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.
Thanks!
First, a characteristic subgroup of a normal subgroup is normal.
Now let $a$ have order $6$. Then $a^5$ also has order $6$. Hence conjugation by any element of $G$ takes $a$ to $a$ or $a^5$, by the assumption of only two elements of order six. Hence $\langle a\rangle $ is invariant under conjugation, hence normal.
Next, $a^2$ has order $3$.
But $\langle a^2\rangle $ is characteristic in $\langle a\rangle $ (since $\langle a\rangle $ is cyclic).
The result follows.