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Let $\langle G; \star, \hat, e\rangle$ be a group. Show that if for all $a\in G$, we have $a\star a=e$, then G is abelian.

What I did:

Assume G is abelian, then we have $a\star b = b\star a \quad \forall a,b\in G$

We also know that $ a = \hat a$, so:

$a\star b = b \star a$

$\Leftrightarrow a\star a \star b = a \star b \star a$

$\Leftrightarrow b = a \star b \star a$

$\Leftrightarrow b \star b = (b \star a ) \star (b \star a)$

$\Leftrightarrow e = e$

and since $e\in G$ we are finished.

I'm very unsure about this proof. I know that there's a way better proof using $\widehat{a\star b} = \hat b \star \hat a$

Does what I did actually work?

xotix
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  • You are trying to prove that the group is abelian, so,you can’t just assume it is. You need to assume the condition, and use it to show the group is abelian. – Ittay Weiss Jan 04 '18 at 13:28
  • @IttayWeiss OP is not assuming the group is abelian. – 5xum Jan 04 '18 at 13:29
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    @5xum They might not really be using it in the proof, but the very first statement is that assumption. – Tobias Kildetoft Jan 04 '18 at 13:29
  • It's correct, but for elegance I would take it the other way, starting from $(ab)(ab)=e$, then right multiplying by $b$, &c. – Bernard Jan 04 '18 at 13:31

2 Answers2

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Your proof seems fine in general, but it needs rewriting.

Writing "assume $G$ is abelian" is not OK if what you are trying to prove is the very fact that $G$ is abelian. In fact, in place of that, I think what you wanted to say is

To prove $G$ is abelian, we need to prove $ab=ba$ for all $a,b\in G$.

After that, the proof is written a little strangely since it looks like you start with the equation $a\cdot b=b\cdot a$. It's technically OK, but for readability, I think it's better to start with something that is known (i.e., $e=e$, and prove that that is equivalent to $ab=ba$.

Alternatively, you can start $ab$ and end with $ba$, i.e. write something like

$$ab=abe=ab(ba)(ba) = abbaba = aeaba = aaba = eba = ba$$

5xum
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For any $a $ in $G $, $a^2=e $. Then, we have for $a, b \in G $: $$(a * a) * (b *b) = e * e = e $$ $$(a*b)*(a*b)=e $$

Hence, using associativity, we get: $$a*(a*b)*b = a*(b*a)*b $$ It follows that $$a^{-1}*a*(a*b)*b * b^{-1} = a^{-1}*a*(b*a)*b*b^{-1} $$ $$e*(a*b)*e = e*(b*a)*e $$

Hence, $a*b = b*a $. Therefore $G $ is an abelian group.