I am working on proving that if a group $G$ is finite, then $G$ has a unique largest solvable normal subgroup.
One of the proofs claims that if $G$ (finite or infinite) has two normal subgroups and they are solvable, say $M,N$ in $G$, then $MN$ is also normal and solvable subgroup in $G$. (I am not sure where we need this fact!)
Then, the proof picks a solvable normal subgroup of $G$ of largest order, say $S$. (I don't know if this is possible; to assume the existence of the subgroup which we need to prove it exists), and show that $S$ contains all solvable normal subgroups of $G$.
My question if if this proof is correct? and if there is another proof?
Best Answer
Well, since $M$ is solvable we have that $MN/N\cong M/(M\cap N)$ is a solvable group. It shows that $MN$ is solvable (because its normal and solvable subgroup $N$ induces a solvable factor).
Now, let $\Gamma$ be the family of all normal and solvable subgroups of $G$. I claim $$H=\bigcup_{N\in\Gamma}N$$ is the largest normal and solvable subgroup of $G$. In fact, given $a,b\in H$, there exist $N,M\in\Gamma$ such that $a\in N$ and $b\in M$. Thus, $ab^{-1}\in MN$. Since $MN$ is a normal and solvable subgroup of $G$, we have that $MN\in\Gamma$ and so $ab^{-1}\in H$. It shows that $H$ is a subgroup of $G$. Normality follows by the definition of $H$. Finally, if $K$ is a normal and solvable subgroup of $G$, by definition we have $K\in\Gamma$ and so $K\subseteq\bigcup_{N\in\Gamma}N=H$, which means that $H$ is the largest normal and solvable subgroup of $G$.