Exercise 1.7.4 (Kernel of the action and stabilizers)

Proof. Let $G$ be a group acting on a set $A$ and fix some $a\in A$.

(a)

By definition (see Example 1 p. 43), the kernel of the action is the set $$K=\{g\in G\mid \forall <!--\; FUNCTION\; APPLICATION\; -->a\in A,g\cdot a=a\}.$$

Consider the map

$$\phi \{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & S_A\hfill \\ \hfill g\hfill & \hfill \mapsto \hfill & {\phi }_g,\hfill \end{array}\text{where}{\phi }_g\{\begin{array}{ccc}\hfill A\hfill & \hfill \to \hfill & A\hfill \\ \hfill a\hfill & \hfill \mapsto \hfill & {\phi }_g(a)=g\cdot a.\hfill \end{array}$$

It is proved in the text (p. 42) that $\phi$ is a homomorphism.

Note that $K=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$: for all $g\in G$,

$$\begin{array}{llll}\hfill g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& ⟺\phi (g)={\mathrm{Id}}_A& \hfill & \\ \hfill & ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->a\in A,g\cdot a=a& \hfill & \\ \hfill & ⟺g\in K.& \hfill & \end{array}$$

Therefore

$$K=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ).$$

By Exercise 1.6.14, this kernel $K$ is a subgroup of $G$.

(b)

Put $G_a=\{g\in G\mid g\cdot a=a\}$ (the stabilizer of $a$ in $G$). Then

• Since $1_G\cdot a=a$, $1_A\in G_a$, so $G_a\ne \varnothing$.
• If $g,h\in G_a$, $g\cdot a=a$ and $h\cdot a=a$, thus $(\mathit{gh})\cdot a=g\cdot (h\cdot a)=g\cdot a=a$, so $\mathit{gh}\in G_a$.
• If $g\in G_a$, then $a=g\cdot a$, thus $g^{-1}\cdot a=g^{-1}\cdot (a\cdot a)=(g{g}^{-1})\cdot a=1_G\cdot a=a$. So $g^{-1}\in G_a$.

This shows that $G_a$ is a subgroup of $G$.

Note that $K=\underset{a\in A}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}{G}_a$. This gives a new proof that $K$ is a subgroup of $G$.