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?