TITLE: If $G$ is a group with order $364$, then it has a normal subgroup of order $13$
QUESTION [10 upvotes]: 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?

REPLY [7 votes]: After m.k.'s proof another one. There is indeed exactly one $7$-sylow subgroup, let's call is $S$: it is normal in $G$. There is also exactly one Sylow $13$-group in $G/S$ because the number divides $4$ and has residue $1$ modulo $13$. Take some generator of this unique Sylow $13$-subgroup of $G/S$. It has exactly $7$ antecedents in $G$. Also, any Sylow $13$-subgroup of $G$ intersects $S$ trivially so projects isomorphically onto the unique Sylow Subgroup of $G/S$. Each Sylow $13$-subgroup of $G$ thus contains one of the seven antecedents mentioned above, and two distinct Sylow $13$-subgroups of $G$ must intersect trivially, thus there areat most $7$ Sylow $13$-subgroups in $G$. But there can only be $1$ or $14$ so there is exactly one and it is normal.