If every nonidentity element in a group is of order $2$, the group is abelian

Question

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

Answer 1

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^{-1}=ab$, but $(ab)^{-1}=b^{-1}a^{-1}=ba$ by using twice again the remark.

Answer 2

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).