Subgroup of solvable group is solvable (not using tower definition)

I'm studying for an exam, and I'm having problems proving that subgroups of solvable groups are solvable. I want to use this definition of solvability:

A group $G$ is solvable if and only if $G$ is abelian or there is a normal subgroup $H$ such that $1<|H|<|G|$ with both $H$ and $G/H$ solvable

The proof I have goes like this:

Proof: Any subgroup of an abelian group is abelian and normal, so assume $G$ is not abelian. Let $K < G$. Since $G$ is solvable, there is a subgroup $H$ such that both $H$ and $G/H$ are solvable.

Define $H' = H \cap K$. $H'$ is a normal subgroup of $K$ (simple verification). Now, by induction on the size of $H$, $H' < H$ is solvable as $|H'| < |H|$.

Now, my problem is with showing $K/H'$ is solvable. I have in my notes the map $aH' \to aH$ with $a \in K$, but then becomes unclear so I'm not sure what to do.

Could anyone point me into the right direction? Thanks in advance.

Note: We did learn the tower definition as well, but the aforementioned one is the primary one we used. My apologies if this is a duplicate, as I wasn't able to find any here using the definition I want.

• Are you assuming that $G$ is finite? – Derek Holt May 11 '14 at 13:31
• @DerekHolt It isn't clear, but I believe so, yes. We were working toward showing that $S_5$ was not solvable. – Lost May 11 '14 at 19:45

Are you assuming your group $G$ is finite? In this case we can proceed by induction on $|G|$. You should add in your proof "Since $G$ is solvable, there is a [normal] subgroup $H$ such that both $H$ and $G/H$ are solvable [and $1<|H|<|G|$]." You have three cases:
-- if $H\subseteq K$, by inductive hypothesis, $K/H$ is a solvable subgroup of $G/H$ (where $|G/H|<|G|$ since $1<|H|$), and the fact that both $H$ and $K/H$ are solvable gives $K$ solvable;
-- if, $K\subseteq H$, then by inductive hypothesis $K$ is solvable (use $|H|<|G|$);
-- in the remaining case, you have $|K\cap H|<|H|$, so, by inductive hypothesis $K\cap H$ is solvable. Furthermore, $K/(K\cap H)\cong KH/H$ is a subgroup of $G/H$ and it is therefore solvable (use that $|G/H|<|G|$).
• We should be given that we used induction to show $H'$ was solvable. Thanks, I see now, I was trying to work directly with $K/H'$. One question: Why is $|K \cap H| < |H|$ a separate case? Isn't it covered by both of the other cases? – Lost May 11 '14 at 8:39
• Well $H\subseteq K$ is equivalent to say $K\cap H=H$, while $K\subseteq H$ is the same as saying that $K\cap H=K$. For the third condition to be verified, neither of the first two needs to be verified, so it is not covered. With this I'm not saying that you cannot find a way to do this proof with less than three cases! – Simone May 11 '14 at 8:49