To show this we use Sylow's third theorem. The number of Sylow -subgroups is congruent to mod and divides since . Note that is prime. This means the number of Sylow 13-subgroups is or . If the number of Sylow -subgroups was then be cyclic (I assume you know this result) and therefore abelian. Thus, the number of Sylow -subgroups is .
We will use a well-known result: If is a finite group and is a subgroup then the number of subgroups conjugate to is (the index of the normalizer in the group). However, by Sylow's 2nd theorem the number of conjugates to (Sylow -subgroup) are in fact all the Sylow -subgroups. Thus, . We know that by above. Thus, . But with thus .