Exercise 4.1.1 (Kernel of an action)

Question

Let $G$ act on the set $A$. Prove that if $a,b \in A$ and $b = g \cdot a$ for some $g \in G$, then $G_b = gG_ag^{-1}$ ($G_a$ is the stabilizer of $a$). Deduce that if $G$ acts transitively on $A$ then the kernel of the action is $\bigcap_{g\in G} gG_ag^{-1}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $G$ act on the set $A$, and let $a,b \in A$ such that $b = g \cdot a$ for some $g \in G$.
Then, for all $h \in G$,
\begin{align*}
h \in G_b &\iff h \in G_{g\cdot a}\
&\iff h \cdot (g\cdot a)= g \cdot a \
&\iff (g^{-1} h g) \cdot a = a\
&\iff g^{-1} h g \in G_a\
&\iff h \in gG_ag^{-1}.
\end{align*}
Therefore
\begin{align}
G_b = gG_ag^{-1}.
\end{align}

\bigskip
Let $\varphi : G 	o S_A$ denote the associate homomorphism of the action, i.e., $\varphi(g)(x) = g\cdot x$ for all $x \in A$.



Then, for all $h \in G$,
\begin{align*}
h \in \ker(\varphi) & \iff \varphi(h) = \mathrm{Id}_A\
&\iff \forall b \in A, \ \varphi(h)(b) = b\
&\iff \forall b\in A, \ h\cdot b= b\
&\iff h \in \bigcap_{b\in A} G_b,
\end{align*}
so (as seen p. 113)
$$\ker(\varphi) = \bigcap_{b\in A} G_b.$$

Let $a$ be a fixed element in $A$ (this is possible, since $A 
e \varnothing$). Since $G$ acts transitively on $A$, every $b \in A$ is in the orbit of $a$, so $b = g \cdot a$ for some $g \in G$. Therefore
$$\{G_b \mid b \in A\} = \{G_{ga} \mid g \in G\}$$
(with some repetitions, since distinct $g$ can produce the same $b = g\cdot a$, but this two sets are equal: every $G_b$ is equal to $G_{ga}$ for some $g \in G$, and conversely, every $G_{ga}$ is equal to $G_b$, where $gb = g\cdot a \in A$).

 So
$$\ker(\varphi) = \bigcap_{b\in A} G_b = \bigcap_{g \in G} G_{g\cdot a}.$$

By equality (1), $G_{g\cdot a} =  g G_a g^{-1}$. Hence
$$\ker(\varphi) =  \bigcap_{g \in G} g G_a g^{-1}.$$
\end{proof}

Exercise 4.1.1 (Kernel of an action)

Answers

Comments

Exercise 4.1.1 (Kernel of an action)

Answers

Comments

Add answer