If $g^2 = e$ for all $g \in G$, then $G$ is abelian

Question

Let $G$ be a group. Prove that $g^2 = e$ for all $g \in G$, then $G$ is abelian. ($e$ is the identity element.)

My Solution: Let $a,b \in G$. Then $a(ab)b = a^2b^2 = e^2 =e$. Now I tried to reverse $ab$ in the brackets to get the same solution to show that the group is also commutative but I was not able to do so.

This is one of the most duplicated group questions I know of. @Ron.J.Adams Please use the search function. I'm 100% certain you'd have found an answer before you posted your question. — rschwieb, Dec 29 '14 at 21:52
What @rschwieb said. And almost everytime an answer with cancellation is given instead of the more aesthetic $ab = a(abab)b=(aa)ba(bb)=ba$. — Myself, Dec 29 '14 at 21:53

score 8 · Answer 1 · answered Dec 29 '14 at 21:43

8

Let $a,b \in G$ be arbitrary elements. Notice:

$$abab = (ab)^2 = e = a^2 b^2 = aabb$$

Cancellation gives $ba = ab$.

answered Dec 29 '14 at 21:43

ILoveMath

10,999

What I don't get is how $(ab)^2 = e$. – George J. Adams Dec 29 '14 at 21:47
Because $ab$ is in the group since $a,b$ are in the group. And the property $g^2 = e$ holds foe very element of $g$. In particular, it must hold for $g = ab$ – ILoveMath Dec 29 '14 at 21:48
I didn't know that you could do this. So like shouldn't G have the multiplicative property for this. – George J. Adams Dec 29 '14 at 21:51
$$a,b \in G \implies ab \in G$$ – ILoveMath Dec 29 '14 at 21:51
@Ron.J.Adams what is this "multiplicative property"? Does a group not satisfy it by definition? – Ben Grossmann Dec 29 '14 at 23:29

score 1 · Answer 2 · answered Dec 29 '14 at 21:47

1

we have that $ab ab= e$. Thus (multiply by $b$ two sides) $abab^2=b$ or $aba=b$. In the same way $aba^2=ba$ or $ab=ba.$

answered Dec 29 '14 at 21:47

Leox

8,354

If $g^2 = e$ for all $g \in G$, then $G$ is abelian

2 Answers2