Vtyjkyuk

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

This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.

(Please note that $e$ in the question is the group's identity.)

Here's my attempt though...

First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...

So, I begin by trying to play around with the elements of the group based on their definition...

$$(g_2g_1)^r=e$$
$$(g_2g_1g_2g_2^-1)^r=e$$
$$(g_2g_1g_2g_2^-1g_2g_1g_2g_2^-1...g_2g_1g_2g_2^-1)=e$$

I assume that the $g_2^-1$'s and the $g_2$'s cancel out so that we end up with something like,

$$g_2(g_1g_2)^rg_2^-1=e$$
$$g_2^-1g_2(g_1g_2)^r=g_2^-1g_2$$

Then ultimately...

$$g_1g_2=e$$

I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.

Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.

Hint: Take $(ab)^2=1$ and multiply both sides on the right with $b$, then again on the right with $a$.

For any $g, h in G$, consider the element $gcdot hcdot hcdot g.~$
Since $g^2 = gcdot g= e$ for all $g in G$, we find that
$$gcdot hcdot hcdot g = gcdot(hcdot h)cdot g = gcdot ecdot g = gcdot g = e.$$
But, $gcdot h$ has unique inverse element $gcdot h$, while we have just proved that $(gcdot h)cdot (hcdot g) = e$, and so it must be that $gcdot h = hcdot g$ for all $g, h in G$, that is, $G$ is an abelian group.

Whenever you have a condition $g^2=e$ in a group, it's equivalent to $g=g^-1$ (multiply both sides by $g^-1$).

In this case, it applies to every element of the group, so you can add or remove inverses from any expression freely. So the proof is simply $$ab=(ab)^-1=b^-1a^-1=ba.$$

Hint: Note that $g_1g_2=g_2g_1$ if and only if $g_1g_2g_1^-1g_2^-1=e$ (Why?), and that $g^-1=g$ for all $gin G$ (Why?).

Proof: let for all $a,b$ in group $G$.
claim that To show $ab=ba$ a commutative.
By using a fact that $acdot a=bcdot b=(ab)cdot(ab)=e$.
since $(ab)^2=a^2cdot b^2=ecdot e=e$. We have $abcdot ab=e$. Multiplying on the right by $ba$, we obtain
beginalign
abcdot abcdot ba &= ecdot ba\
abcdot a(bcdot b)cdot a &= ba\
abcdot acdot b^2cdot a &=\
abcdot acdot ecdot a &=\
abcdot acdot a &=\
abcdot e &=\
ab &= ba,
endalign
for all $a,b$ in $G$. since $G$ is abelian group. This is proved last.

Another proof is by contradiction.

Let G be a group with operation *. You want to show that:
$(forall g in G:g^2=e)implies(Gtext AbelianLeftrightarrow forall x,y in G: x*y = y*x)$.

(where $g^2$ is shorthand for $g*g$)

Suppose by contradiction that the group is not Abelian, i.e. that ($exists x,yin G: x*yneq y*x)$. Now multiply on the left by $x$ and on the right by $y$.

You get $x^2*y^2 ne (xy)^2$. But then it means that $e*e neq e$ which is a contradiction.

$(ab)^2=e
$($because$ $a,b in G, abin G$, due to the closure property of group axiom )$implies
(ab)(ab)=etag 1$
Pre multiply $a$ on both sides also post multiply $b$ on both sides. The equation (1) becomes,
$aababb=abtag2$($because$ by associativity and self invertible property.
$ba=abtag3$

Then for all $a,b in G$: $$ab = (bb)ab(aa) = b(baba)a = ba.$$