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Group of order $2^n m$ ($m$ odd) with a cyclic Sylow $2$-subgroup has a characteristic subgroup of order $m$.

I'd like to approach this by induction, but I can't see how to go about it. I'm stuck just on the case where $n=1$.

Tengu
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puns
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  • See also https://math.stackexchange.com/questions/2404906/elements-of-odd-order-form-a-subgroup-when-the-sylow-2-subgroup-is-cyclic (the fact that the subgroup is characteristic rather than just normal follows from it having order coprime to its index). – Tobias Kildetoft Oct 30 '17 at 09:31

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By Cayley's theorem, $G$ of order $2m$ has an embedding in $S_{2m}$ where all non-identity elements have no fixed points. In particular elements of order $2$ have $m$ cycles of length $2$ and so are odd permutations. Thus the elements of $G$ mapping to even permutations form an index $2$ subgroup of $G$, which is clearly characteristic.

Angina Seng
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