## 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.

• 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

Thanks!

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