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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

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It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$

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