A group theory problem

I have some difficulties with the following problem:

Let $G$ be a group with a composition serie $\mathcal G$; Let $A$ and $B$ be maximal normal subgroups of $G$; if $A'$ is a maximal normal subgroup of $A$ and $B'$ is a maximal normal subgroup of $B$ prove that $\left<A',B'\right>$ is a subnormal subgroup of $G$.

The situation is the following:

Which is the strategy to approach this problem (it is enough only a hint for the moment)? I'm looking for a solution for several days... :(

Edited: I should prove the porposition in the case "$G$ is a group (infinite order is allowed) with a composition serie... ". There is no need that $G$ is finite.

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Hint: look at the normal subgroup $AB$. What can you say about that one? Where does it lie in your diagram? –  Nicky Hekster Oct 10 '12 at 19:42
In this case $AB=G$ because $A$ and $B$ are maximal. –  Dubious Oct 10 '12 at 20:21
I think the following outline argument will work, although there might be a quicker way. $AB' = A$ or $G$. If $AB'=A$ then $B' = A \cap B$, etc, so assume $AB'=G$. If $A'$ is normal in $G$ then so is $\langle A',B' \rangle$ so assume not. So $[A',B'] \not\le A'$. If the simple group $A/A'$ has prime order, then $A'[A',B'] = A$, so $\langle A',B' \rangle = G$. So assume $A/A'$ is nonabelian simple and let $C$ be the core of $A'$ in $G$. Then $A/C$ is a direct product of copies of $A/A'$ and is the unique minimal normal subgroup of $G/C$. So $BC=B'C=G$, hence $\langle A',B' \rangle = G$. –  Derek Holt Oct 11 '12 at 11:35
I think my argument still works. The fact that $G$ has a composition series ensures that $A/C$ is a direct product of finitely many copies of the (possibly infinite) simple group $A/A'$. –  Derek Holt Oct 12 '12 at 19:06
What I said is also still true, no need for finiteness. Just an induction on the length of the series. –  user641 Oct 12 '12 at 21:53

1. the lattice of subnormal subgroups is modular: $$U\subset W \implies (U,V\cap W) = (U,V)\cap W$$
Update: ..hmmm.. I might have misunderstood the question. So, we don't know yet that $\langle A',B'\rangle$ belongs to this lattice...