TITLE: Proof that a group is abelian if every square is the identity.
QUESTION [13 upvotes]: I need to prove that for $(\mathbf G, \circ)$ a group, if for every $a\in\mathbf G$ $$a\circ a=e$$ where $e$ is identity element of that group, then the group is abelian group.$$\\$$My proof:
$$\text{We know that:}\\a\circ e=a=e\circ a\\b\circ e=b=e\circ b\\b\circ b=e=a\circ a\text{, so if}\\a\circ a=b\circ b\text{, then}\\a\circ (e\circ a)=(e\circ b)\circ b\\a\circ(b\circ b\circ a)=(a\circ a\circ b)\circ b\\(a\circ b)\circ (b\circ a)=(a\circ a)\circ(b\circ b)=e\circ e=e\\(a\circ b)\circ (b\circ a)=e\Rightarrow a\circ b=b\circ a$$
And I'm not really sure of the last step. Is this proof proper?

REPLY [3 votes]: The last step is not good: there are groups where you have elements $x,y$ satisfying $xyyx = e$ and $xy \neq yx$, so on its own, knowing $abba = e$ is not sufficient to conclude $ab = ba$.