Let $G$ be a finite non-abelian group and $k$ be the number of cyclic subgroups of order $2$. Why is every prime divisor of $k$ a prime divisor of the order of $G$?
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Consider $G$ acting on the set of order 2 subgroups via conjugation... – Aug 29 '13 at 17:52
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Of course, as soon as you show some work of your own. – Aug 29 '13 at 17:58
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Oh wait, I think I have misread your question. What you're asking doesn't seem to be true, unless I am still misreading it. Where is this problem from? – Aug 29 '13 at 18:07
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A counterexample: $G={\mathbb Z}_2\times {\mathbb Z}_2\times H$, where $2$ and $3$ do not divide $|H|$.
Boris Novikov
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@Steve D: Yes, but OP could add an additional constraint: $G$ is not $2$-group. :-) – Boris Novikov Aug 29 '13 at 19:12