Kernel of an action through left multiplication is

Question

I am trying to prove that the kernel (K) of an action by G through left multiplication on the set of distinct cosets of a subgroup $H$ is normal. I am aware that the map $\phi:G\rightarrow S_{G}$ is a homomorphism and the kernel of this map is the same as that of the action and hence must be normal. But I would prefer to use the following argument: If $x\in G$ and $g\in K$, then for any coset $aH$, $(xgx^{-1})aH=aH$. I also know that $xgx^{-1}\in H$. Notwithstanding, I am unable to manipulate the LHS. Would be grateful if some direction could be given.

Answer 1

We want to evaluate $xgx^{-1}aH$. Since $g$ fixes any coset, $gx^{-1}aH=x^{-1}aH$. Thus $$xgx^{-1}aH=x(gx^{-1}aH)$=x(x^{-1}aH)=aH$$ This completes your argument that the kernel is normal.

Answer 2

The kernel of an action $\mathcal{A}\colon G\times X \to X$ is just the kernel of the equivalent homomorphism $\lambda\colon G\to\operatorname{Sym}(X)$, and the kernel of every homomorphism from $G$ to another group is normal in $G$.

In general, it is given by:

\begin{alignat}{1} \operatorname{ker}\lambda &= \{g\in G\mid\lambda_g=\iota_X\} \\ &= \{g\in G\mid\lambda_g(x)=\iota_X(x), \forall x \in X\} \\ &= \{g\in G\mid g\cdot x=x, \forall x \in X\} \\ &= \{g\in G\mid g\in \operatorname{Stab}(x), \forall x \in X\} \\ &= \{g\in \operatorname{Stab}(x), \forall x \in X\} \\ &= \bigcap_{x\in X}\operatorname{Stab}(x) \\ \tag 1 \end{alignat}

In your case, $X=\{gH, g\in G\}$ and $\operatorname{Stab}(gH)=\{g'\in G\mid g'gH=gH\}$.

Lemma. $\operatorname{Stab}(gH)=gHg^{-1}$.

Proof.

\begin{alignat}{1} g'\in \operatorname{Stab}(gH) &\Rightarrow \exists h,h'\in H\mid g'gh=gh' \\ &\Rightarrow \exists h,h'\in H\mid g'=gh'(gh)^{-1}=gh'h^{-1}g^{-1} \\ &\Rightarrow \exists h''\in H\mid g'=gh''g^{-1} \\ &\Rightarrow g'\in gHg^{-1} \\ \end{alignat}

and thence $\operatorname{Stab}(gH)\subseteq gHg^{-1}$. Viceversa:

\begin{alignat}{1} g'\in gHg^{-1} &\Rightarrow g'g \in gH \\ &\Rightarrow g'gH\subseteq gH \\ \end{alignat}

Now:

\begin{alignat}{1} gH\subseteq g'gH &\iff \forall h \in H, \exists h'\in H\mid gh=g'gh' \\ &\iff \forall h \in H, \exists h'\in H\mid gh=(gh''g^{-1})gh'=gh''h' \\ &\iff h'=(gh'')^{-1}gh=h''^{-1}g^{-1}gh=h''^{-1}h \\ \end{alignat}

Since $h''$ exists by hypothesis ("Viceversa..."), such a $h'$ exists, and then indeed:

$$gH\subseteq g'gH$$

Therefore, $g'\in gHg^{-1} \Rightarrow g'gH=gH \Rightarrow g'\in \operatorname{Stab}(gH) \Rightarrow gHg^{-1}\subseteq \operatorname{Stab}(gH)$.

$\Box$

By $(1)$ and the Lemma, we have finally:

$$\operatorname{\ker}\lambda = \bigcap_{gH\in G/H}\operatorname{Stab}(gH)= \bigcap_{g\in G}\operatorname{Stab}(gH)=\bigcap_{g\in G}gHg^{-1}$$