Show $\phi (a)=\phi (x)$ iff $a^{-1}x \in N$ iff $aN=xN$

Question

disclaimer: This is not a homework question, it's purely a question to reinforce my understanding:

Let $\phi :G \rightarrow H$ be a homomorphism of groups with kernel N.

$ \forall a,x \in G$ show that:

$\phi (a)=\phi (x)$ iff $a^{-1}x \in N$ iff $aN=xN$

here $aN$ denotes { $an: n \in N$ }

I would like to know how to show this but im getting stuck. I've done some work, but im not getting anywhere.

Thanks!

Answer 1

$$\phi(ax^{-1})=\phi(a)\phi(x^{-1})=\phi(a)\phi(x)^{-1}=\phi(a)\phi(a)^{-1}=e_H$$

By definition this means the argument on the far left is in $N$, i.e. $ax^{-1}\in N$.

For the last equivalence, note that $ax^{-1}\in N$ is an equivalence relation.

• Reflexivity: $aa^{-1}\in N\iff e_G\in N$ but $N$ is a subgroup, hence $e_G\in N$ by assumption.

• Symmetry: $ax^{-1}\in N\implies (ax^{-1})^{-1}=xa^{-1}\in N$ since $N$ is a group and is therefore closed under inverses.

• Transitivity: $ax^{-1}\in N, xy^{-1}\in N\implies (ax^{-1})(xy^{-1})=a(xx^{-1})y^{-1}=ay^{-1}\in N$ since $N$ is a subgroup, hence is closed under the group operation.