2 Tricks to prove Every group with an identity and x*x = identity is Abelian - Fraleigh p. 48 4.32

Solution 1:

A key idea is clearer this way: $\ x\cdot x = e\,\Rightarrow\, x \color{#c00}{\overset{(1)}=} x^{-1}\, $ so $\ ab \color{#c00}{\overset{(1)}=} \color{blue}{(ab)^{-1}}\! \color{#0a0}{\overset{(2)}=} b^{-1} a^{-1}\! \color{#c00}{\overset{(1)}=} b a\ \ \,$ QED

Now it is clearer where the proof "comes from", namely it arises by "overlapping" the identities $x\! \color{#c00}{\overset{(1)}=} x^{-1}$ and $\,(xy)^{-1}\color{#0a0}{\overset{(2)}=} y^{-1} x^{-1}$ i.e. by discovering some term $\,\color{blue}{(ab)^{-1}}\!$ where both identities apply. Then, just as we did above, we can rewrite the term in two ways, giving a possibly new equality.

This is a widely applicable method of deriving consequences of identities, i.e. by "unifying" terms of both so that both identities apply. In fact in some cases it can be used to algorithmically derive all of the consequences, so yielding algorithms for deciding equality, e.g. see the Knuth-Bendix equational completion algorithm and the Grobner basis algorithm, and see George Bergman's classic paper The Diamond Lemma in Ring Theory.

Solution 2:

Let $G$ be a group, to show it is abelian is the same as to show that for $a,b\in G$, one has $ab=ba$, which is the same as $aba^{-1}b^{-1}=e$.

Now in this specific group, $a^{-1}=a$ and $b^{-1}=b$, so the last equation becomes $(ab)(ab)=e$. That is why we consider $(a*b)*(a*b)$.

Solution 3:

We are just given that $x^2=e$ for all $x\in G$. I am not a superhuman and right now I am guessing and wanting just to manipulate the elements. You want to see $ab=ba$ for all $a,b\in G$. Let's multiply both sides from the right(or left) by $ab$ for example. We get: $$(ab)*(ab)=(ab)*(ba)$$ Since $(ab)^2=e$, so $$(ab)*(ba)=e\to a*b^2*a=e\to e=e$$ This is a good point to start and we understand why the first key was given to us. That's it.