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If a group $G$ has an abelian subgroup of index $2$, then $G$ must be solvable.

Thoughts:

I know that if $N$ is a subgroup of a group $G$ with $[G:N]=2$, then $N$ is normal.

Context:

Prof. Derek Holt makes the claim in a comment on this answer.

Please help :)

Shaun
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    There is essentially nothing to show here. If you understand the definition of solvability, this is almost a tautology. (We have a normal abelian subgroup, and the quotient has order 2, so is abelian.) – verret May 16 '18 at 01:16
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    ${e} \triangleleft N \triangleleft G$ is a normal series such that each factor is abelian. – Antoine Giard May 16 '18 at 01:17
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    As I said in the other post, such groups are not only solvable, they have the stronger property of being polycyclic. There are effective algorithms for computing within polycyclic groups. For finite groups polycyclic is equivalent to solvable, but for infinite groups there are solvable, and even abelian, groups that are not polycyclic, such as ${\mathbb Z}^\infty$. – Derek Holt May 16 '18 at 09:57

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In order to close the question, here is @AntoineGiard's comment:

$\{e\} \triangleleft N \triangleleft G$ is a normal series such that each factor is abelian.

Shaun
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