[proofplan] The stabiliser of the base coset consists exactly of those group elements whose left multiplication leaves the coset $H$ unchanged. We prove the equality by two inclusions. Elements of $H$ stabilise $H$ because multiplying the subgroup $H$ on the left by one of its own elements leaves $H$ unchanged; conversely, if $gH = H$, then $g = ge$ lies in $gH = H$ because $e \in H$. [/proofplan] [step:Identify the stabiliser condition for the base coset] Let $e \in G$ denote the identity element of $G$. Since $H \le G$, the subgroup axiom gives $e \in H$, and hence the base coset satisfies $H = eH$. By the definition of the [stabiliser](/page/Stabiliser) for the given [group action](/page/Group%20Action), the stabiliser of the point $H \in G/H$ is \begin{align*} G_H = \{g \in G : g \cdot H = H\}. \end{align*} For the left action on left cosets, $g \cdot H = gH$. Therefore \begin{align*} G_H = \{g \in G : gH = H\}. \end{align*} [/step] [step:Show every element of $H$ fixes the base coset] Let $h \in H$. We prove $hH = H$ as subsets of $G$. First, if $x \in hH$, then by the definition of a [left coset](/page/Left%20Coset), there exists $k \in H$ such that $x = hk$. Since $h \in H$, $k \in H$, and $H$ is closed under the group multiplication inherited from $G$, we have $hk \in H$. Hence $x \in H$, so $hH \subset H$. Conversely, if $y \in H$, then $h^{-1} \in H$ because $H$ is closed under inverses. Since $h^{-1}y \in H$, we may write \begin{align*} y = h(h^{-1}y), \end{align*} which belongs to $hH$. Hence $H \subset hH$. Thus $hH = H$. Since $h \in H$ was arbitrary, every element of $H$ belongs to $G_H$, and therefore \begin{align*} H \subset G_H. \end{align*} [guided] Let $h \in H$. To prove that $h$ stabilises the base coset, we must prove that the left coset $hH$ is equal to $H$. Equality of sets means two inclusions. For the first inclusion, take an element $x \in hH$. By definition of the left coset $hH$, there is an element $k \in H$ such that \begin{align*} x = hk. \end{align*} Because $H$ is a subgroup of $G$, it is closed under the multiplication of $G$. Since both $h$ and $k$ lie in $H$, their product $hk$ lies in $H$. Therefore $x \in H$, and we have shown \begin{align*} hH \subset H. \end{align*} For the reverse inclusion, take an element $y \in H$. We want to express $y$ as $h$ multiplied by an element of $H$, because that is exactly what it means for $y$ to lie in $hH$. Since $h \in H$ and $H$ is closed under inverses, $h^{-1} \in H$. Since $h^{-1} \in H$ and $y \in H$, closure under multiplication gives $h^{-1}y \in H$. Now \begin{align*} y = e y = (hh^{-1})y = h(h^{-1}y), \end{align*} using the identity law, the inverse law, and associativity in $G$. Since $h^{-1}y \in H$, this representation shows $y \in hH$. Hence \begin{align*} H \subset hH. \end{align*} Combining the two inclusions gives $hH = H$. Therefore $h \cdot H = H$ under the left action on $G/H$, so $h \in G_H$. Since the choice of $h \in H$ was arbitrary, we conclude \begin{align*} H \subset G_H. \end{align*} [/guided] [/step] [step:Show every stabilising element lies in $H$] Let $g \in G_H$. From the description of the stabiliser established above, $gH = H$. Since $e \in H$, the element $ge$ belongs to $gH$. By the identity law in $G$, $ge = g$. Hence $g \in gH$. Since $gH = H$, it follows that $g \in H$. Thus every element of $G_H$ lies in $H$, so \begin{align*} G_H \subset H. \end{align*} [/step] [step:Combine the two inclusions] The previous two steps established \begin{align*} H \subset G_H \end{align*} and \begin{align*} G_H \subset H. \end{align*} By equality of sets through mutual inclusion, we obtain \begin{align*} G_H = H. \end{align*} This is the stabiliser of the base coset for the left action of $G$ on $G/H$. [/step]