ and $3$. Hence there is not in this case a new type.

If the group of order $8$ is not self-conjugate, the self-conjugate group of order $4$ may be either or . In either case, if $C$ is an operation of order $3$, it must be one of two conjugate operations while  is self-conjugate. Hence there are two new types respectively given by

(iii) Let a group of order $8$ be an Abelian group defined by

If it is self-conjugate, and if the group of order $24$ is not the direct product of groups of orders $8$ and $3$, an operation $D$ of order $3$ must transform the $7$ operations of order $2$ among themselves; and it must therefore be permutable with one of them. Now the relations

Now if

Hence there is in this case a single type. It is the direct product of and , where

If the group of order $8$ is not self-conjugate, the self-conjugate group of order $4$ may be taken to be ; and $D$ being an operation of order $3$, there is a single new type given by

(iv) Let a group of order $8$ be a non-Abelian group defined by

If the sub-group of order $8$ is not self-conjugate, the self-conjugate sub-group of order $4$ is cyclical, and each of its operations must be permutable with $C$. Hence again we get a single new type, given by

(v) Lastly, let a sub-group of order $8$ be a non-Abelian group defined by

This contains one cyclical and two non-cyclical sub-groups of order $4$. If it is self-conjugate, the group of order $24$ must therefore be the direct product of groups of orders $8$ and $3$; and there is no new type.

If the sub-group of order $8$ is not self-conjugate, and the self-conjugate sub-group of order $4$ is the cyclical group , then $A$ must be permutable with an operation $C$ of order $3$, and there is a single new type given by

If the self-conjugate sub-group of order $4$ is not cyclical, it may be taken to be . If $C$ is permutable with each operation of this sub-group, there is a single type given by

If $C$ is not permutable with every operation of the self-conjugate sub-group, it must transform $A^2$, $B$, $A^{2}B$ among themselves and we may take

Now is self-conjugate, and therefore $A$ must transform $C$ into another operation of order $3$ contained in this sub-group. Hence

The only values of $x$, $y$, $z$ which are consistent with the previous relation

The last new type is therefore defined by

When $B$ is eliminated between these relations, it will be found that the only independent relations remaining are

It is a good exercise to verify that these form a complete set of defining relations for the group. (Compare Ex. 1, § 35.)

There are therefore, in all, fifteen distinct types of group of order $24$. The last of these is the only type, which has neither a self-conjugate sub-group of order $8$, nor one of order $3$. The reader should satisfy himself, as an exercise, that, in the ten cases where the group is not a direct product of groups of orders $8$ and $3$, the defining relations which we have given are self-consistent. This is of course an essential part of the investigation, and it may, in more complicated cases, involve some little difficulty. We have omitted the verification here, where it is very easy, for the sake of brevity.

It is to be noticed that the last type obtained gives an example, and indeed the simplest possible, of Theorem V, § 82. Thus in of order $8$, $A^2$ is a self-conjugate operation and $B$ is not. In the group of order $24$, the operations $A^2$ and $B$ are conjugate; and $C$ is an operation, of order prime to $2$, such that $A^2$, $C^{-1}{A}^{2}C$, $C^{-2}{A}^2{C}^2$ generate three mutually permutable sub-groups.

A discussion similar to that of the present section (but simpler, since in each case the number of types is smaller), will verify the following table²⁷:—

This table, taken with the results of Chapter V, gives the number of distinct types of groups for all orders less than $32$.

85. As a second example, we will discuss the various distinct types of group of order $60$.

A group of order $60$ must, by Sylow’s theorem, contain either $1$ or $6$ cyclical sub-groups of order $5$.

We will first suppose that a group $G$ of order $60$ contains a single cyclical sub-group of order $5$, which is necessarily self-conjugate. There will then be four operations of order $5$ in $G$; and we may deal with two sub-cases according as these operations are or are not self-conjugate.

(i) Suppose that each operation of order $5$ is self-conjugate. There must then be either $1$ or $3$ sub-groups of order $4$. If there is only one, it must be permutable with an operation of order $3$; and then $G$ contains a sub-group of order $12$. If there are three, it follows, by Theorem II, Cor. II (§ 78), that some pair of them must have a common sub-group of order $2$. But (Theorem III, § 80) this sub-group of order $2$ must be permutable with some operation of order prime to $2$, which is not permutable with a sub-group of order $4$. This operation must be of order $3$; hence in this case also there must be a sub-group of order $12$. Thus then the group is, with either alternative, the direct product of a group of order $5$ and a group of order $12$. Now there are five distinct types of group of order $12$; there are therefore five distinct types of group of order $60$ which contain self-conjugate operations of order $5$.

(ii) Suppose that $S$ is an operation of order $5$ which is not self-conjugate. If $T$ is an operation which is not permutable with $S$, then

It follows that $x$ must be either $2$ or $4$. If $x$ is $2$ for every operation $T$ which is not permutable with $S$, then $S$ and $S^4$ form a complete set of conjugate operations; as also do $S^2$ and $S^3$. If $x$ is $4$ for any operation $T$, the four operations $S$, $S^2$, $S^3$, $S^4$ form a single conjugate set.

First, let $S$ be one of two conjugate operations; it must then be self-conjugate in a sub-group of order $30$, and by Sylow’s theorem this sub-group must contain a single sub-group of order $3$. It will therefore be given by

according as $B$ is or is not a self-conjugate operation; $S$ in either case being permutable with both $A$ and $B$.

If the sub-groups of order $4$ are cyclical, $G$ must contain an operation $A_1$ of order $4$ whose square is $A$; and $A_1$ must transform $S$ into its inverse and into itself. The latter condition clearly cannot be satisfied if the self-conjugate sub-group of order $30$ is of type ($\beta$). Hence we have two types given by

If the sub-groups of order $4$ are not cyclical, $G$ must contain an operation $A\prime$ of order $2$, which is permutable with $A$; and $A\prime$ transforms $S$ into its inverse and  into itself. In this case, if the self-conjugate sub-group is of type ($\alpha$), there are two types given by

If the self-conjugate sub-group is of type ($\beta$), there is a single type in which the last two of the preceding equations are replaced by

Secondly, let the operations of order $5$ form a single conjugate set. The sub-groups of order $4$ must then be cyclical since $G$ contains an operation $T$, such that $T^4$ is the lowest power of $T$ which is permutable with $S$. Also $S$ is permutable in a sub-group of order $15$. This sub-group must be self-conjugate; and therefore $G$ contains a self-conjugate sub-group of order $3$. Let

(iii) Next, suppose that $G$ contains $6$ conjugate sub-groups of order $5$. No operation of order $3$ can be permutable with an operation of order $5$, and therefore by Sylow’s theorem there must be $10$ conjugate sub-groups of order $3$. Hence $G$ contains $24$ operations of order $5$ and $20$ operations of order $3$. If any one operation of order $5$ were permutable with an operation of order $2$, all its powers would be permutable with the same operation, and therefore, since the sub-groups of order $5$ form a single conjugate set, every operation of order $5$ would be permutable with an operation of order $2$. The group would then contain at least $24$ operations of order $10$. This is clearly impossible, since the sum of the numbers of operations of orders $3$, $5$ and $10$ would be greater than the order of the group. Hence the sub-group of order $10$, which contains self-conjugately a sub-group of order $5$, must be of the type

In a similar way, we shew that a sub-group of order $6$, which contains self-conjugately a sub-group of order $3$, is of the type

Since no operation of order $3$ or $5$ is permutable with an operation of order $2$, it follows (Theorem III, § 80) that no two sub-groups of order $4$ can have a common operation other than identity. Hence there must be $5$ sub-groups of order $4$; for if there were $3$ or $15$, some of them would necessarily have common operations. Each sub-group of order $4$ is therefore contained self-conjugately in a sub-group of order $12$. Such a sub-group of order $12$ can contain no self-conjugate operation of order $2$, since $G$ contains no operation of order $6$. Hence the sub-groups of order $4$ are non-cyclical, and the $3$ operations of order $2$ in any sub-group of order $4$ are conjugate operations in the sub-group of order $12$ containing it. This sub-group must therefore be of the type

The $5$ sub-groups of order $4$ contain therefore $15$ distinct operations of order $2$; and these form a conjugate set. We have already seen that the $20$ operations of order $3$ form a conjugate set, and that the $24$ operations of order $5$ form two conjugate sets of $12$ each. Hence the $60$ operations of the group are distributed in $5$ conjugate sets, containing respectively $1$, $12$, $12$, $15$ and $20$ operations. It follows at once (§ 27, p. 111) that the group, when it exists, is simple.

A sub-group of order $12$, the existence of which has been proved, must be one of $5$ conjugate sub-groups; and, since the group is simple, no operation can transform each of these into itself. Hence if the $5$ conjugate sub-groups

Ex. 1. If $p$, $q$, $r$ are distinct primes, shew that a group of order $p^{2}qr$, which contains a self-conjugate operation of order $r$, must be the direct product of two groups of orders $p^{2}q$ and $r$ respectively.

Ex. 2. Shew that there is a single type of group of order $84$ which contains $28$ sub-groups of order $3$; and determine its defining relations.

Ex. 3. If $p^{\alpha }$ ($\alpha >1$) is the highest power of $p$ which divides the order of $G$, and if $1+kp$ be the number of sub-groups of $G$ of order $p^{\alpha }$, shew that (i) if $1+kp<p^2$, (ii) if a group of order $p^x$ is cyclical and $1+kp<p^{\alpha }$, $G$ is composite. (Maillet, Comptes Rendus, CXVIII, (1894), p. 1188.)

86. A remarkable and important extension of Sylow’s theorem has recently been given by Herr Frobenius²⁸. In the theorem, as stated above, $p^{\alpha }$ is the highest power of a prime $p$ that divides the order of a group. Herr Frobenius shews that, if $p^k$ is any power of a prime $p$ that divides the order of a group, the number of sub-groups of order $p^k$ is congruent to unity, . These groups do not however, in general, form a single conjugate set.

For a group whose order is a power of $p$ higher than $p^k$, this result has been already proved in § 61. Suppose now that

Now it is easy to see that $x$ is not less than unity, and that every one of these $p^x$ sub-groups belongs to the second class. For suppose first that $x$ is zero, so that $G_k$ is transformed into itself by every operation of $G_{\alpha }$. Then is a sub-group of $G$ whose order is a power of $p$; and since $G_k$ is not contained in $G_{\alpha }$, the order of this sub-group is not less than $p^{\alpha +1}$. This is impossible, since $p^{\alpha +1}$ is not a factor of the order of $G$. Secondly, if $P^{-1}{G}_{k}P$ were contained in $G_{\alpha }$, $P$ being some operation of $G_{\alpha }$, then $G_k$ would be contained in $P{G}_{\alpha }{P}^{-1}$ or in $G_{\alpha }$, contrary to supposition. If the sub-groups of the second class are not thus exhausted, and if $G\prime$ is a new one, we may on transforming $G\prime$ by the operations of $G_{\alpha }$ form a fresh set of $p^{x\prime }$ sub-groups of the second class which are distinct from each other and from the previous set; and this process may be continued till the second class is exhausted. The number of sub-groups in the second class is therefore a multiple of $p$. Hence:—

THEOREM VI. If $p^k$, where $p$ is prime, divides the order of a group $G$, the number of sub-groups of $G$ of order $p^k$ is congruent to unity, .

87. If $p$ is a prime which divides the order of a group $G$, it immediately follows from the foregoing theorem that the number of operations of $G$, which satisfy the relation

This result has been generalized by Herr Frobenius²⁹ in the following form:—

THEOREM VII. If $n$ is a factor of the order $N$ of a group $G$, the number of operations of $G$, including identity, whose orders are factors of $n$, is a multiple of $n$.

It is easy to verify the truth of this theorem directly for small values of $N$; and we may therefore assume it true for every group whose order is less than that of the given group $G$. Again, when $n$ is equal to $N$, the theorem is obviously true. If then, on the assumption that the theorem is true for all factors of $N$ which are greater than $n$, we shew that it is true for $n$, the general truth of the theorem will follow by induction.

If $p$ is any factor of $\frac{N}{n}$, we assume that the number of operations of $G$ whose orders divide $np$ is a multiple of $np$, and therefore also of $n$; and we have to shew that the number of operations of $G$, whose orders divide $np$ and do not divide $n$, is also a multiple of $n$. Let this set of operations be denoted by $A$. If $p^{\lambda -1}$ is the highest power of $p$ that divides $n$, the order of every operation of the set $A$ must be equal to or be a multiple of $p^{\lambda }$. Let $P$ be one of these operations and $m$ its order. Then, if is the number of integers less than and prime to $m$, the cyclical sub-group  contains  operations of order $m$, and each of these belongs to the set $A$. If these do not exhaust the set, let $P\prime$ of order $m\prime$ be another operation belonging to it. Then no one of the  operations of order $m\prime$, contained in the cyclical sub-group , can be identical with any of the preceding set of  operations, since $P\prime$ itself is not contained in that set; at the same time, the new set of  operations all belong to $A$. This process may be continued till the set $A$ is exhausted. Now $m$, $m\prime$, … are all divisible by $p^{\lambda }$, and therefore , , … are all divisible by . Hence the number of operations in the set $A$ is a multiple of $p^{\lambda -1}$.

If now $n=p^{\lambda -1}s$, where by supposition $s$ is relatively prime to $p$, it remains to shew that the number of operations in the set $A$ is a multiple of $s$. For this purpose, let $P$ be any operation of $G$ of order $p^{\lambda }$; and let those operations of $G$, which are permutable with $P$, form a sub-group $H$ of order $p^{\lambda }r$. The number of operations of $H$ whose orders divide $s$ is the same as the number whose orders divide $t$, where $t$ is the greatest common measure of $r$ and $s$. Now the order of  is less than $N$, and therefore we may assume that the number of operations of  whose orders divide $t$ is a multiple of $t$, say $kt$. $H$ therefore contains $kt$ operations of the form $PT$, where $P$ and $T$ are permutable and the order of $T$ divides $s$. Now $P$ is one of a set of $\frac{N}{p^{\lambda }r}$ conjugate operations in $G$; and corresponding to each of these, there is a similar set of $kt$ operations. Moreover no two of these operations can be identical; for we have seen (§ 16) that, if $m$ and $n$ are relatively prime, an operation of order $mn$ of a group $G$, can be expressed in only one way as the product of two permutable operations of $G$ of orders $m$ and $n$.

The complete set of $\frac{N}{p^{\lambda }r}kt$ operations of the form $PT$ belongs to $A$; if $A$ is not thus exhausted, its remaining operations can be divided into similar sets. Now $N$ is divisible by both $r$ and $s$, and therefore by their least common multiple $\frac{rs}{t}$. Hence $\frac{Nkt}{p^{\lambda }r}$ is divisible by $s$, and therefore also the number of operations in the set $A$ is divisible by $s$. The number of operations in $A$, being divisible both by $p^{\lambda -1}$ and by $s$, is therefore divisible by $n$. Hence finally, the number of operations of $G$, whose orders divide $n$, is a multiple of $n$; and the theorem is proved.

Corollary I. Let $N_n$ be the number of operations of $G$ whose orders divide $n$, and suppose that

This process may be repeated to shew that, for each value of $\beta$ which is not greater than $\alpha$, we have

Moreover it is easy to see that the reasoning holds when $p^{\alpha }$ is the highest power of $p$ that divides the order of $G$, provided that then the sub-groups of $G$, of order $p^{\alpha }$, are cyclical.

Suppose now that

Then we may prove as above, first, that $G$ contains operations of each of the orders $p_1^{{\alpha }_1}$, $p_2^{{\alpha }_2}$, …, $p_{n-1}^{{\alpha }_{n-1}}$; and secondly that, for each value of $r$ from $2$ to $n$,

The equation

Again, since $G$ contains a self-conjugate sub-group of order $p_n^{{\alpha }_n}$ and also operations of order $p_{n-1}^{{\alpha }_{n-1}}$, it must contain a sub-group of order $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$. If it had more than one sub-group of this order, it would contain more than $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$ operations whose orders divide $p_{n-1}^{{\alpha }_{n-1}}{p}_n^{{\alpha }_n}$. But since