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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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Seems fine to me :) – carlosayam Aug 27 '14 at 7:57
good attempt !!! – Hirak Aug 27 '14 at 8:02
It is fine. You may show is using reversal rule also. $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba$$ – Shodharthi Aug 27 '14 at 8:03
Is "non-trivially" a common word + used correctly? – DracoMalfoy Aug 27 '14 at 8:03
@LaxmikantMishra: D'oh. I was at $abba = aea = aa = e$, therefore $ba = (ab)^{-1} = ab$, but yours is better. – Steve Jessop Aug 27 '14 at 12:12
up vote 5 down vote accepted

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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