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I need a hint for the next problem: We have a group $G$ of order 56. Up to isomorphism, there is an unique group of order 56 which doesn't contain normal Sylow $7$-subgroups and the Sylow $2$-subgroup is elementary abelian. I must prove that this group doesn't contain a subgroup of order 28.

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Hint: First show that a subgroup $H$ of order $28$ is normal. Moreover, by Sylow's theorems, $H$ has a unique subgroup $S$ of order $7$. This subgroup is therefore characteristic as a subgroup of $H$. From these two facts, what can you say about $S$ as a subgroup of $G$?

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  • $\begingroup$ Thanks a lot, really forgot the properties of the characteristic subgroups. $\endgroup$ – Alchimist Aug 13 '14 at 8:02

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