[proofplan] We prove equality of the two subsets of $G$ by proving both inclusions. If an element fixes the translated point $g \cdot x$, conjugating it by $g^{-1}$ gives an element fixing $x$. Conversely, if an element fixes $x$, then conjugating it by $g$ gives an element fixing $g \cdot x$. [/proofplan] [step:Show that every element fixing $g \cdot x$ lies in $gG_xg^{-1}$] Let $h \in G_{g \cdot x}$. By definition of the stabilizer, \begin{align*} h \cdot (g \cdot x) = g \cdot x. \end{align*} Using associativity of the [group action](/page/Group%20Action), this is \begin{align*} (hg) \cdot x = g \cdot x. \end{align*} Acting on both sides by $g^{-1}$ gives \begin{align*} (g^{-1}hg) \cdot x = x. \end{align*} Thus $g^{-1}hg \in G_x$. Therefore there exists $k \in G_x$, namely $k = g^{-1}hg$, such that \begin{align*} h = gkg^{-1}. \end{align*} Hence $h \in gG_xg^{-1}$, and so \begin{align*} G_{g \cdot x} \subseteq gG_xg^{-1}. \end{align*} [guided] We want to prove the inclusion \begin{align*} G_{g \cdot x} \subseteq gG_xg^{-1}. \end{align*} So take an arbitrary element $h \in G_{g \cdot x}$. The meaning of this membership is that $h$ fixes the point $g \cdot x$ under the action: \begin{align*} h \cdot (g \cdot x) = g \cdot x. \end{align*} The group action satisfies the compatibility law \begin{align*} a \cdot (b \cdot y) = (ab) \cdot y \end{align*} for all $a,b \in G$ and all $y \in X$. Applying this law with $a = h$, $b = g$, and $y = x$, we rewrite the left-hand side as \begin{align*} (hg) \cdot x = g \cdot x. \end{align*} Now we act on both sides by $g^{-1}$. On the left, the action law gives \begin{align*} g^{-1} \cdot ((hg) \cdot x) = (g^{-1}hg) \cdot x. \end{align*} On the right, the same action law and the identity property give \begin{align*} g^{-1} \cdot (g \cdot x) = (g^{-1}g) \cdot x = e \cdot x = x, \end{align*} where $e \in G$ is the identity element. Therefore \begin{align*} (g^{-1}hg) \cdot x = x. \end{align*} This says exactly that $g^{-1}hg \in G_x$. Define $k := g^{-1}hg$. Then $k \in G_x$, and multiplying the equality $k = g^{-1}hg$ on the left by $g$ and on the right by $g^{-1}$ gives \begin{align*} h = gkg^{-1}. \end{align*} Thus $h$ has the required form with $k \in G_x$, so $h \in gG_xg^{-1}$. Since $h \in G_{g \cdot x}$ was arbitrary, we conclude \begin{align*} G_{g \cdot x} \subseteq gG_xg^{-1}. \end{align*} [/guided] [/step] [step:Show that every conjugate of an element fixing $x$ fixes $g \cdot x$] Let $h \in gG_xg^{-1}$. Then there exists $k \in G_x$ such that \begin{align*} h = gkg^{-1}. \end{align*} Since $k \in G_x$, we have \begin{align*} k \cdot x = x. \end{align*} Using associativity of the group action and $g^{-1}g = e$, where $e \in G$ is the identity element, we compute \begin{align*} h \cdot (g \cdot x) &= (gkg^{-1}) \cdot (g \cdot x) \\ &= (gkg^{-1}g) \cdot x \\ &= (gk) \cdot x \\ &= g \cdot (k \cdot x) \\ &= g \cdot x. \end{align*} Therefore $h \in G_{g \cdot x}$. Hence \begin{align*} gG_xg^{-1} \subseteq G_{g \cdot x}. \end{align*} [/step] [step:Combine the two inclusions to identify the stabilizer] The two inclusions proved above give \begin{align*} G_{g \cdot x} \subseteq gG_xg^{-1} \quad\text{and}\quad gG_xg^{-1} \subseteq G_{g \cdot x}. \end{align*} By extensional equality of subsets of $G$, \begin{align*} G_{g \cdot x} = gG_xg^{-1}. \end{align*} This is the desired [stabilizer conjugation formula](/theorems/5002). [/step]