0
$\begingroup$

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

|cite|improve this question|||||
$\endgroup$
1
$\begingroup$

  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 $xyaH$.

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.