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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 $\left$ is a subnormal subgroup of $G$.

The situation is the following:

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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?2012-10-10
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    In this case $AB=G$ because $A$ and $B$ are maximal.2012-10-10
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    How can I use that $G$ ha a composition serie?2012-10-10
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    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 $\langle A',B' \rangle = G$.2012-10-11
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    Joins of subnormal subgroups are always subnormal. I am confused, I guess you want an argument that avoids that fact?2012-10-12
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    I have edited the question.2012-10-12
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    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 $A/A'$.2012-10-12
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    What I said is also still true, no need for finiteness. Just an induction on the length of the series.2012-10-12

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