Chapter IX.
On Substitution Groups: Primitive and Imprimitive Groups.

W e have seen that the symbols permuted by the operations of an intransitive substitution group may be divided into sets, such that every substitution of the group permutes the symbols of each set among themselves. For a transitive group the symbols must, from this point of view, be regarded as forming a single set. It may however in particular cases be possible to divide the symbols permuted by a transitive group into sets in such a way, that every substitution of the group either interchanges the symbols of any set among themselves or else changes them all into the symbols of some other set. That this may be possible, it is clearly necessary that each set shall contain the same number of symbols.

In the present Chapter we shall discuss those properties of transitive groups which depend on their possessing or not possessing the property indicated.

121. Definition. When the symbols operated on by a transitive substitution group can be divided into sets, each set containing the same number of distinct symbols and no symbol occurring in two different sets, and when the sets are such that all the symbols of any set are either interchanged among themselves or changed into the symbols of another set by every substitution of the group, the group is called imprimitive. When no such division into sets is possible, the group is called primitive. The sets of symbols which are interchanged by an imprimitive group are called imprimitive systems.

A simple example of an imprimitive group is given by group VII of § 17. An examination of the substitutions of this group will shew that they all either transform the systems of symbols $xyz$ and $abc$ into themselves or else interchange them, and that the same is true of the systems $xa$, $yb$, $zc$; so that, in this case, the symbols may be divided into two distinct sets of imprimitive systems.

It follows at once, from the definition, that an imprimitive group cannot be more than simply transitive. For if it were doubly transitive, it would contain substitutions changing any two symbols into any other two, and of these the first pair might be chosen from the same imprimitive system and the second pair from distinct systems.

The question as to whether a given group can be expressed as a transitive group of given degree, and the further question as to whether such a representation of the group, when possible, is imprimitive or primitive, finds its complete solution in the following investigation due to Herr Dyck⁴⁹.

122. In § 20 it was shewn how any group $G$ of order $N$ could be represented as a substitution group of $N$ symbols. This form of the group, defined as the regular form, is simply transitive; for all its substitutions except identity displace all the symbols, and therefore there must be just one substitution changing a given symbol into any other. Let us now suppose that $N=\mu \nu$, and that $G$ has a sub-group $H$ of order $\mu$, consisting of the operations

The tableau representing the group as a substitution group of $N$ symbols will, in terms of these symbols, take the form given on the following page. Every symbol in this tableau is of the form

Since

Hence the substitution of $G$, represented by the line beginning with $S_{m\prime }{T}_{n\prime }$, changes the set of symbols $S_p{T}_n$ into the set $S_p{T}_k$ in some sequence or other.

Every substitution of the group therefore changes the symbols of each of the $\nu$ sets, into which the first line of the tableau is divided, either into themselves or into the symbols of some other of the $\nu$ sets. Hence:—

THEOREM I. If a group of order $\mu \nu$ contains a sub-group of order $\mu$, the regular form of the group will be imprimitive in such a way that the $\mu \nu$ symbols may be divided into $\nu$ imprimitive systems of $\mu$ symbols each.

The converse of this theorem is also true. For if the $\mu \nu$ symbols

by whose permutations the group can be expressed in regular form, are divisible into $\nu$ imprimitive systems of $\mu$ each, let

123. Let us now represent the imprimitive system

This group is isomorphic with the original group $G$; and if no substitution of the original group interchanges among themselves the symbols of each imprimitive system, the isomorphism must be simple. Now a substitution of $G$, which does not change any imprimitive system into another, must, if it exists, be a substitution $S\prime$ of the sub-group $H$, which is constituted by

THEOREM II. If $H$, of order $\mu$, is a sub-group of $G$ of order $\mu \nu$, and if no self-conjugate sub-group of $G$ is contained in $H$, then $G$ can be expressed as a transitive group of degree $\nu$.

That the converse of this theorem is true is immediately obvious.

124. If the sub-group $H$ of $G$ is contained in a greater sub-group $K$ of order $\mu \nu \prime$, where $\nu =\nu \prime \nu \prime \prime$, the operations of $K$ consist of the sets

The regular form of the group has $\nu \prime \nu \prime \prime$ imprimitive systems corresponding to the sub-group $H$, and $\nu \prime \prime$ imprimitive systems corresponding to the sub-group $K$; the above method of representing $K$ shews that each of the latter systems contain $\nu \prime$ complete systems of the former set.

Now it has just been proved that, if $H$ contains no self-conjugate sub-group of $G$, the group can be represented as a transitive substitution group of the symbols

But from the division of the symbols in the regular form of the group into $\nu \prime \prime$ imprimitive systems, it follows that the set of symbols

THEOREM III. If a group $G$ of order $\mu \nu \prime \nu \prime \prime$ has a sub-group $H$ of order $\mu$, which contains no self-conjugate sub-group of $G$; and if $H$ is contained in a sub-group $K$ of $G$ of order $\mu \nu \prime$; then the representation of $G$ as a transitive group of degree $\nu \prime \nu \prime \prime$ $inrespectofH$ is imprimitive, the $\nu \prime \nu \prime \prime$ symbols being divisible into $\nu \prime \prime$ systems of $\nu \prime$ symbols each.

Corollary I. A transitive group of order $\mu \nu$ and degree $\nu$ will be primitive if, and only if, a sub-group of order $\mu$ that keeps one symbol fixed is a maximum sub-group.

Corollary II. A group, which contains other self-conjugate operations besides identity, cannot be represented in primitive form.

For if a sub-group $H$ of order $\mu$ contains a self-conjugate operation, the group (of order $\mu \nu$) cannot be represented as a transitive group of degree $\nu$ in respect of $H$; and if $H$ contains none of the self-conjugate operations, and is not a self-conjugate sub-group, it cannot be a maximum sub-group.

In particular, a group whose order is the power of a prime cannot be represented as a primitive group.

Corollary III. An Abelian group when represented as a transitive substitution group, must be in regular form.

Corollary IV. A simple group can always be represented in primitive form.

125. Every possible representation of a group as a transitive substitution group is given by the method of the preceding paragraphs. There is another method of dealing with the same problem which we may shortly consider here in view of its utility in many special cases, though it does not in general lead to all possible modes of representation. Let

If $S$ is any operation of $G$, then

Moreover, since there are operations of $G$ which transform $H_1$ into each of the other sub-groups (or operations) of the conjugate set, the substitution group is transitive in the $\nu$ symbols. The substitution group will be simply isomorphic with $G$ if, and only if, there is no operation of $G$ which transforms each of the $\nu$ sub-groups into itself. Now the only operations of $G$ which transform $H_1$ into itself are the operations of $I_1$; and hence the substitution group will be simply isomorphic with $G$ if, and only if, the conjugate set of sub-groups

It has been seen (§ 123) that, when this condition is satisfied, $G$ can be represented as a transitive substitution group whose degree is $\nu$, the ratio of the orders of $G$ and $I_1$. That the form there obtained is identical with the form obtained in the present paragraph may be easily verified. Thus in the earlier form, the substitution corresponding to $S$ is

In the present mode of representation, the substitution corresponding to $S$ can be written in the form

It should be noticed that, if $G$ contains self-conjugate operations other than identity, these operations necessarily occur in $I_1$; and therefore in such a case the present method cannot lead to a substitution group which is simply isomorphic with $G$. In any case, if $K$ is the greatest self-conjugate sub-group of $G$ contained in $I_1$, the substitution group is simply isomorphic with $\frac{G}{K}$.

126. As an illustration of the preceding paragraphs, we will determine the different modes in which the alternating group of degree $5$ can be represented as a transitive group.

The only cyclical sub-groups contained in $G_{60}$, the alternating group of degree $5$, are groups of orders $2$, $3$ and $5$; and of each of these cyclical sub-groups there is a single conjugate set.

The non-cyclical sub-groups may be determined as follows. The lowest possible order for such a sub-group is $4$; since this is the highest power of $2$ that divides $60$, there is a single conjugate set of sub-groups of order $4$. The next lowest possible order is $6$. Now no operation of order $3$ is permutable with an operation of order $2$, as the group contains no operations of order $6$; on the other hand, every sub-group of order $3$ is permutable with an operation of order $2$; thus

Hence finally, since each of the sub-groups leads to a transitive representation of the group, $G_{60}$ can be represented as a transitive substitution group in $30$, $20$, $15$, $12$, $10$, $6$ and $5$ symbols, and in one distinct form in each case. The second method, as given in § 125, does not lead to all these modes of representation. The group will be found to contain $15$ conjugate operations (or sub-groups) of order $2$: $10$ conjugate sub-groups and $20$ conjugate operations of order $3$: $5$ conjugate sub-groups of order $4$: $6$ conjugate sub-groups and two sets of $12$ conjugate operations, each of order $5$: $10$ conjugate sub-groups of order $6$: $6$ conjugate sub-groups of order $10$: and $5$ conjugate sub-groups of order $12$. Hence, by using the second method, the representation of the group as transitive in $30$ symbols would be missed.

Since a sub-group of order $2$ is contained in sub-groups of orders $4$, $6$, $10$ and $12$, the $30$ symbols permuted by $G_{60}$, when it is expressed as a transitive group of degree $30$, can be divided into sets of imprimitive systems, containing respectively $2$, $3$, $5$ and $6$ symbols each. Similarly, when $G_{60}$ is represented as a transitive group of degree $20$, $15$ or $12$, it is imprimitive. When expressed as a group of order $10$, $6$ or $5$, it is primitive.

127. As a further illustration, we shall determine all the distinct forms of imprimitive groups of degree $6$. Let $G$ be such a group, and $H$ that sub-group of $G$ which interchanges among themselves the symbols of each imprimitive system.

We will first suppose that there are two systems of three symbols each, viz. 

In this case, $H$ cannot consist of the identical operation only; for there must be a substitution changing $1$ into $2$, and this must permute $1$, $2$ and $3$ among themselves, and therefore also $4$, $5$ and $6$ among themselves.

Let $H$ contain substitutions which leave $4$, $5$ and $6$ unchanged. These must (§ 114) form a self-conjugate sub-group of $H$, which will be either

Since $G$ has substitutions interchanging the systems, $H$ must similarly contain

In the first alternative, $H$ is the group

In the second alternative, $H$ contains the self-conjugate sub-group

Now if $H$ is of order $2^2\cdot 3^2$, it is necessarily of the form (i). If it is of order $2\cdot 3^2$, it must contain a substitution of order $2$ which transforms the sub-group just given into itself. This may be taken, without loss of generality, to be ; and then $H$ is the group

If $H$ is of order $3^2$, it is the group

Next, let $H$ contain no substitutions which leave $4$, $5$ and $6$ unchanged. Then $H$ is simply isomorphic with a group of degree $3$, and therefore it must be of the form

Now $\frac{G}{H}$ is of order $2$; therefore $G$ must have a substitution of order $2$ or $4$, which interchanges the systems. If the order of $H$ is odd, this substitution must be of order $2$. When the substitution is of order $2$, we may, without loss of generality, take it to be or . If $H$ is of the form (i), (ii), or (iv), we get for $G$, in each case, the same group whichever of these substitutions we take. When $H$ is of the form (iii), we get two groups which are easily seen to be conjugate in the symmetric group. These we do not regard as distinct. When $H$ is of the form (v), we get two distinct groups, one of which is simply isomorphic with the symmetric group of three symbols, while the other is a cyclical group. In these two latter cases, the symbols can be divided into three imprimitive systems of two each.

If the substitution which interchanges the systems is of order $4$, its square must occur in $H$; we may therefore take it to be . When $H$ is of form (i), this gives the same group as before; but when $H$ is of either of the forms (ii) or (iv), we get new forms for $G$. There are therefore eight distinct forms of groups of degree $6$, in which the symbols form two imprimitive systems of three symbols each.

Secondly, suppose that there are three systems of symbols, containing two each, viz. 

The self-conjugate sub-group $H$ is of order $2^3$, $2^2$, $2$ or $1$. Corresponding to the first three cases, the forms of $H$ are easily seen to be

Again, since $\frac{G}{H}$ interchanges the three systems, it must be simply isomorphic with a group of degree $3$, and its order is therefore either $3$ or $6$. First, let its order be $3$. It must then contain a substitution of order $3$ which, without loss of generality, may be taken to be ; this gives, with the three above forms of $H$, three distinct forms for $G$. The form of $G$ corresponding to the form (iii) of $H$ is, however, the same as one of those already determined.

If $\frac{G}{H}$ is of order $6$, $G$ must contain as a self-conjugate sub-group one of the three groups just obtained. Also if $\frac{G}{H}$ were cyclical, there would be a substitution in $G$, not belonging to $H$ and permutable with . This is clearly impossible, and therefore $G$ must contain a substitution which transforms into its inverse. We may take this to be or . With the form (i) for $H$, these two substitutions lead to the same group. When $H$ is of the form (ii), they give two distinct forms for $G$. When $H$ is of the form (iii), $G$ admits the imprimitive systems $1$, $3$, $5$, and $2$, $4$, $6$.

Lastly, if $H$ is the identical operation, $G$ is necessarily of order $6$; no new forms can arise.

There are therefore five distinct forms of groups of degree $6$, in which the symbols form three imprimitive systems of two symbols each but do not at the same time form two imprimitive systems of three symbols each.

128. An actual test to determine whether any transitive group is primitive or imprimitive may be applied as follows. Consider the effect of the substitutions of the group $G$ on $r$ of the symbols which are permuted transitively by it. Those substitutions, which permute the $r$ symbols, say

Suppose now, if possible, that the group contains a substitution $S\prime$, which changes some of the $a$’s into $b$’s, and the remainder into new symbols. We may assume that $S\prime$ changes $a_1$ into $b_1$, and $a_2$ into a new symbol $c_2$. Among the set $HS$ there is at least one substitution, $T$, which changes $a_1$ into $b_1$. Hence $TS{\prime }^{-1}$ changes $a_1$ into itself and some new symbol into $a_2$. This however contradicts the supposition that every substitution, which changes $a_1$ into one of the set of $a$’s, belongs to $H$. Hence no substitution such as $S\prime$ can belong to $G$; and every substitution, which changes one of the $a$’s into one of the $b$’s, must change all the $a$’s into $b$’s.

If the substitutions of the group are not thus exhausted, there must be another set of $r$ symbols

THEOREM IV. If, among the symbols permuted by a transitive group, it is possible to choose a set such that every substitution of the group, which changes a chosen symbol of the set either into itself or into another of the set, permutes all the symbols of the set among themselves; then the group is imprimitive, and the set of symbols forms an imprimitive system.

Corollary I. If $a_1$, $a_2$, …, $a_r$ are a part of the symbols permuted by a primitive group, there must be substitutions of the group, which replace some of this set of symbols by others of the set, and the remainder by symbols not belonging to the set.

Corollary II. A sub-group of a primitive group, which keeps one symbol unchanged, must contain substitutions which displace any other symbol.

If the sub-group $H$, that leaves $a_1$ unchanged, leaves every symbol of the set $a_1$, $a_2$, …, $a_r$ unchanged, then $H$ must be transformed into itself by every substitution which changes any one of these symbols into any other. Every substitution, which changes one of the set into another, must therefore permute the set among themselves; and the group, contrary to supposition, is imprimitive.

129. It has already been seen that, in particular cases, it may be possible to distribute the symbols, which are permuted by an imprimitive group, into imprimitive systems in more than one way. When this is possible, suppose that two systems which contain $a_1$ are

and that the symbols common to the two systems are

A substitution of the group, which changes $a_1$ into $a\prime$, must change $a_1$, $a_2$, …, $a_r$ into $r$ symbols of that system of the first set which contains $a\prime$, while it changes the system of the second set that contains $a_1$ into itself. Hence the latter system contains at least $r$ symbols of that system of the first set in which $a\prime$ occurs. By considering the effect of the inverse substitution, it is clear that the system

Suppose now that $b_1$ is any symbol which is not contained in either of the above systems. A substitution that changes $a_1$ into $b_1$ must change the two systems into two others, which have $r$ symbols

Further, from the mode in which the set $b_1$, $b_2$, …, $b_r$ has been obtained, any operation, which changes one of the symbols $a_1$, $a_2$, …, $a_r$ into one of the symbols $b_1$, $b_2$, …, $b_r$, must change all the symbols of the first set into those of the second. Hence the symbols operated on by the group can be divided into systems of $r$ each, by taking together the sets of $r$ symbols which are common to the various pairs of the two given sets of imprimitive systems; and the group is imprimitive in regard to this new set of systems of $r$ symbols each. Hence:—

THEOREM V. If the symbols permuted by a transitive group can be divided into imprimitive systems in two distinct ways, $m$ being the number of symbols in each system of one set and $n$ in each system of the other; and if some system of the first set has $r$ symbols in common with some system of the second set; then (i) $r$ is a factor of both $m$ and $n$, and (ii) the symbols can be divided into a set of systems of $r$ each, in respect of which the group is imprimitive.

This result may also be regarded as an immediate consequence of Theorem III, § 124. For if $H_1$ and $H_2$ are two sub-groups of $G$ each of which contains the sub-group $R$, and if $K$ is the greatest common sub-group of $H_1$ and $H_2$, then $K$ contains $R$. Now if $R$ contains no self-conjugate sub-group of $G$, then $G$ can be represented as a transitive group whose degree is the order of $G$ divided by the order of $R$. If the respective orders of $H_1$ and $H_2$ are $m$ times and $n$ times the order of $R$, the symbols can be divided in two distinct ways into sets of imprimitive systems, the systems containing $m$ and $n$ symbols respectively. Also, if the order of $K$ is $r$ times the order of $R$, then $r$ is a factor of $m$ and of $n$; by considering the sub-group $K$, the symbols may be divided into a set of systems which contain $r$ symbols each.

It might be expected that, just as we can form a new set of imprimitive systems by taking together the symbols which are common to pairs of systems of two given sets, so we might form another new set of systems by combining all the systems of one set which have any symbols in common with a single system of the other set. A very cursory consideration will shew however that this is not in general the case. In fact, it is sufficient to point out that, with the notation already used, the number of symbols in such a new system would be $\frac{mn}{r}$; and this number is not necessarily a factor of the degree of the group. Also, even if this number is a factor of the degree of the group, it will not in general be the case that the symbols so grouped together form an imprimitive system.

130. We may now discuss, more fully than was possible in § 106, the form of a self-conjugate sub-group of a given transitive group. Such a sub-group must clearly contain one or more operations displacing every symbol operated on by the group. For if every operation of the sub-group keeps the symbol $a_1$ unchanged, then since it is self-conjugate, every operation will keep $a_2$, $a_3$, …, unchanged: and the sub-group must reduce to the identical operation only.

Suppose now, if possible, that $H$ is an intransitive self-conjugate sub-group of a transitive group $G$; and that $H$ permutes the $n$ symbols of $G$ in the separate transitive sets $a_1$, $a_2$, …, $a_{n_1}$; $b_1$, $b_2$, …, $b_{n_2}$; …. If $S$ is any operation of $G$ which changes $a_1$ into $b_1$, then, since

Moreover every operation of $G$, since it transforms $H$ into itself, must either permute the symbols of any set among themselves, or it must change them all into the symbols of some other set. Hence $G$ must be imprimitive, and $H$ must consist of those operations of $G$ which permute the symbols of each imprimitive system among themselves.

Conversely, when $G$ is imprimitive, it is immediately obvious that those operations of $G$, if any such exist, which permute the symbols of each of a set of imprimitive systems among themselves, form a self-conjugate sub-group. Hence:—

THEOREM VI. A self-conjugate sub-group of a primitive group must be transitive; and if an imprimitive group has an intransitive self-conjugate sub-group, it must consist of the operations which permute among themselves the symbols of each of a set of imprimitive systems.

If $G$ is an imprimitive group of degree $mn$, and if there are $n$ imprimitive systems of $m$ symbols each, then we have seen in § 123 that $G$ is isomorphic with a group $G\prime$ of degree $n$. In particular instances, it may at once be evident, from the order of $G$, that this isomorphism cannot be simple. For example, if the order of $G$ has a factor which does not divide $n!$, this is certainly the case: and more generally, if it is known independently that $G$ cannot be expressed as a transitive group of degree $n$, then $G$ must certainly be multiply isomorphic with $G\prime$. In such instances the self-conjugate sub-group of $G$, which corresponds to the identical operation of $G\prime$, is that intransitive self-conjugate sub-group, which interchanges among themselves the symbols of each imprimitive system.

If $G$ is soluble, a minimum self-conjugate sub-group of $G$ must have for its order a power of a prime. Also, if $G$ has an intransitive self-conjugate sub-group, it must have an intransitive minimum self-conjugate sub-group. Hence if $G$ is soluble and has intransitive self-conjugate sub-groups, the symbols permuted by $G$ must be capable of division into imprimitive systems, such that the number in each system is the power of a prime.

131. Let $G$ be a $k$-ply transitive group of degree $n$ ($k>2$), and let $G_r$ be that sub-group of $G$ which keeps $r$ ($<k$) given symbols unchanged, so that $G_r$ is -ply transitive in the remaining $n-r$ symbols. Also, let $H$ be a self-conjugate sub-group of $G$, and let $H_r$ be that sub-group of $H$ which keeps the same $r$ symbols unchanged; so that $H_r$ is the common sub-group of $H$ and $G_r$. Since every operation of $G_r$ transforms both $H$ and $G_r$ into themselves, every operation of $G_r$ must be permutable with $H_r$; i.e. $H_r$ is a self-conjugate sub-group of $G_r$. Now, if $r=k-2$, $G_{k-2}$ is doubly transitive in the $n-k+2$ symbols on which it operates; it is therefore primitive. Hence, unless $H_{k-2}$ consists of the identical operation only, it must be transitive in the $n-k+2$ symbols.

If $H_{k-2}$ is the identical operation, $H$ contains no operation, except identity, which displaces less than $n-k+3$ symbols. Suppose, first, that $H$ contains operations, other than identity, which leave one or more symbols unchanged. Then, since $H$ is a self-conjugate sub-group and $G$ is $k$-ply transitive, it may be shewn, exactly as in § 110, that $H$ must contain operations displacing not more than $2k-2$ symbols. Hence $H_{k-2}$ can consist of the identical operation alone, only if

When this inequality holds, we have seen (p. 470) that $G$ contains the alternating group. Hence in this case, if $G$ does not contain the alternating group, it follows that $H_{k-2}$ is transitive in the $n-k+2$ symbols on which it operates.

Since $H$ is self-conjugate and $G$ is $k$-ply transitive, $H$ must contain a sub-group conjugate to $H_{k-2}$ which keeps any other $k-2$ symbols unchanged. Hence $H_{k-3}$ must be doubly transitive in the $n-k+3$ symbols on which it operates; and so on. Finally, if $G$ is not the symmetric group (the alternating group, being simple, contains no self-conjugate sub-group) $H$ must be -ply transitive.

Suppose, next, that $H$ contains no operation, except identity, which leaves any symbol unchanged. Then if, with the notation of § 110, $j=k-1$ for every operation of $H$, the argument there used does not apply. For it is impossible to choose the operation $T$ so that $c_k$ is a symbol which does not occur in $S$.

The self-conjugate sub-group $H$ contains a single operation changing a given symbol $a_1$ into any other symbol $a_r$. Also $G$ contains operations which leave $a_1$ unchanged and change $a_r$ into any other symbol $a_s$. Hence the operations of $H$, other than identity, form a single conjugate set in $G$; and therefore $H$ must be an Abelian group of order $p^m$ and type ; $p$ being a prime. Further, since $G$ is by supposition at least triply transitive, it must contain operations which transform any two operations of $H$, other than identity, into any other two. If $p$ were an odd prime, and $P_1$ and $P_2$ were two of the generating operations of $H$, it follows that $G$ would have an operation $S$ such that

THEOREM VII. A self-conjugate sub-group of a $k$-ply transitive group of degree $n$ $$

, is in general at least -ply transitive⁵⁰. The only exception is that a triply transitive group of degree $2^m$ may have a self-conjugate sub-group of order $2^m$.

132. For the further discussion of the self-conjugate sub-groups of a primitive group, it is necessary to consider in what forms the direct product of two groups can be represented as a transitive group.

Let $G$ be the direct product of two groups $H_1$ and $H_2$, and suppose that $G$ can be represented as a transitive group of degree $n$. When $G$ is thus represented, we will suppose that $H_1$ is transitive in the $n$ symbols that $G$ permutes. We have seen in § 107 that every substitution of $n$ symbols, which is permutable with each of the substitutions of a group transitive in the $n$ symbols, must displace all the $n$ symbols. It follows that every substitution of $H_2$ must displace all the $n$ symbols on which $G$ operates; and that the order of $H_2$ is equal to or is a factor of $n$.

If the order of $H_2$ is equal to $n$, then $H_2$ is transitive in the $n$ symbols, so that the order of $H_1$ cannot be greater than $n$. In this case, $H_1$ and $H_2$ must (§ 107) be two simply isomorphic groups of order $n$, which have no self-conjugate operations except identity. Further, if $H_1$ and $H_2$ in this case are not simple groups, let $K$ be a self-conjugate sub-group of $H_1$. Since every operation of $H_2$ is permutable with every operation of $H_1$, $K$ is a self-conjugate sub-group of $G$. Now the order of $K$ is less than $n$, the degree of $G$; therefore $K$ is intransitive and $G$ is imprimitive. On the other hand, we have seen (loc. cit.) that, if $H_1$ and $H_2$ are simple, the sub-group of $G$ that keeps one symbol fixed is a maximum sub-group: and therefore $G$ is primitive. Hence:—

THEOREM VIII. If the direct product $\Gamma$ of $H_1$ and $H_2$ can be represented as a transitive group of degree $n$, in such a way that $H_1$ and $H_2$ are transitive sub-groups of $\Gamma$, then $H_1$ and $H_2$ must be simply isomorphic groups of order $n$, which have no self-conjugate operations except identity. When this condition is satisfied, $\Gamma$ will be primitive if, and only if, $H_1$ and $H_2$ are simple.

133. Suppose now that a primitive group $G$, of degree $n$, has two distinct minimum self-conjugate sub-groups $H_1$ and $H_2$. Then every operation of $H_1$ (or $H_2$) is permutable with $H_2$ (or $H_1$), and $H_1$, $H_2$ have no common operation except identity. Hence (§ 34) the group , which we will call $\Gamma$, is the direct product of $H_1$ and $H_2$. Now $\Gamma$ is a self-conjugate sub-group of $G$: it is therefore transitive in the $n$ symbols which $G$ permutes. Also $H_1$ and $H_2$, being self-conjugate sub-groups of $G$, are transitive. Hence, by Theorem VIII (§ 132), $H_1$ and $H_2$ are simply isomorphic, and $n$ is equal to the order of $H_1$. Moreover, since $H_1$ is a minimum self-conjugate sub-group of $G$ which contains no self-conjugate operations except identity, it must (Theorem IV, § 94) be either a simple group of composite order, or the direct product of several simply isomorphic simple groups of composite order. It follows that $G$ cannot have two distinct minimum self-conjugate sub-groups unless the degree of $G$ is equal to or is a power of the order of some simple group of composite order.

134. Let now $\Gamma$ be a minimum self-conjugate sub-group of a doubly transitive group $G$, and suppose that $\Gamma$ is the direct product of the $\alpha$ simply isomorphic simple groups

Since $G$ is primitive, $\Gamma$ is transitive. If $H_1$ is a cyclical group of prime order $p$, the order of $\Gamma$ is $p^{\alpha }$; therefore the degree of $\Gamma$, or what is the same thing, the degree of $G$, is $p^{\alpha }$.

If $H_1$ is a simple group of composite order, and if $\alpha >2$, then (§ 132) $H_1$ cannot be transitive. The intransitive systems of $H_1$, since they form a set of imprimitive systems for $\Gamma$, must each contain the same number $m$ of symbols. If $m$ is less than the order of $H_1$, a sub-group of $H_1$ which leaves unchanged one symbol of one intransitive system will leave unchanged one symbol of each intransitive system. Now we shall see, in § 136, that the operations of an imprimitive self-conjugate sub-group of a doubly transitive group must displace all or all but one of the symbols. Hence $m$ cannot be less than the order of $H_1$. We may similarly shew that, if $m$ is equal to the order of $H_1$, some of the operations of $\Gamma$ must keep more than one symbol fixed; and therefore, if $\alpha >2$, the group assumed cannot exist. If $\alpha =2$, $H_1$ may be transitive. But in this case certainly contains operations which leave more than one symbol unchanged; and again the group assumed cannot exist. Hence finally no doubly transitive group can contain a minimum self-conjugate sub-group of the type assumed.

No general law can be stated regarding self-conjugate sub-groups of simply transitive primitive groups; but for groups which are at least doubly transitive the preceding results may be summed up as follows:—

THEOREM IX. A group $G$ which is at least doubly transitive either must be simple or must contain a simple group $H$ as a self-conjugate sub-group. In the latter case no operation of $G$, except identity, is permutable with every operation of $H$. The only exceptions to this statement are that a triply transitive group of degree $2^m$ may have a self-conjugate sub-group of order $2^m$; and that a doubly transitive group of degree $p^m$, where $p$ is a prime, may have a self-conjugate sub-group of order $p^m$.

Corollary. If a primitive group is soluble, its degree must be the power of a prime⁵¹.

In fact, if a group is soluble, so also is its minimum self-conjugate sub-group. The latter must be therefore an Abelian group of order $p^{\alpha }$: and since this group must be transitive, its order is equal to the degree of the primitive group.

135. As illustrating the occurrence of an imprimitive self-conjugate sub-group in a primitive group, we will construct a primitive group of degree $36$ which has an imprimitive self-conjugate sub-group. For this purpose, let⁵²

and

so that is an intransitive group of degree $36$, the symbols being interchanged in $6$ transitive systems containing $6$ symbols each. This group is simply isomorphic with

Also let

Then

and

and is similar to and simply isomorphic with .

Now it may be directly verified that $S$ and $A$ are, each of them, permutable with $JSJ$ and $JAJ$; and therefore every operation of the group is permutable with every operation of the group . Also these two groups can have no common operation, since the symbols into which changes any given symbol are all distinct from those into which change it. Hence is the direct product of and ; it is therefore a group of order $3600$. It is also, from its mode of formation, a transitive group of degree $36$; and it interchanges the symbols in two and only two distinct sets of imprimitive systems, of which

Now $J$ does not interchange the $36$ symbols in either of these systems, and therefore it cannot occur in . Further

136. We have seen that, with a single exception, a $k$-ply transitive group ($k>2$) cannot have an imprimitive self-conjugate sub-group; while the example in the preceding section illustrates the occurrence of an imprimitive self-conjugate sub-group in a simply transitive primitive group. We will now consider, from a rather different point of view, the possibility of an imprimitive self-conjugate sub-group in a doubly transitive group. Let $G$ be a doubly transitive group of degree $mn$, and let $H$ be an imprimitive self-conjugate sub-group of $G$. Suppose that $m$ is the smallest number, other than unity, of symbols which occur in an imprimitive system; and let

Now $a_{m+1}$ may be any symbol not contained in the original system. Hence it must be possible to distribute the $mn$ symbols into sets of imprimitive systems of $m$ each, such that every pair of symbols occurs in one system and no pair in more than one system. This implies that $mn-1$ is divisible by $m-1$, or that $n-1$ is divisible by $m-1$.

Consider now a substitution of $H$ which leaves $a_1$ unchanged. It must permute among themselves the remaining $m-1$ symbols of each of the $\frac{mn-1}{m-1}$ systems in which $a_1$ occurs. If $a_r$ is any other symbol, a similar statement applies to it. Now no two systems have more than one symbol in common. Hence every substitution, which leaves both $a_1$ and $a_r$ unchanged, must leave all the symbols unchanged. The sub-group $H$ is therefore such that each of its substitutions displaces all or all but one of the symbols. Moreover the substitutions, which leave $a_1$ unchanged, permute among themselves the remaining symbols of each system in which $a_1$ occurs; therefore the order of $H$ must be $mn\mu$, where $\mu$ is a factor of $m-1$.

It has been seen in § 112 that, for certain values of $mn$, groups satisfying these conditions actually exist. The doubly transitive groups, of degree $p^m$ and order  there obtained, have imprimitive self-conjugate sub-groups of orders and $p^m$.

137. Ex. 1. Shew that a transitive group of order $N$ and degree $2n$, which is imprimitive in respect of two systems of $n$ symbols each, must have a self-conjugate sub-group of order $\frac{1}{2}N$.

Ex. 2. Shew that an imprimitive group of order $N{p}^2$ and degree $np$, where $p$ is an odd prime, which has $p$ imprimitive systems of $n$ symbols each, must contain a self-conjugate sub-group whose order is a multiple of $p$.

Ex. 3. Shew that, if $n$ is the smallest number of symbols in which a group $G$ can be represented as a transitive group, then $n^{\alpha }$ is the smallest number of symbols in which the direct product of $\alpha$ groups, simply isomorphic with $G$, can be represented as a transitive group.

Ex. 4. Prove that if a group $G$, of order $N$, represented as a transitive substitution group, is imprimitive only when its degree is $N$; then either $G$ is Abelian, or $N$ is the product of two distinct primes. (Dyck.)

Ex. 5. Shew that if, in the tableau of p. 527, the sub-group

Shew also that, if each square compartment of the tableau is regarded as a single symbol, the permutations of these symbols given by the tableau represents the group $\frac{G}{H}$ in regular form. (Dyck.)