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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 $\phi(xy)=\phi(x)\phi(y)$.

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  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 $xyaH$.

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  • $\begingroup$ Thank you! Can I ask one more question? Does it mean that $\ker(\phi(x)):=\lbrace x \mid (xa)H=aH \rbrace$ ? $\endgroup$ – user112358 Dec 3 '15 at 6:50
  • $\begingroup$ @Lewis: you need the appropriate quantifiers. It means the kernel consists of $x$ such that $xaH = aH$ for all $a \in G$. From here, you can show that the kernel is in fact the intersection $\bigcap_{g \in G} gHg^{-1}$ of all conjugates of $H$. $\endgroup$ – Qiaochu Yuan Dec 3 '15 at 6:50

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