# In a group of order 400, is the normalizer of one of the 16 Sylow 5-subgroups itself?

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!

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