If $G$ is non-abelian group of order 6, it is isomorphic to $S_3$ Let $G$ be a non-abelian group of order $6$ with exactly three elements of order $2$. Show that the conjugation action on the set of elements of order $2$ induces an isomorphism. 
I just need to show that the kernel of the action is trivial. Not sure how to go about doing that. I think maybe a proof by contradiction but I can't find a contradiction. I would think it would violate "non-abelian-ness" of the group. Thanks for any help!
 A: Here is a hands-on method.
Note that $G$ has an element $a$ of order $3$ hence at least two as $a^2$ has order $3$, but can't have an element of order $6$ or it would be cyclic and hence abelian.
Suppose the elements of order $2$ are $b,c,d$, then the elements of the group are $1,a,a^2,b,c,d$. No element of order $3$ can commute with any element of order $2$ else the product would have order $6$
Now $ab\neq ba$ implies both $aba^{-1} \neq b$ and $bab^{-1} \neq a$ - so neither the elements of order $3$ nor those of order $2$ can have a trivial action.
A: If you Sylow Theorems, you get 1 Sylow 3-subgroup and either 1 or 3 Sylow 2-subgroup(s). 
If G has 1 Sylow 2-subgroup, it must be normal since it is a unique subgroup of given order. So we can take direct product of these two subgroups, which is isomorphic to G. But since each Sylow subgroups here are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$, we have that $G \cong \mathbb{Z}_2 \times \mathbb{Z}_3$, which contradicts that $G$ is not abelian.
So G has 3 Sylow 2-subgroups. Now you can explicitly list out elements of $G$ to see why it is isomorphic to $S_3$
A: Hint: Suppose $x\in G$ is an element of the kernel of the action, i.e. fixes the three involutions under conjugation. What do you know about the group generated by the three involution, and what does that tell you about $x$?
