**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.1**

Solution: First we prove that $\mathsf{stab}_G(b) = g \mathsf{stab}_G(a) g^{-1}$.

($\subseteq$) If $x \in \mathsf{stab}_G(b)$, then $x \cdot b = b$. Now $$g^{-1}xg \cdot a = g^{-1}x \cdot b = g^{-1} \cdot b = a,$$ so that $g^{-1}xg \in \mathsf{stab}_G(a)$. Hence $x \in g \mathsf{stab}_G(a) g^{-1}$.

($\supseteq$) Let $x \in g \mathsf{stab}_G(a) g^{-1}$. Then $$gxg^{-1} \cdot b = gx \cdot a = g \cdot a = b,$$ so that $gxg^{-1} \in \mathsf{stab}_G(b)$. Hence $g \mathsf{stab}_G(a) g^{-1} \subseteq \mathsf{stab}_G(b)$.

Suppose now that the action of $G$ on $A$ is transitive; that is, for all $a,b \in A$, there exists $g \in G$ such that $b = g \cdot a$. We show that, if $K$ is the kernel of the action, $K = \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$.

($\subseteq$) Let $x \in K$. Then $x \cdot a = a$ for all $a \in A$. Let $g \in G$. Then $$g^{-1}xg \cdot a = g^{-1} \cdot (x \cdot (g \cdot a)) = g^{-1} \cdot (g \cdot a) = g^{-1}g \cdot a = 1 \cdot a = a,$$ so that $g^{-1}xg \in \mathsf{stab}_G(a)$ for all $g \in G$. Thus $x \in g \mathsf{stab}_G(a) g^{-1}$ for all $g \in G$, hence $x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$.

($\supseteq$) Let $x \in \bigcap_{g \in G} g \mathsf{stab}_G(a) g^{-1}$, and let $b \in A$. Because the action of $G$ on $A$ is transitive, we have $b = h \cdot a$ for some $h \in G$. Now $x = hyh^{-1}$ for some $y \in \mathsf{stab}_G(a)$, thus $$x \cdot b = hyh^{-1} \cdot b = hy \cdot a = h \cdot a = b.$$ Hence $x$ stabilizes $b$; since $b \in A$ is arbitrary, $x \in K$.