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Given a subgroup $H\leq G$, for $G$ finite. Let $G$ act on the left cosets of $H$ by left multiplication. What is the kernel of this group action?

Since the left cosets are themselves constructed by left multiplication, wouldn't the kernel of this map just be equal to $H$? Am I missing something?

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2 Answers 2

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Consider the action of $g$ on $g'H$. Then: $g$ acts trivially on $g'H$ for every $g' \in G$ $\leftrightarrow $ $gg'H=g'H$ for every $g'$. Taking $g'=$identity element, you see $g \in H$. But unless $H$ is normal, $h \in H$: $hg'H \ne g'H$ does not hold for all $h \in H$ and $g'\in G$. For example take $H=\{\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix},n \in \Bbb N\}$, $h=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$,$g'=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Then $g'^{-1}hg'Id_n=\begin{pmatrix} -5 & -8 \\ 9/2 & 7 \end{pmatrix} \notin H$.

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Could you perhaps recommend a small example where I can see that $hg'H\neq g'H$? So that I might gain some intuition for this. –  Ron Jeremy Nov 25 '12 at 20:52
ok I see what you mean, thanks! –  Ron Jeremy Nov 25 '12 at 21:16

The kernel is what is called the core of $\,H\,$ = the largest normal subgroup of $\,G\,$ contained in the subgroup $\,H\,$ , and it equals

$$cor(H):=\bigcap_{g\in G}H^g=\bigcap_{g\in G}g^{-1}Hg$$

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