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Given a normal subgroup H of G, H and G/H are solvable. Then is G solvable? I know the converse is true... but I have no idea for our statement.

Hobin. J
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Yes, you are correct. For the proof, write down an abelian tower for $G/H$ starting at $1$ and ending at $G/H$. One can show that this lifts to an abelian tower starting at $H$ and ending at $G$. The abelian tower for $H$ completes the required tower.

Alex G.
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  • Thanks to you, I proved this problem. :) Proving the problem, I used "The third group isomorphism theorem"! Anyway, thank you!! – Hobin. J Jun 06 '14 at 05:48