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