Originally Posted by

**topspin1617** No, because it's not true: just take an arbitrary subgroup of $\displaystyle S_3$ of order 2, you can show that it isn't normal.

But the fact that a group of exponent 2 is abelian is nearly trivial. Let $\displaystyle G$ be such a group, so that $\displaystyle x^2=e$, for all $\displaystyle x\in G$.

Suppose $\displaystyle a,b\in G$ are arbitrary. Consider the relation

$\displaystyle (ab)^2=e$

$\displaystyle abab=e$

$\displaystyle ababb^{-1}=eb^{-1}$

$\displaystyle aba=b^{-1}$

$\displaystyle abaa^{-1}=b^{-1}a^{-1}$

$\displaystyle ab=b^{-1}a^{-1}$.

But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to

$\displaystyle ab=ba$

and, since these elements were arbitrary, this implies that $\displaystyle G$ is abelian.