### 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 on a finite set is nothing more or less than a homomorphism from to , the group of permutations of . As has order , the image of under this homomorphism has order at most ; in fact, it has order dividing , by Lagrange's theorem. We thus obtain a normal subgroup of index dividing as the kernel of this homomorphism. The kernel is a proper normal subgroup if and only if the action of on is non-trivial.

### Prerequisites

Basic group theory

### Example 1

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

Proof: We have an action of by right multiplication on the set of right cosets of , that is, a homomorphism from to . Now let be the kernel of ; this is automatically a normal subgroup. The cardinality of is precisely the index of in , so , and hence divides . The subset of that fixes the coset is precisely itself, so is contained in . Hence is the normal subgroup we require.

Indeed, it can be seen that the given above is in fact the largest normal subgroup of to be contained in , known as the core of in , in the sense that it contains all normal subgroups of that are contained in . This is because the stabiliser of the coset is the conjugate of , and given any such that is normal in , we have . So acts trivially on the right cosets of , and is thus contained in .

### Example 2

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

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

Let be a group with normal (characteristic) subgroups and , such that contains with finite, and such that the commutator of and is not contained in . Then has a proper normal subgroup of index dividing .

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

If more is known about the structure of , 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 under the action of its automorphism group have size at most , then must act on a single equivalence class non-trivially, giving a proper normal (characteristic) subgroup of index dividing .

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

## Comments

## This looks nice, but I for

Sat, 18/04/2009 - 00:47 — gowersThis 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 — taoHere'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 — wiltonFor me, it seems unnatural to say this without mentioning the symmetric group. I would give the proof as follows. The action of on is the same thing as a homomorphism . If is the kernel of this homomorphism then is isomorphic to a subgroup of and so of order dividing . It's clear that is contained in (indeed, it's precisely the intersection of with its conjugates).

## I've gone ahead and written

Sun, 19/04/2009 - 00:20 — gowersI'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 — gowersI 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 ReidI'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)?