Group of exponent $2$. When I have a group $G$ of exponent $2$ and I know that all elements in $G-\{e\}$ are conjugated, is it right that $G$ is of order $2$?
My try: For $g,h \in G - \{e\}$ the conjugation assumption implies $$ga=ah$$ but since $\exp G=2$ then $G$ is abelian and therefore $$ga=ha$$
It follows that $g=h$.

Note that if $x, y \in G$. Then $yx = y^{-1}x^{-1}= (xy)^{-1} = xy$ by the hypothesis. Hence $G$ having an exponent 2 implies that $G$ is abelian.
Choose $a, b \ G\backslash\{1 \}$. 
Then there exists a $g$ such that $b = gag^{-1}$. 
Then $h = ba^{-1} = gag^{-1}a^{-1} = 1$ since $G$ is abelian.
It follows that $a = b$ and we have the result as it was promised.  
