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This is a question from an exam in an undergraduate group-theory course:

Given a group $G$ and normal subgroups $N,K\trianglelefteq G$ such that $G/K,G/N$ are solvable, prove that $G/(K\cap N)$ is also solvable.

I have seen a partial solution that uses the finite lengths of the derived series' of $G/K$ and $G/N$ (let's call them $n$ and $m$) to show $G^{(m+n)}\leq K\cap N$ and then used the third isomorphism theorem, but it was very unclear and I couldn't understand how to construct a normal series that would show that $G/(K\cap N)$ is indeed solvable.

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Hint: $G/(K \cap N)$ can be homomorphically embedded in $G/K \times G/N$. And use that a product of solvable groups is again solvable and that subgroups of solvable groups are again solvable.

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  • $\begingroup$ Thank you. It seems like: $g(K\cap N)\mapsto(gK,gN)$ does the trick, right? If so this is much simpler and clearer than the solution I have seen. $\endgroup$ – Dean Gurvitz Jul 25 '18 at 8:24
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    $\begingroup$ Yes that is correct! The point is that from the sheer definition of solvable groups (with a subnormal series with abelian factor groups) one can derive lemma's and theorems that capture the original definition but are much easier to work with. $\endgroup$ – Nicky Hekster Jul 25 '18 at 10:19

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