1
$\begingroup$

Prove that left multiplication defines an action of $G$ on the coset space $\frac{G}{H}$ and that the kernel of the associated homomorphism from $G$ to $\operatorname{Aut}\left(\frac{G}{H}\right)$ is the largest normal subgroup contained in $H$.

Let $\sigma: G \times \frac{G}{H} \rightarrow \frac{G}{H}$ denote the group action and $\tilde \sigma: G \rightarrow$ Aut$\frac{G}{H}$ denote the associated homomorphism. I want to show that ker$(\tilde \sigma)$ is a normal subgroup of $G$ and that for any normal subgroup $N \subseteq H$ that is normal to $G$, $N \subseteq$ ker$(\tilde \sigma)$.

I can show the second part but the first part I am completely stuck. So far I have...

Let $g \in$ ker$\tilde \sigma$ and $kH \in \frac{G}{H}$. Then $\sigma_g(kH)=g \cdot kH=kH$. So in particular, $\sigma_g(H) = gH = H$ So $g \in H$. Since $H$ is a subgroup of $G$, $g^{-1} \in H$ so $gHg^{-1} = H$. So $H$ is normal in $G$. Which is not what I am trying to prove at all...

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $g \in \ker(\sigma)$. Then for all $x\in G$ we have $gxH = xH$ which is equivalent to $x^{-1}gxH = H$ which in turn is equivalent to $x^{-1}gx \in H$ for all $x\in G$. For a given coset $xH$, the isotropy group of the action is precisely $xHx^{-1}$ so the previous statement is saying that $g$ lies in every isotropy group i.e., $g$ is in the kernel of the action if and only if $g \in \cap_{x \in G} xHx^{-1}$. Since this is an intersection of conjugates, it follows that it is a normal subgroup. Not sure about the largest normal subgroup statement.

$\endgroup$
1
  • $\begingroup$ If $\;N\lhd G\;,\;\;N\le H\;$ , then for all $\;x\in G\;$ and for all $\;n\in N\;$ we'd have $$x^{-1}nx\in N\le H\iff nxH=xH\iff n\in \ker\sigma\ldots$$ $\endgroup$
    – DonAntonio
    Commented Mar 13, 2014 at 5:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .