Questions regarding the existence of normal subgroup with finite index

Question

If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $N$ whose index in $G$ is finite.

I found a proof of the question here: How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$

I have a few questions regarding this proof:

1. The map $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$ takes an argument from $G$. Why do we have $(aH)$ behind $\phi(x)$? What does the notation mean? Can we write $\phi(x)=xG/H$ instead?

2.How do I show that $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$ is a homomorphism? I don't see immediately that $\phi(xy)=\phi(x)\phi(y)$.

Answer 1

1. It means that $\phi(x)$ is a function, and it takes as input $aH$ and returns as output $xaH$.

2. $\phi(xy)$ takes $aH$ and gives $xyaH$. $\phi(x) \phi(y)$ takes $aH$ and first gives $yaH$, then $xyaH$.