# Solvable groups

1. Jul 19, 2007

### bham10246

I've been working on this problem and I need just a small hint.

Let $A$ and $B$ be solvable subgroups of a group $G$ and suppose that $A\triangleleft G$. Prove that $AB$ is solvable.

My idea:

So we have a chain of normal subgroups of A so that their quotient is abelian. We also have a chain of normal subgroups of B so that their quotient is abelian. Since A is normal in G, should I multiply the normal subgroups $A_i$ in A by B to obtain $B=1*B=A_0 B \triangleleft A_1 B \triangleleft ... \triangleleft A_k B = AB$, but how do we know that $(A_{i+1}B)/(A_i B)$ is abelian?

If I understand this one thing, then I think I can finish the rest of the proof. Thank you!

This is a right approach, right?

2. Jul 19, 2007

### mathwonk

somehow i am tempted to use the fundamental isomorphism theorem about

the structure of (AB)/A.

3. Jul 19, 2007

### bham10246

Thanks, I'll try that. I thought that the above construction $B=1*B=A_0 B \triangleleft A_1 B \triangleleft ... \triangleleft A_k B = AB$ is correct but I'm not even sure that $A_i B$ is normal in $A_{i+1}B$?!