Is the following argument Correct?
Let $G$ be a group such that $g^2=1$ for all $g\in G$. Prove that $G$ is abelian.
Proof. Let $g\in G$ from the axiom of inverse we know that there exists a $g^{-1}\in G$ such that $gg^{-1} = 1$ in addition $g^2 = gg=1$ then by proposition $3.3$ it follows that $g = g^{-1}$.
Now returning to our original proposition. Let $\alpha\in G$ and $\beta\in G$ from corollary $3.4$ it follows that $(\alpha\beta)^{-1} = \beta^{-1}\alpha^{-1}$ but from our argument above it follows that $\alpha\beta = (\alpha\beta)^{-1}$ and $\beta^{-1}\alpha^{-1} = \beta\alpha$ implying $\alpha\beta = \beta\alpha$.
$\blacksquare$
NOTE:
Proposition $3.3$ Given a group $G$ and $a,b\in G$ there exists a unique solutions to the equation $ax=b$ and $ya=b$.
Proposition $3.4$ Given $a,b\in G$ we have $(ab)^{-1} = b^{-1}a^{-1}$.
Yes, it's correct. The sentences need punctuation though.