Let $G$ be a group with exactly $2$ elements of order $6$. Prove that $G$ has a normal subgroup of order $3$.
Since there is an element of order $6$, by Lagrange's Theorem, the order of $G$ must be a multiple of $6$. That means that both $2$ and $3$ are also divisors of the order of $G$, so, again by Cauchy's Theorem, $G$ must contain elements of order $2$ and order $3$ as well, respectively.
I suppose I'm not sure where to proceed from here. How can we use the fact that $G$ has exactly $2$ elements of order $6$ ? Would Sylow Theorems be helpful here at all ? I don't see how - since we don't know the exact order of $G$ here, which is when I'm used to using the Sylow Theorems.
Any help would be appreciated.