# Finite groups have unique largest solvable normal subgroup?

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?

• “Largest” here doesn’t mean “largest order,” it means that it contains all the other solvable normal subgroups. A solvable normal subgroup of largest order always exists but you don’t know a priori that it contains the others or that it’s unique. Commented Oct 15, 2020 at 20:53
• I see. Thank you! @QiaochuYuan Commented Oct 15, 2020 at 20:59

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 $$G$$.
The set of solvable normal subgroups of $$G$$ forms a partially ordered set with respect to inclusion. Because $$G$$ is finite this poset is finite, and it is nonempty because it contains the trivial subgroup. It follows that it has maximal elements, and also that it contains a subgroup of largest order. It remains to show that this subgroup contains all other solvable normal subgroups, i.e. that it is the unique maximal element of the poset.