Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ a group, $H \le G$ and $A= G/H $. Then there exists an action $\phi: G\rightarrow S_A$ such that the kernel is the maximum subgroup normalized by $G$ and contained in $H$.

share|cite|improve this question
Well, what is the simplest operation you can imagine that takes an element of $G$ and a coset in $G/H$ to produce another coset? – Hagen von Eitzen Sep 4 '12 at 6:06
up vote 1 down vote accepted

the homomorfism is well defined for $g\in G$: $\phi_g (aH)=gaH $ for $ aH\in G/H$.

$$\begin{align*} \ker\phi_g&= \{g\in G:\phi_g(aH)= aH\ \forall a\in G \}\\ &=\{g\in G: gaH=aH\ \forall a\in G\}\\ &=\{g\in G:a^{-1}ga \in H\ \forall a\in G\}\\ &=\bigcap\limits_{x\in G}aHa^{-1} \end{align*}$$

next any normal subgroup is in $\ker\phi_g$

share|cite|improve this answer
You still have to show that $\bigcap_{x\in G}aHa^{-1}$ is normal in $G$, and that if $N\le H$ is normal in $G$, then $N\subseteq\bigcap_{x\in G}aHa^{-1}$. – Brian M. Scott Sep 4 '12 at 7:56
but a theorem said that $ker\phi_g$ always is normal at group, and containment results from $N\le G$ – Camilo Acevedo. Sep 4 '12 at 9:57
I’ll grant that it’s pretty obvious that $\ker\varphi_t$ is normal in $G$, but you have to do a little work to justify the claim that if $N\le H$ is normal in $G$, then $N\subseteq_{x\in G}aHa^{-1}$. – Brian M. Scott Sep 4 '12 at 19:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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