Chapter III.
On the Simpler Properties of a Group which are Independent of its Mode of Representation.

I n this chapter we proceed to discuss some of the simplest of the properties of groups of finite order which are independent of their mode of representation. If among the operations of a group $G$ a certain set can be chosen which do not exhaust all the operations of the group $G$, yet which at the same time satisfy all the conditions of § 12 so that they form another group $H$, this group $H$ is called a sub-group of the group $G$. Thus if $S$ be any operation, order $m$, of $G$, the operations

THEOREM I. If $H$ is a sub-group of $G$, the order $n$ of $H$ is a factor of the order $N$ of $G$.

Let

Then the operations

For if

If the $2n$ operations thus obtained do not exhaust all the operations of $G$, let $S_2$ be any operation of $G$ not contained among them.

Then it may be shewn, by repeating the previous reasoning, that the $n$ operations

When $N$ is a prime $p$, the group $G$ can have no sub-group other than one of order unity consisting of the identical operation alone. Every operation $S$ of the group, other than the identical operation, is of order $p$, and the group consists of the operations

23. THEOREM II. The operations common to two groups $G_1$, $G_2$ themselves form a group $g$, whose order is a factor of the orders of $G_1$ and $G_2$.

For if $S$, $T$ are any two operations common to $G_1$ and $G_2$, $ST$ is also common to both groups; and hence the common operations satisfy conditions ($\alpha$) and ($\beta$) of the definition in § 12. But their orders are finite and they must therefore satisfy also condition ($\gamma$), and form a group $g$. Moreover $g$ is a sub-group of both $G_1$ and $G_2$, and therefore by Theorem I its order is a factor of the orders of both these groups.

If $G_1$ and $G_2$ are sub-groups of a third group $G$, then $g$ is also clearly a sub-group of $G$.

The set of operations, that arise by combining in every way the operations of the groups $G_1$ and $G_2$, evidently satisfy the conditions of § 12 and form a group; but this will not necessarily or generally be a group of finite order. If however $G_1$ and $G_2$ are sub-groups of a group $G$ of finite order, the group $g\prime$ that arises from their combination will necessarily be of finite order; it may either coincide with $G$ or be a sub-group of $G$. In either case, the order of $g\prime$ will be a multiple of the orders of $G_1$ and $G_2$.

It is convenient here to explain a notation that enables us to avoid an otherwise rather cumbrous phraseology. Let $S_1$, $S_2$, $S_3$, … be a given set of operations, and $G_1$, $G_2$, … a set of groups. Then the symbol

Thus, for instance, the group $g\prime$ above would be represented by

24. Definition. If $S$ and $T$ are any two operations of a group, the operations $S$ and $T^{-1}ST$ are called conjugate operations; while $T^{-1}ST$ is spoken of as the result of transforming the operation $S$ by $T$.

The two operations $S$ and $T^{-1}ST$ are identical only when $S$ and $T$ are permutable. For if

Two conjugate operations are always of the same order. For

Therefore, if

and conversely, if

The operations $ST$ and $TS$ are always conjugate and therefore of the same order; for

Ex. Shew that the operations $S_1{S}_2\dots S_{n-1}{S}_n$ and $S_r{S}_{r+1}\dots$ $S_n{S}_1\dots S_{r-1}$ are conjugate within the group .

Definition. An operation $S$ of a group $G$, which is identical with all its conjugate operations, is called a self-conjugate operation. Such an operation must evidently be permutable with each of the operations of $G$.

In every group the identical operation is self-conjugate; and in a group, whose operations are all permutable, every operation is self-conjugate. A simple example of a group, which contains self-conjugate operations other than the identical operation, while at the same time its operations are not all self-conjugate, is given by

If all the operations of a group be transformed by a given operation, the set of transformed operations form a group. For if $T_1$ and $T_2$ are any two operations of the group, so that $T_1{T}_2$ is also an operation of the group, then

Now unless $S$ is a self-conjugate operation of $G$, the pairs of operations $T$ and $S^{-1}TS$ will not all be identical when for $T$ the different operations of $G$ are put in succession. Hence the process of transforming all the operations of a group by one of themselves is equivalent to establishing a correspondence between the operations of the group, which exhibits it as simply isomorphic with itself.

Definitions. When $H$ is a sub-group of $G$ and $S$ is any operation of $G$, the groups $H$ and $S^{-1}HS$ are called conjugate sub-groups of $G$.

If $H$ and $S^{-1}HS$ are identical, $S$ is said to be permutable with the sub-group $H$. This does not necessarily involve that $S$ is permutable with each of the operations of $H$.

If $H$ and $S^{-1}HS$ are identical, whatever operation $S$ is of $G$, $H$ is said to be a self-conjugate sub-group of $G$.

A group is called composite or simple, according as it does or does not possess at least one self-conjugate sub-group other than that formed of the identical operation alone.

25. THEOREM III. The operations of a group $G$, which are permutable with a given operation $T$, form a sub-group $H$; and the order of $G$ divided by the order of $H$ is the number of operations conjugate to $T$⁷.

If $R_1$ and $R_2$ are any two operations permutable with $T$, so that

For

Also the $n$ operations thus obtained are the only operations which transform $T$ into $T\prime$; for if

Since

It is convenient to have a symbol to represent the set of operations

THEOREM IV. The operations of a group $G$ which are permutable with a sub-group $H$ form a sub-group $I$, which is either identical with $H$ or contains $H$ as a self-conjugate sub-group. The order of $G$ divided by the order of $I$ is the number of sub-groups conjugate to $H$⁸.

If $S_1$, $S_2$ are any two operations of $G$ which are permutable with $H$, then

If now $\Sigma$ is any operation of $G$ not contained in $I$, it may be shewn, exactly as in the proof of Theorem III, that the operations $I\Sigma$ and no others transform $H$ into a conjugate sub-group $H\prime$ which is not identical with $H$; and therefore that the number of sub-groups in the conjugate set to which $H$ belongs is the order of $G$ divided by the order of $I$.

The operations of $G$ which are permutable with $H\prime$ may also be shewn to form the group ${\Sigma }^{-1}I\Sigma$.

It is perhaps not superfluous to point out that two distinct conjugate sub-groups may have some operations in common with one another.

26. Let $S_1$ be any operation of $G$, and

Similarly, if $H_1$ is any sub-group of $G$, and

No sub-group of $G$ can contain operations belonging to every one of the complete sets of conjugate operations of $G$. For if such a sub-group $H$ existed, the complete set of conjugate sub-groups, to which $H$ belongs, would contain all the operations of $G$. Let $m$ be the order of $H$ and $n$ ($\ge m$) the order of the sub-group $I$ formed of those operations of $G$ which are permutable with $H$. Then $H$ is one of $\frac{N}{n}$ conjugate sub-groups, each of which contains $m$ operations. The identical operation is common to all these sub-groups, and they therefore cannot contain more than

27. If a group contains self-conjugate operations, it must contain self-conjugate sub-groups. For the cyclical sub-group generated by any self-conjugate operation must be self-conjugate. The only exception to this statement is the case of the cyclical groups of prime order. Every operation of such a group is clearly self-conjugate; but since the cyclical sub-group generated by any operation coincides with the group itself, there can be no self-conjugate sub-group⁹.

If every operation of a group is not self-conjugate, or, in other words, if the operations of a group are not all permutable with each other, the totality of the self-conjugate operations forms a self-conjugate sub-group. For, if $S_1$ and $S_2$ are permutable with every operation of the group, so also is $S_1{S}_2$.

THEOREM V. The operations common to a complete set of conjugate sub-groups form a self-conjugate sub-group.

It is an immediate consequence of Theorem II that the operations common to a complete set of conjugate sub-groups form a sub-group. Also the set of conjugate sub-groups, when transformed by any operation of the group, is changed into itself. Hence their common sub-group must be self-conjugate.

It may of course happen that the identical operation is the only one which is common to every sub-group of the set.

Corollary. The operations permutable with each of a complete set of conjugate sub-groups form a self-conjugate sub-group.

For, if the operations permutable with the sub-group $H$ form a sub-group $I$, the operations permutable with every sub-group of the conjugate set to which $H$ belongs are the operations common to every sub-group of the conjugate set to which $I$ belongs.

Further, the operations which are permutable with every operation of a complete set of conjugate sub-groups form a self-conjugate sub-group.

THEOREM VI. If $T_1$, $T_2$, …, $T_r$ are a complete set of conjugate operations of $G$, the group , if it does not coincide with $G$, is a self-conjugate sub-group of $G$; and it is the self-conjugate sub-group of smallest order that contains $T_1$.

Since the operations $T_1$, $T_2$, …, $T_r$ are merely rearranged in a new sequence when the set is transformed by any operation of $G$, it follows that

In exactly the same way it may be shewn that, if

The theorem just proved suggests a process for determining whether any given group is simple or composite. To this end, the groups corresponding to each set of conjugate operations in the group are formed. If any one of them differs from the group itself, it is a self-conjugate sub-group and the group is composite; but if each group so formed coincides with the original group, the latter is simple. If the order of $T_1$ contains more than one prime factor and if $T_1^m$ is of prime order, it is easy to see that the distinct operations of the set $T_1^m$, $T_2^m$, …, $T_r^m$ form a complete set of conjugate operations, and that the group contains the group . Hence practically it is sufficient to form the groups for all conjugate sets of operations whose orders are prime.

With the notation of § 26 (p. 106), the order of any self-conjugate sub-group of $G$ must be of the form $m_{\alpha }+m_{\beta }+\dots$; for if the sub-group contains any given operation, it must contain all the operations conjugate with it. Moreover one at least of the numbers $m_{\alpha }$, $m_{\beta }$, … must be unity, since the sub-group must contain the identical operation. It may happen that the numbers $m_r$ are such that the only factors of $N$ of the form $m_{\alpha }+m_{\beta }+\dots$, one of the $m$’s being unity, are $N$ itself and unity. When this is the case, $G$ is necessarily a simple group. It must not however be inferred that, if $N$ has factors of this form, other than $N$ itself and unity, then $G$ is necessarily composite.

If $G_1$ and $G_2$ are sub-groups of $G$, it has already been seen (§ 23) that the operations common to $G_1$ and $G_2$ form a sub-group $g$ of $G$; and it is now obvious that, when $G_1$ and $G_2$ are self-conjugate sub-groups, so also is $g$. Moreover the group is a self-conjugate sub-group unless it coincides with $G$. For

Again, with the same notation, if $T_1$ is an operation of $G$ not contained in the self-conjugate sub-group $G_1$, and if $T_1$, $T_2$, …, $T_r$ is a complete set of conjugate operations, the group is a self-conjugate sub-group, unless it coincides with $G$.

Definitions. If $G_1$, a self-conjugate sub-group of $G$, is such that the group

If $H$ is a sub-group of $G$, and if, for every operation $S$ of $G$ which does not belong to $H$, the group coincides with $G$, $H$ is said to be a maximum sub-group of $G$.

28. Definition. When a correspondence can be established between the operations of a group $G$ and the operations of a group $G\prime$, whose order is smaller than the order of $G$, such that to each operation $S$ of $G$ there corresponds a single operation $S\prime$ of $G\prime$, while to the operation $S_p{S}_q$ there corresponds the operation $S\prime S\prime$, the group $G$ is said to be multiply isomorphic with the group $G\prime$.

THEOREM VII. If a group $G$ is multiply isomorphic with a group $G\prime$, then (i) the operations of $G$, which correspond to the identical operation of $G\prime$, form a self-conjugate sub-group of $G$; (ii) to each operation of $G\prime$ there correspond the same number of operations of $G$; and (iii) the order of $G$ is a multiple of the order of $G\prime$.

Let

Again, to the operation $T^{-1}{S}_{p}T$ of $G$ corresponds the operation $T{\prime }^{-1}\cdot 1\cdot T\prime$, that is, the identical operation of $G\prime$. Hence, whatever operation of $G$ is taken for $T$,

The sub-group $\Gamma$ of $G$ formed of the operations

Again, if $T$ and $T_1$ are two operations of $G$ which correspond to the operation $T\prime$ of $G\prime$, the operation $T^{-1}{T}_1$ corresponds to the identical operation of $G\prime$, and therefore belongs to $\Gamma$. Hence the operations that correspond to $T\prime$ are all contained in the set $T\Gamma$. The operations of this set are all distinct and equal in number to the order of $\Gamma$. Hence if $n$ is the order of $\Gamma$, to each operation of $G\prime$ there correspond $n$ operations of $G$.

Finally, since to each operation of $G$ there corresponds only one of $G\prime$, while to each operation of $G\prime$ there correspond $n$ of $G$, the order of $G$ is $n$ times the order of $G\prime$.

To any sub-group of $G\prime$ of order $\mu$, there corresponds a sub-group of $G$ of order $\mu n$. For if $T\prime T\prime$ forms one of the set

It should be noticed that no correspondence is thus established between a sub-group of $G$ which does not contain $\Gamma$ and any sub-group of $G\prime$.

29. The relation of multiple isomorphism between two groups can be presented in a manner rather different from that of the last paragraph. Let $G$ be any composite group, and $\Gamma$ a self-conjugate sub-group of $G$ consisting of the operations

Then, as in § 22, the operations of $G$ can be arranged in the scheme

Now $\Gamma$ being a self-conjugate sub-group of $G$, it follows that

If now $G$ and $\Gamma$ coincide with the groups of the preceding paragraph which are represented by the same symbols, then $G\prime$ of the preceding paragraph must be simply isomorphic with the group whose operations are

It follows that a group $G\prime$ with which a group $G$ is multiply isomorphic, in such a way that to the identical operation of $G\prime$ there corresponds a given self-conjugate sub-group $\Gamma$ of $G$, is completely defined (as an abstract group) when $G$ and $\Gamma$ are given. This being so it is natural to use a symbol to denote directly the group thus defined in terms of $G$ and $\Gamma$. Herr Hölder¹⁰ has introduced the symbol

It may not be superfluous to notice that the symbol $\frac{G}{\Gamma }$ has no meaning¹¹, unless $\Gamma$ is a self-conjugate sub-group of $G$. Moreover, it may happen that $G$ has two simply isomorphic self-conjugate sub-groups $\Gamma$ and $\Gamma \prime$. When this is the case, there is no necessary relation between the factor-groups $\frac{G}{\Gamma }$ and $\frac{G}{\Gamma \prime }$ (except of course that their orders are equal); in other words, the type of the factor-group $\frac{G}{\Gamma }$ depends on the actual self-conjugate sub-group of $G$ which is chosen for $\Gamma$ and not merely on the type of $\Gamma$.

Further, though in relation to its definition by means of $G$ and $\Gamma$ we call $\frac{G}{\Gamma }$ a factor-group of $G$, we may without ambiguity, since the symbol represents a group of definite type, omit the word factor and speak of the group $\frac{G}{\Gamma }$.

It is also to be observed that $G$ has not necessarily a sub-group simply isomorphic with $\frac{G}{\Gamma }$. This may or may not be the case.

30. If $G$ is multiply isomorphic with $G\prime$ so that the self-conjugate sub-group $\Gamma$ of $G$ corresponds to the identical operation of $G\prime$, it was shewn, at the end of § 28, that to any self-conjugate sub-group of $G\prime$ there corresponds a self-conjugate sub-group of $G$ containing $\Gamma$. Hence, unless $\frac{G}{\Gamma }$ is a simple group, $\Gamma$ cannot be a maximum self-conjugate sub-group of $G$. If $g_1$ is any self-conjugate sub-group of $\frac{G}{\Gamma }$, and $G_1$ the corresponding (necessarily self-conjugate) sub-group of $G$, containing $\Gamma$, we may form the factor-group $\frac{G}{G_1}$, and determine again whether this group is simple or composite. By continuing this process a maximum self-conjugate sub-group of $G$, containing $\Gamma$, must at last be reached.

31. Though $\frac{G}{\Gamma }$ is completely defined by $G$ and $\Gamma$, where $\Gamma$ is any given self-conjugate sub-group of $G$, the reader will easily verify that $G$ is not in general determined when $\Gamma$ and $\frac{G}{\Gamma }$ are given.

We shall have in the sequel to consider the solution of this problem in various particular cases. There is, however, in every case one solution of it which is immediately obvious. We may take any two groups $G_1$ and $G_2$, simply isomorphic with the given groups $\Gamma$ and $\frac{G}{\Gamma }$, such that $G_1$ and $G_2$ have no common operation except identity, while each operation of one is permutable with each operation of the other. The group , formed by combining these two, is clearly such that is simply isomorphic with $\frac{G}{\Gamma }$; it therefore gives a solution of the problem.

Definition. If two groups $G_1$, $G_2$ have no common operation except identity, and if each operation of $G_1$ is permutable with each operation of $G_2$, the group is called the direct product of $G_1$ and $G_2$.

Ex. If $H$, $h$ are self-conjugate sub-groups of $G$, and if $h$ is contained in $H$, so that $\frac{H}{h}$ is a self-conjugate sub-group of $\frac{G}{h}$, shew that the quotient of $\frac{G}{h}$ by $\frac{H}{h}$ is simply isomorphic with $\frac{G}{H}$.

32. If $H$ is a self-conjugate sub-group of $G$, of order $n$, and if $H\prime$ is a self-conjugate sub-group of $G\prime$, of order $n\prime$, and if $\frac{G}{H}$ and $\frac{G\prime }{H\prime }$ are simply isomorphic, a correspondence of the most general kind may be established between the operations of $G$ and $G\prime$. To every operation of $G$ (or $G\prime$) there will correspond $n\prime$ (or $n$) operations of $G\prime$ (or $G$), in such a way that to the product of any two operations of $G$ (or $G\prime$) there corresponds a definite set of $n\prime$ (or $n$) operations of $G\prime$ (or $G$). Let

For, if $h\prime$and $h\prime$ are any two operations of $H\prime$, the set of operations $S\prime h\prime S\prime h\prime$includes $n\prime$ distinct operations only, namely those of the set $S\prime S\prime H\prime$. Hence to the product of any given operation of the set $S_{r}H$ by any given operation of the set $S_{s}H$, there corresponds the set of $n\prime$ operations $S\prime S\prime H\prime$; at the same time the product of the two given operations belongs $($in consequence of the isomorphism between $\frac{G}{H}$ and $\frac{G\prime }{H\prime }$$)$ to the set $S_r{S}_{s}H$. The same statements clearly hold when we interchange accented and unaccented symbols.

We still speak of $G$ and $G\prime$ as isomorphic groups, and the correspondence between their operations is said to give an $n$-to-$n\prime$ isomorphism of the two groups. We shall return to this general form of isomorphism in dealing with intransitive substitution groups.

33. Definition. Two groups $G$ and $G\prime$ are said to be permutable with each other when the distinct operations of the set $S_{i}S\prime$, where for $S_i$ every operation of the group $G$ is put in turn and for $S\prime$ every operation of the group $G\prime$, coincide with the distinct operations of the set $S\prime S_i$ except possibly as regards arrangement.

If the two groups $G$ and $G\prime$ are permutable, the group must be of finite order. For, by the definition, every operation

Then every operation of the set $S_{i}S\prime$ can clearly be expressed in the form

If every operation of $G$ is permutable with $G\prime$, then $g$ must be a self-conjugate sub-group of $G$. For $G$ and $G\prime$ are transformed, each into itself, by any operation of $G$; and therefore their common sub-group $g$ must be transformed into itself by any operation of $G$.

Moreover those operations of $G$, which are permutable with every operation of $G\prime$, form a self-conjugate sub-group of $G$. For if $T$ is an operation of $G$, which is permutable with every operation $S\prime$ of $G\prime$, so that

If $G$ is a simple group, $g$ must consist of the identical operation only; and either all the operations of $G$, or none of them, must be permutable with every operation of $G\prime$.

A special case is that in which the two groups $G\prime$ and $G$ are respectively a self-conjugate sub-group $\Gamma$ and any sub-group $H$ of some third group; for then every operation of $H$ is permutable with $\Gamma$. If $H$ is a cyclical sub-group generated by an operation $S$ of order $n$, and if $S^m$ is the lowest power of $S$ which occurs in $\Gamma$, then $m$ must be a factor of $n$. For if $m\prime$ is the greatest common factor of $m$ and $n$, integers $x$ and $y$ can be found such that

Now

THEOREM VIII. If an operation $S$, of order $n$, is permutable with a group $\Gamma$, and if $S^m$ is the lowest power of $S$ which occurs in $\Gamma$; then $m$ is a factor of $n$, and $\frac{n}{m}$ is a factor of the order of $\Gamma$.

The operations of  can clearly be distributed in the sets

34. A still more special case, but it is most important, is that in which the two groups are both of them self-conjugate sub-groups of some third group. If in this case the two groups are $G$ and $H$, while $S$ and $T$ are any operations of the two groups respectively, then

Consider now the operation $S^{-1}{T}^{-1}ST$. Regarded as the product of $S^{-1}$ and $T^{-1}ST$ it belongs to $G$, and regarded as the product of $S^{-1}{T}^{-1}S$ and $T$ it belongs to $H$. Every operation of this form therefore belongs to the common group of $G$ and $H$. If $G$ and $H$ have no common operation except identity, then

THEOREM IX¹². If every operation of $G$ transforms $H$ into itself and every operation of $H$ transforms $G$ into itself, and if $G$ and $H$ have no common operation except identity; then every operation of $G$ is permutable with every operation of $H$.

Corollary. If every operation of $G$ transforms $H$ into itself and every operation of $H$ transforms $G$ into itself, and if either $G$ or $H$ is a simple group; then $G$ and $H$ have no common operation except identity, and every operation of $G$ is permutable with every operation of $H$.

For, by § 33, if $G$ and $H$ had a common sub-group, it would be a self-conjugate sub-group of both of them; and neither of them could then be simple, contrary to hypothesis. Consequently, the only sub-group common to $G$ and $H$ is the identical operation.

35. If $N$ and $N\prime$are the orders of two permutable groups $G$ and $G\prime$, the $NN\prime$ equations expressing every operation of the form $SS\prime$, where $S$ belongs to $G$ and $S\prime$ to $G\prime$, in the form $\Sigma \prime \Sigma$, where $\Sigma$ and $\Sigma \prime$ belong to $G$ and $G\prime$ respectively, are never all independent. In particular, if

When, in fact, these conditions are satisfied, it is always possible by a series of steps to express any operation $S\prime S$ in the form $\Sigma \Sigma \prime$.

Ex. 1. Shew that, in the group whose defining relations are

Ex. 2. Shew that the cyclical group generated by the substitution is permutable with the group

Ex. 3. If $g_1$ and $g_2$ are the orders of the groups $G_1$ and $G_2$, $\gamma$ the order of their greatest common sub-group and $g$ the order of , shew that

Ex. 4. If $G_1$ and $G_2$ are two sub-groups of $G$ of orders $g_1$ and $g_2$, and $S$ any operation of $G$, prove that the number of distinct operations of $G$ contained in the set $S_{1}S{S}_2$, when for $S_1$ and $S_2$ are put in turn every pair of operations of $G_1$ and $G_2$ respectively, is $\frac{g_1{g}_2}{\gamma }$; $\gamma$ being the order of the greatest sub-group common to $S^{-1}{G}_{1}S$ and $G_2$.

If $T$ is any other operation of $G$, shew also that the sets $S_{1}S{S}_2$ and $S_{1}T{S}_2$ are either identical or have no operation in common. (Frobenius.)

Ex. 5. If a group $G$ of order $mn$ has a sub-group $H$ of order $n$, and if $n$ has no prime factor which is less than $m$, shew that $H$ must be a self-conjugate sub-group. (Frobenius.)