## Subgroups and quotient groups of solvable groups are solvable

Prove that subgroups and quotient groups of solvable groups are solvable.

We prove some lemmas first.

Lemma 1: Let be a group and let with normal in . Then is normal in . Proof: Let . Then because and .

Lemma 2: Let , , and be groups with normal in . Then is normal in and there is an injective homomorphism . Proof: Let be the natural projection. Certainly , and if , we have , and so . By the First Isomorphism Theorem, the induced homomorphism is injective.

Lemma 3: Let be a group and such that is normal in and is normal, and . Then is normal. Proof: Let . Then .

Let be a solvable group. Then there exists a subnormal series such that is abelian for all .

Let be a subgroup. By Lemma 1 we have . Moreover, by Lemma 2 we have abelian. Thus is solvable.

Let be a normal subgroup. Now is normal by Lemma 3, so that . Moreover, is abelian by the Third Isomorphism Theorem. Hence is solvable.

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## Comments

why is N(K/N) = N again, thanks

I think you’re talking about Lemma 3. is not true, and I’m not seeing it here. Where is that step?

By the way, with the benefit of hindsight I can see that Lemma 3 is just part of the Lattice Isomorphism Theorem.

in lemma 3 where it says a(N(K/N)) = a(N/K), I’m just beginning to understand the idea of a coset

Eh… looking more closely, this solution is messed up. I’m going to have to rewrite it later when I have more time. For now I’ll mark it ‘incomplete’.

So sorry for the inconvenience!

hey so is that operation well defined the one where aN K/N = aK/N,

i know aN is a coset of N and K/N is the set of all cosets ( can’t assume it has a group structure since N is not assumed to be normal), so how do you multiply them. thanks

I haven’t gotten around to fixing this one yet. It the (implicit, now explicit) assumption that , then is in fact normal in , and , so it works.

The ‘mod ‘ notation is a little misleading here. really means a set of equivalence classes in with respect to the relation if and only if .

Again, this will eventually be rewritten.

the indices are wrong, H_i is a subgroup of H_{i+1}

Thanks!

In lemma2, you can define and obviously . Then by the 1st isom thm, the result follows. We don’t have to check well-definedness and injectivity.

Thanks- that’s much better.

In the text, the definition of a solvable group requires just a subnormal series, not a composition series. And this definition is in wikipedia, too. Is it a mistake the phrase “composition series” in your post?

That was a mistake on my part. However, I think either works in this case, since any composition series of a solvable group will serve as a witness to solvability (i.e., have abelian intermediate quotients) and any witness to solvability (i.e. subnormal series with abelian intermediate quotients) can be refined to a composition series.

I wrote an answer to the case of quotients groups. But it’s a little long and actually I wanted to write a post in wordpress, so I wrote it in my wordpress. (First post lol!) I’ll be very glad if you read it, thanks!

http://gobiree.wordpress.com/2011/11/29/quotient-groups-of-a-solvable-group-are-solvable/