2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are you assuming that $G$ is finite? $\endgroup$ – Derek Holt May 11 '14 at 13:31
  • $\begingroup$ @DerekHolt It isn't clear, but I believe so, yes. We were working toward showing that $S_5$ was not solvable. $\endgroup$ – Lost May 11 '14 at 19:45
2
$\begingroup$

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 $|G/H|<|G|$).

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Lost May 11 '14 at 8:39
  • $\begingroup$ 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! $\endgroup$ – Simone May 11 '14 at 8:49
  • $\begingroup$ Oh, yes, of course. I shouldn't be doing math late at night. Yes, but this is a clear and straightforward proof. $\endgroup$ – Lost May 11 '14 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.