Let be a prime and a positive integer. Prove that if is a group of order then every subgroup of index is normal in . Deduce that every group of order has a normal subgroup of order .

Let be a group of order and a subgroup with . Now acts on the conjugates by conjugation, since and . Moreover, under this action we have . By a previous theorem, we have , a prime.

If , then and we have ; moreover, has exactly conjugates in . Let be the permutation representation induced by the action of on the conjugates of , and let be the kernel of this representation. Now . By the first isomorphism theorem, the induced map is injective, so that divides . Note, however, that is a power of and that the only powers of that divide are and . So is or . If , then so that for all ; then and we have , a contradiction. Now suppose . Again by a previous theorem we have , so that , hence . Again, this implies that is normal so that for all , and we have , a contradiction. Thus .

If , then . That is, for all ; thus is normal.

Now let be a group of order . By Cauchy’s Theorem, contains an element of order . Then , hence is normal.

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## Comments

Do you mean “stab(H)” is stabilizer of H? And how we can say if [G:stab(H)]=p, then [stab(H):H]=1?

1. Yes.

2. Since is prime and because we have .

Hey,

Your blog is really interesting because it contains so many examples and results about groups and rings which I am studying.

I also have a following problem, but I haven’t solved it yet. I hope you can give me an idea for that.

Problem: Let G be a group (maybe not finite) and H is a subgroup of G such that for all Prove that $H$ is normal in G.

Thank you in advance.

Hanh.

Try this:

Let . Note that , since . Now , so that . Thus , and we have . Similarly, . Thus for all , and is normal.

Wow! Great!

Thank you very much for the solution. It has taken me for 3 months to struggle the problem, but I did not know how to do that.

Best,

Hanh.

We can use corollary 5 in sec 4.2 straight away right?

Indeed you can.

🙂, Thanks for the reply.