Proof that a group is abelian if every square is the identity. 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?
 A: The proof is valid if you meant $a^2=e$ for all $a\in G$ at the beginning. Here's a variant:
We can read $(ab)^2=e$ as $ab=(ab)^{-1}$. Thus from the general rule of computing product inverses
$$
ab=b^{-1}a^{-1}.
$$
But also $a^2=b^2=e$, so that $a^{-1}=a$ and $b^{-1}=b$. Plugging in the previous displayed formula we get
$$
ab=ba.
$$ 
A: 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 $ab = ba$.
A: Your proof is correct as others pointed out (although the last line might require further clarification). Let me provide a shorter way though:
\begin{align}
a \circ b &= \underbrace{(b\circ b)}_{e} \circ (a \circ b) \circ \underbrace{(a\circ a)}_{e} \\
&= b\circ \underbrace{((b \circ a) \circ( b \circ a))}_{e}\circ a \\
&= b \circ a 
\end{align}
A: The third line is incorrect. You cannot get that $b\circ b = e$ (consider $\mathbb{Z}_4^+$, where $2 + 2 = 0$ but $1+1 \neq 0$).
A: Your proof is nearly there. You have shown that $(a \circ b)$ is the inverse of $(b \circ a)$.
Now you can multiply by $(b \circ a)$ and combine that with the knowledge that $(b \circ a)\circ(b \circ a)=e$ :
\begin{align}
(a \circ b)\circ(b \circ a) &= e \\
(a \circ b)\circ(b \circ a)\circ(b \circ a) &= e\circ(b \circ a) \\
(a \circ b)\circ e &= (b \circ a) \\
(a \circ b) &= (b \circ a) \\
\end{align}
