A way of getting proper normal subgroups of small index

This article is incomplete. More detailed examples needed, as well as relevant definitions and general discussion of the idea.

Quick description

Given a group, it can be useful to know that there is a proper normal subgroup of small index, where 'small' is defined in terms of some finite invariant of the group. To obtain such a result, it suffices to show that the group has a non-trivial action on a sufficiently small finite set. These may arise from the internal structure of the group.

An action of a group $G$ on a finite set $X$ is nothing more or less than a homomorphism $\theta$ from $G$ to $\mathrm{Sym}(X)$ , the group of permutations of $X$ . As $\mathrm{Sym}(X)$ has order $|X|!$ , the image of $G$ under this homomorphism has order at most $|X|!$ ; in fact, it has order dividing $|X|!$ , by Lagrange's theorem. We thus obtain a normal subgroup of index dividing $|X|!$ as the kernel of this homomorphism. The kernel is a proper normal subgroup if and only if the action of $G$ on $X$ is non-trivial.

Prerequisites

Basic group theory

Example 1

Let $G$ be a group with a subgroup $H$ such that the index $H|$ of $G$ in $H$ is finite. Then there is a normal subgroup of $G$ contained in $H$ of index dividing $H|!$ .

Proof: We have an action of $G$ by right multiplication on the set $g \in G \}$ of right cosets of $H$ , that is, a homomorphism $\theta$ from $G$ to $\mathrm{Sym}(X)$ . Now let $N$ be the kernel of $\theta$ ; this is automatically a normal subgroup. The cardinality of $X$ is precisely the index of $H$ in $G$ , so $H|!$ , and hence $N|$ divides $H|!$ . The subset of $G$ that fixes the coset $H$ is precisely $H$ itself, so $N$ is contained in $H$ . Hence $N$ is the normal subgroup we require.

Indeed, it can be seen that the $N$ given above is in fact the largest normal subgroup of $G$ to be contained in $H$ , known as the core of $H$ in $G$ , in the sense that it contains all normal subgroups of $G$ that are contained in $H$ . This is because the stabiliser of the coset $Hg$ is the conjugate $g^{-1}Hg$ of $H$ , and given any $K \leq H$ such that $K$ is normal in $G$ , we have $K = K^g \leq H^g$ . So $K$ acts trivially on the right cosets of $H$ , and is thus contained in $N$ .

Example 2

We say a subgroup $H$ of a group $G$ is characteristic if $H = H^\alpha$ for every automorphism $\alpha$ of $G$ . This is a stronger property than normality, and is useful primarily for the following reason: if $H$ is a subgroup of $G$ , and $K$ is a characteristic subgroup of $H$ , then every automorphism of $G$ that sends $H$ to itself induces an automorphism of $H$ , and hence also sends $K$ to itself. In particular, if $H$ is normal in $G$ then $K$ is normal in $G$ , and if $H$ is characteristic in $G$ then $K$ is characteristic in $G$ (so the property of being a characteristic subgroup is transitive).

In addition, it can be said that 'any subgroup $K$ uniquely specified by a family $\mathcal{K}$ of characteristic subgroups is itself characteristic', even if $\mathcal{K}$ was chosen arbitrarily. This is because any automorphism of $G$ will preserve $\mathcal{K}$ element-wise, so the specification of $K$ is unchanged by applying the automorphism.

Let $G$ be a group with normal (characteristic) subgroups $A$ and $B$ , such that $A$ contains $B$ with $B|$ finite, and such that the commutator $[G,A]$ of $G$ and $A$ is not contained in $B$ . Then $G$ has a proper normal subgroup of index dividing $B|!$ .

Proof: Since $A$ and $B$ are normal in $G$ , we have an action of $G/B$ by conjugation on the quotient $A/B$ . Since $[G,A]$ is not contained in $B$ (which is the same as saying that $A/B$ is not in the centre of $G/B$ ), this action is non-trivial. By mapping $G$ onto its quotient $G/B$ and composing this with the action of $G/B$ on $A/B$ , we get a non-trivial homomorphism $\theta$ from $G$ to $\mathrm{Sym}(A/B)$ . The kernel $N$ of $\theta$ is therefore a proper normal (characteristic) subgroup of $G$ of index at most $B|!$ . If $A$ and $B$ are characteristic, then $N$ is characteristic, since it has been specified uniquely by our choice of $A$ and $B$ .

If more is known about the structure of $A/B$ , then the bound on the index of the proper normal subgroup can be improved. For instance, if it is known that all equivalence classes of elements of $A/B$ under the action of its automorphism group have size at most $k$ , then $G/B$ must act on a single equivalence class non-trivially, giving a proper normal (characteristic) subgroup of index dividing $k!$ .

For an example of how to use this in infinite group theory, suppose $A$ is a group with a characteristic subgroup $B$ of finite index that does not contain the derived subgroup $[A,A]$ of $A$ (this ensures that $A/B$ is non-abelian, so $A$ itself acts non-trivially on $A/B$ by conjugation). Now suppose $G$ is any group containing $A$ as a normal (characteristic) subgroup, but of arbitrary index. Then the argument above applies, and we obtain a proper normal (characteristic) subgroup $N$ of $G$ of finite index. Moreover, if $A$ has a descending sequence of characteristic subgroups $A_i$ of finite index such that each factor $A_i/A_{i+1}$ is non-abelian, then we obtain a corresponding strictly descending sequence of normal (characteristic) subgroups of $G$ .

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Comments

This looks nice, but I for

Sat, 18/04/2009 - 00:47 — gowers

This looks nice, but I for one will not understand it without some help with basic definitions. Here are the ones I don't know: non-central section (or any kind of section for that matter), derived subgroup. I sort of half remember what a characteristic subgroup is but wouldn't mind being reminded.

Incidentally, when this article is in a more finished state, it will also be a useful contribution to the page on how to use group actions.

Normal subgroups

Sat, 18/04/2009 - 01:20 — tao

Here's a simple result in the same spirit: if H is a subgroup of G of index n, then there is a normal subgroup K of G contained in H of index dividing n!. Proof: G acts by left multiplication on the coset space G/H, which has cardinality n. There are only n! possible ways any group element can act here, so the set of elements which act trivially is a subgroup of index dividing n!. But one easily checks that this subgroup is a normal subgroup that is contained in H.

Normal subgroups

Sat, 18/04/2009 - 23:04 — wilton

For me, it seems unnatural to say this without mentioning the symmetric group. I would give the proof as follows. The action of $G$ on $G/H$ is the same thing as a homomorphism $G\to\mathrm{Sym}(G/H)\cong S_n$ . If $K$ is the kernel of this homomorphism then $G/K$ is isomorphic to a subgroup of $S_n$ and so of order dividing $n!$ . It's clear that $K$ is contained in $H$ (indeed, it's precisely the intersection of $H$ with its conjugates).

I've gone ahead and written

Sun, 19/04/2009 - 00:20 — gowers

I've gone ahead and written up this example as a way of trying to get started an article on proving results by letting a group act on a finite set. I think it's a nice one. (I came across it last year when giving supervisions on a first Cambridge course in group theory: the lecturer had set it as a question.)

Some basic definitions

Sat, 18/04/2009 - 16:06 — Ben Fairbairn (not verified)

Tim wrote:

"I for one will not understand it without some help with basic definitions. Here are the ones I don't know: non-central section (or any kind of section for that matter), derived subgroup. I sort of half remember what a characteristic subgroup is but wouldn't mind being reminded."

Let G be a group.

A SECTION of G is simply a homomorphic image of a subgroup of G (or phrased differently if there is a subgroup H

I see that you too have

Sun, 19/04/2009 - 01:12 — gowers

I see that you too have written up the example that Terence Tao suggested. I think this is absolutely fine: the example can now be found in two genuinely distinct ways: one by somebody who wants to find a normal subgroup of small index (who will naturally arrive at this article) and the other by somebody who wants to understand why people are interested in group actions and how they manage to use them to solve problems that are not ostensibly about actions (who will be led to the article on proving results by letting a group act on a finite set. In general, I am all in favour of the same example popping up in more than one place, and wouldn't even be unduly disturbed if it was simply cut and pasted from one article to another, though that has not happened in this case.

I've rewritten the article in

Sun, 19/04/2009 - 01:14 — Colin Reid

I've rewritten the article in a way that hopefully assumes less background of the reader, and also included Terence's example. This is indeed a very good example of what I am talking about. Is it a bad thing to have it written up here and in Tim's article, or should there just be a link to the most detailed instance of an example (Tim's in this case)?