1. ## Group

Hi people,

(G;T) is a group , "e" is it's identity element.

I must show that if $\ \ (\forall x \in G) x^2=e$, (G;T) is an abelian group.

2. Originally Posted by bhitroofen01
Hi people,

(G;T) is a group , "e" is it's identity element.

I must show that if $\ \ (\forall x \in G) x^2=e$, (G;T) is an abelian group.

Hint: $ab=ba \Leftrightarrow aba^{-1}b^{-1} = e$, and note that $(ab)^2=e$.

3. Originally Posted by bhitroofen01
Hi people,

(G;T) is a group (e is it's identity element),

I must show that : $\ (\forall x \in G)\ \ x^2=e \Longrightarrow (G;T)$ is commutative.

Hint: $\forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy$ ...

Tonio

4. $\forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=yxyx=(yx)^2$

5. Originally Posted by bhitroofen01
$\forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=yxyx=(yx)^2$

How do you justify the 3rd equality above?? Think: $x^2=e\Longleftrightarrow x=x^{-1}$

Tonio

6. $\forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=xyy^{-1}x^{-1}$

7. But why $aba^{-1}b^{-1} = e \Leftrightarrow ab=ba$

8. Originally Posted by bhitroofen01
But why $aba^{-1}b^{-1} = e \Leftrightarrow ab=ba$
$aba^{-1}b^{-1}\implies ab=e\left(a^{-1}b^{-1}\right)^{-1}=e\left(ba\right)=ba$