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Prove that if $g^2=e$ for all g in G then G is Abelian.

This is how I proved it:

Abelian means that the following axioms hold: Associativity, Existence of Identity and inverse elements, commutativity.

1) Associativity:

For some element $h \in G$, we have (hg)g = h(gg) = h. Therefore holds

2) Existence of identity

From definition: $g^2 = I_G$

3) Existence of inverse

As G is already a group, thus there exists a $g^{-1}$ such that $g^{-1} \cdot g = 1_G$

4) Commutativity

$hg^2 = g^2h = I_Gh = h$

Thus commutativity holds.

Is this proof correct?

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marked as duplicate by Thomas Andrews, Cameron Buie, Did, Ross Millikan, Micah Dec 6 '12 at 16:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

By hypothesis, you already know that $G$ is a group. Therefore, you only need to prove commutativity. And your answer is not very clear. –  Thibaut Dumont Dec 6 '12 at 15:51
Yeah, it is not clear why $(hg)g=h(gg)=h$ shows associativity, but as @ThibautDumont noted, $G$ being a group means associativity is a given. –  Thomas Andrews Dec 6 '12 at 15:54
@ThomasAndrews In that link you put, why does she start off by saying $(g_2g_1)^r = e$. How can she justify that? Or is r all even powers? –  Kaish Dec 6 '12 at 16:07
@Kaish Don't read the question as an answer. The answers are the answers in the linked post. –  Thomas Andrews Dec 6 '12 at 16:13

3 Answers 3

up vote 4 down vote accepted

You're given that $G$ is a group, so you don't need to show that $G$ is a group. What you need to show is that for $g,h \in G$ we have $gh=hg$. You don't exactly show this. You might consider showing that $ghg^{-1}h^{-1}=1_G$ and think about what $g^{-1}$ is in this group and also consider what you can say about $(gh)^2$.

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I understand that I just show gh = hg. But why do you look at $ghg^{-1}h^{-1}$? Why do you decide to look at that? Why do we start to look at $(gh)^2$? –  Kaish Dec 6 '12 at 16:04
Standard approach to show abelian is to show that a commutator (the 4-product) equals identity, as this is the same as $ab$ commuting. Just write it down. –  gnometorule Dec 6 '12 at 16:09
So if I'm trying to prove a group is abelian, if I show that, for some elements a,b,c,d, (abcd) = Identity, then that automatically proves commutatitivity? –  Kaish Dec 6 '12 at 16:14
Of course not: notice the special form $aba^{-1}b^{-1}=e$. Use $(ab)^{-1}=b^{-1}a^{-1}$, and right-multiply. –  gnometorule Dec 6 '12 at 16:18

Let $g, h \in G$. Since $g = g^{-1}$ and $h = h^{-1}$, $ghg^{-1}h^{-1} = (gh)^2 = 1$. Hence $gh = hg$.

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How do you get that last line? How does $(gh)^2 = 1$ show that gh = hg? –  Kaish Dec 6 '12 at 16:09
@Kaish Because $gh(hg)^{-1} = ghg^{-1}h^{-1} = (gh)^2 = 1$. –  Makoto Kato Dec 6 '12 at 17:00

Show $gh=k$, all 3 different, from conditions. So $hg=m$ as well. Multiply the 2 equations; note what follows for $k, m$; use conditions given again; and conclude.

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