Let G be any group, and $\displaystyle N \triangleleft G$. Show that $\displaystyle aba^{-1}b^{-1}\in N, \forall a,b\in N \Longleftrightarrow G/N $is abelian
Let G be any group, and $\displaystyle N \triangleleft G$. Show that $\displaystyle aba^{-1}b^{-1}\in N, \forall a,b\in N \Longleftrightarrow G/N $is abelian
Hint: Prove that $\displaystyle aba^{-1}b^{-1}=1 \forall a,b \in G \Leftrightarrow G$ is abelian.
do you mean $\displaystyle aba^{-1}b^{-1}=e$
what is the reason?
1 and e are both commonly used to denote the identity element.
What does this equality tell you? By definition, $\displaystyle ab=ba \: \forall a, b \in G \Leftrightarrow G$ is abelian. Can you get this from the equation I gave you above?