Problem statement:
Let $G$ be any group and recall $\mathcal{S}(G)$ denotes the set of subgroups of $G$. Let $G \times \mathcal{S}(G) \to \mathcal{S}(G)$ be given by $$g \cdot H = gHg^{-1}$$ Prove that this is an action. Prove that $stab_G(H) = N(H)$ the normalizer of $H$ and for which groups $G$ is this action transitive?
- Let $e \in G$ be the identity element. Then we have $$e \cdot H = eHe^{-1} = eHe = H$$ Let $g,k \in G$ then we have $$g \cdot (k \cdot H) = g \cdot (kHk^{-1}) = gkHk^{-1}g^{-1} = gkH(gk)^{-1} = gk \cdot H$$ Thus $g \cdot H = gHg^{-1}$ is an action.
- Now, the normalizer of $H$ is defined to be $$N(H) = \{g \in G \mid gHg^{-1} = H \}$$ The stabilizer of $H$ in $G$ is given by $$\text{stab}_G(H) = \{g \in G \mid g \cdot H = H \} = \{ g \in G \mid gHg^{-1} = H \}$$ Hence by inspection of these definition we conclude they are equal.
- This action is transitive only for abelian groups. This is because any subgroup of an abelian group is normal and hence for any $H \in \mathcal{S}(G)$ we have $$gHg^{-1} = H\ \forall g \in G$$ Thus, $\forall H \in \mathcal{S}(G)$, $\exists g \in G$ such that $$g \cdot H = gHg^{-1} = H$$.
How does this look? Particularly parts $2$ and $3$. Part $2$ just seemed too simple and for part $3$ I believe I am correct but could be overlooking something or missing some nuance.
Parts 1 and 2 are correct, but part 3 is totally wrong. I don't think you have the correct definition of transitive.