Hi people, (G;T) is a group , "e" is it's identity element. I must show that if $\displaystyle \ \ (\forall x \in G) x^2=e$, (G;T) is an abelian group. Can you help me please???
Follow Math Help Forum on Facebook and Google+
Originally Posted by bhitroofen01 Hi people, (G;T) is a group , "e" is it's identity element. I must show that if $\displaystyle \ \ (\forall x \in G) x^2=e$, (G;T) is an abelian group. Can you help me please??? Hint: $\displaystyle ab=ba \Leftrightarrow aba^{-1}b^{-1} = e$, and note that $\displaystyle (ab)^2=e$.
Originally Posted by bhitroofen01 Hi people, (G;T) is a group (e is it's identity element), I must show that :$\displaystyle \ (\forall x \in G)\ \ x^2=e \Longrightarrow (G;T)$ is commutative. Can you help me please??? Hint: $\displaystyle \forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy$ ... Tonio
$\displaystyle \forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=yxyx=(yx)^2$
Originally Posted by bhitroofen01 $\displaystyle \forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=yxyx=(yx)^2$ How do you justify the 3rd equality above?? Think: $\displaystyle x^2=e\Longleftrightarrow x=x^{-1}$ Tonio
$\displaystyle \forall\,x,y\in G\,,\,\,e=(xy)^2=xyxy=xyy^{-1}x^{-1}$
But why $\displaystyle aba^{-1}b^{-1} = e \Leftrightarrow ab=ba$
Originally Posted by bhitroofen01 But why $\displaystyle aba^{-1}b^{-1} = e \Leftrightarrow ab=ba$ $\displaystyle aba^{-1}b^{-1}\implies ab=e\left(a^{-1}b^{-1}\right)^{-1}=e\left(ba\right)=ba$
View Tag Cloud