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 G be a group and let H,K,N \leq G with N normal in H. Then N \cap K is normal in H \cap K. Proof: Let a \in H \cap K. Then a(N \cap K) = aN \cap aK = Na \cap Ka = (N \cap K)a because a \in H and a \in K. \square

Lemma 2: Let A, B, and C be groups with B normal in A. Then B \cap C is normal in A \cap C and there is an injective homomorphism \varphi : (A \cap C)/(B \cap C) \rightarrow A/B. Proof: Let \pi : A \cap C \rightarrow A/B be the natural projection. Certainly B \cap C \subseteq \mathsf{ker}\ \pi, and if x \in \mathsf{ker}\ \pi, we have x \in B, and so x \in B \cap C. By the First Isomorphism Theorem, the induced homomorphism \varphi : (A \cap C)/(B \cap C) \rightarrow A/B is injective. \square

Lemma 3: Let G be a group and H,K,N \leq G such that N is normal in G and K \leq H is normal, and N \leq K \leq H \leq G. Then K/N \leq H/N is normal. Proof: Let a \in H. Then (aN)(K/N) = aK/N = Ka/N = (K/N)(aN). \square

Let G be a solvable group. Then there exists a subnormal series 1 = H_1 \vartriangleleft H_2 \vartriangleleft \cdots \vartriangleleft H_{k-1} \vartriangleleft H_k = G such that H_i/H_{i-1} is abelian for all 1 < i \leq k.

Let K \leq G be a subgroup. By Lemma 1 we have 1 = H_1 \cap K \vartriangleleft H_2 \cap K \vartriangleleft \cdots \vartriangleleft H_{n-1} \cap K \vartriangleleft H_n \cap K = K. Moreover, by Lemma 2 we have (H_i \cap K)/(H_{i-1} \cap K) \leq H_i/H_{i-1} abelian. Thus K is solvable.

Let N \leq G be a normal subgroup. Now H_{i+1}N/N \leq H_iN/N is normal by Lemma 3, so that 1 = H_1N/N \vartriangleleft H_2N/N \vartriangleleft \cdots \vartriangleleft H_{k-1}N/N \vartriangleleft H_kN/N = G/N. Moreover, (H_iN/N)/(H_{i-1}N/N) \cong H_iN/H_{i-1}N is abelian by the Third Isomorphism Theorem. Hence G/N is solvable.

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  • neal  On July 11, 2011 at 11:09 pm

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

    • nbloomf  On July 11, 2011 at 11:41 pm

      I think you’re talking about Lemma 3. N(K/N) = N 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.

  • neal  On July 12, 2011 at 12:01 am

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

    • nbloomf  On July 18, 2011 at 1:16 pm

      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!

  • neal  On October 1, 2011 at 7:07 pm

    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

    • nbloomf  On October 1, 2011 at 7:55 pm

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

      The ‘mod N‘ notation is a little misleading here. K/N really means a set of equivalence classes in K with respect to the relation a \sim b if and only if b^{-1}a \in N.

      Again, this will eventually be rewritten.

  • 1111  On November 27, 2011 at 12:59 am

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

    • nbloomf  On November 28, 2011 at 10:30 am


  • Gobi Ree  On November 29, 2011 at 3:27 am

    In lemma2, you can define \phi : A \cap C \to A/B and obviously \ker \phi = B \cap C. Then by the 1st isom thm, the result follows. We don’t have to check well-definedness and injectivity.

    • nbloomf  On November 29, 2011 at 8:56 am

      Thanks- that’s much better.

  • Gobi Ree  On November 29, 2011 at 8:06 am

    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?

    • nbloomf  On November 29, 2011 at 8:56 am

      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.

  • Gobi Ree  On November 29, 2011 at 9:50 am

    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!

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