If $G$ is a group with order $364$, then it has a normal subgroup of order $13$

I'm a bit stuck into this problem.

If $G$ is a group with order $364$, then it has a normal subgroup of order $13$.

I have tried to use Sylow III but all I could conclude was that the $7$-subgroup of Sylow in $G$ is normal and that the number of $13$-Subgroups of Sylow are $1$ or $14$. Some people told me that I can do this through "counting elements" but I don't know how I could do that. Can someone help me?

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Counting elements normally goes like this: if I had $14$ Sylow $13$-subgroups, then I'd have $14 \cdot 12 = 168$ distinct elements of order $13$. Normally you can then say that this number is too large, although nothing obvious pops out here. –  Dylan Moreland May 4 '12 at 20:40
I don't think counting elements is the best appraoch to this problem. Use the fact that the Sylow 7-subgroup is normal to deduce that it must be centralized by an element of order 13, and then use that fact to rule out the possibility of 14 Sylow 13-subgroups. –  Derek Holt May 4 '12 at 20:42

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