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In a group $G$ of order $400 = 2^4 \cdot 5^2$, assume there are sixteen Sylow 5-subgroups. Let $P_5$ be one of them. Is the normalizer $N_G(P_5)=P_5$?

I think this is true, as the order of $P_5$ is 25, and 400/25=16, which coincides with the index of the normalizer, $[G:N_G(P_5)] = n_5=16$. Then I concluded that $N_G(P_5)=P_5$, but it seems like odd to me that the normalizer of a Sylow p-subgroup be itself. Thanks!

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This happens. The normalizer of a Sylow $2$-subgroup of $S_3$ is itself as well, for example.

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