Jtdylktuy

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 $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$

Hint: Note that $g_1g_2=g_2g_1$ if and only if $g_1g_2g_1^{-1}g_2^{-1}=e$ (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
begin{align}
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,
end{align}
for all $a,b$ in $G$. since $G$ is abelian group. This is proved last.

Alternatively, the map $$begin{align*}f:G&rightarrow G\x&mapsto x^{-1}(=x)end{align*}$$ is an automorphism of $G$ and so $G$ is Abelian!

Another proof is by contradiction.

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

(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.

By construction:

$quadbegin{align*}
(ab)(ab) &= e = a(bb)a \
require{cancel}cancel{(ab)}(ab) &= cancel{(ab)}(ba) qquadtext{by associativity followed by cancellation}\
ab &= ba
end{align*}$

Hence, the group is Abelian.

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

Let $a,bin G$:
begin{align}
abcdot (ab)^{-1} &= abcdot b^{-1}a^{-1}\
&= abcdot baquadquad text{(since each element in $G$ is self inverse)}\
&= a(b^{2})a\
&= acdot ecdot a\
&= a^{2}\
&=e
end{align}
This shows $(ab)^{-1}=ba$. Now $(ab)^2=e$, so $abcdot ab=abcdot ba=e$, which by the cancellation law gives $ab=ba$ for all $a,bin G$, since $a$ and $b$ were arbitrary. Hence $G$ is abelian as required.

$$xyx^{-1}y^{-1}=(xyx^{-1})^2x^2(x^{-1}y^{-1})^2$$

It is easier to solve. Let be c=a○b, then c○c=e, (a○b)○(a○b)=e.
Then multiply the left and right parts of the right by b and a: (a○b)○(a○b)○b=e○b → (a○b)○(a○b○b)=e○b → (a○b)○(a○e)=e○b → (a○b)○a=b → (a○b)○a○a=b○a → (a○b)○e=b○a. the end.