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For any group $G$ of order $36$ and any subgroup $H$ of $G$ of order $4$, is $H$ contained in $Z(G)$?

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Welcome to Math StackExchange. People will be able to give you better help if you edit your question to include some information about what you've tried so far and where you're finding yourself stuck. – Kevin Carlson Oct 27 '12 at 16:17

There is a nontrivial operation of $C_3$ on the Klein four group $C_2^2$. This gives us a group $((C_2^2)\rtimes C_3)\times C_3$ where $C_2^2$ is not in the center.

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