This is one of my first proofs about groups. Please feed back and criticise in every way (including style & language).
Axiom names (see Wikipedia) are italicised. $e$ denotes the identity element.
Let $(G, \cdot)$ be a group.
We assume that every element is its inverse.
It remains to prove that our group is commutative.
Non-trivially, $\textit{associativity}$ implies that parentheses are unnecessary.
Therefore, we do not use parentheses,
we will not use $\textit{associativity}$ explicitly.
By $\textit{identity element}$, $G \ne \emptyset$.
Now, let $a, b \in G$.
By assumption, $$aa = e \text{ and } bb = e. \quad \text{(I)}$$
By $\textit{closure}$, $ab \in G$.
So, by assumption, $$abab = e.\quad \text{(II)}$$
It remains to prove that $ab = ba$.
\begin{equation*}
\begin{split}
ab &= aeb && \quad\text{by }\textit{identity element} \\
&= aababb && \quad\text{by (II)} \\
&= ebabb && \quad\text{by (I)} \\
&= ebae && \quad\text{by (I)} \\
&= bae && \quad\text{by }\textit{identity element} \\
&= ba && \quad\text{by }\textit{identity element}
\end{split}
\end{equation*}
QED
Best Answer
It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$