# Proof that group is commutative if every element is its inverse (feedback wanted)

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

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