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!
Best Answer
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.