The Stacks project

Proof. The implications (5) $\Rightarrow $ (4) and (2) $\Rightarrow $ (5) are clear.

Assume (3). Let $X$ be a connected object of $\mathcal{C}$ and let $*' \to H(X)$ be a morphism. Since $H(X)$ is connected by (3) we see that $*' \to H(X)$ is an isomorphism. Hence the $G'$-set corresponding to $H(X)$ has exactly one element, which means the $G$-set corresponding to $X$ has one element which means $X$ is isomorphic to the final object of $\mathcal{C}$, in particular there is a map $* \to X$. In this way we see that (3) $\Rightarrow $ (4).

If (1) is true, then the functor $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}$ is fully faithful: in this case a map of $G$-sets commutes with the action of $G$ if and only if it commutes with the action of $G'$. Thus (1) $\Rightarrow $ (2).

If (1) is true, then for a $G$-set $X$ the $G$-orbits and $G'$-orbits agree. Thus (1) $\Rightarrow $ (3).

To finish the proof it suffices to show that (4) implies (1). If (1) is false, i.e., if $h$ is not surjective, then there is an open subgroup $U \subset G$ containing $h(G')$ which is not equal to $G$. Then the finite $G$-set $M = G/U$ has a transitive action but $G'$ has a fixed point. The object $X$ of $\mathcal{C}$ corresponding to $M$ would contradict (3). In this way we see that (3) $\Rightarrow $ (1) and the proof is complete. $\square$

Proof. We may replace $H$ and $H'$ by the canonical functors $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$. Then we are saying that the action of $G''$ on every $G$-set is trivial if and only if the homomorphism $G'' \to G$ is trivial. This is clear. $\square$

Proof. By Lemmas 58.4.1 and 58.4.2 we may assume that $H$ is fully faithful, $h$ is surjective, $H' \circ H$ maps objects to disjoint unions of the final object, and $h \circ h'$ is trivial. Let $N \subset G'$ be the smallest closed normal subgroup containing the image of $h'$. It is clear that $N \subset \mathop{\mathrm{Ker}}(h)$. We may assume the functors $H$ and $H'$ are the canonical functors $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$.

Suppose that (2) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, the action of $G'$ factors through $G$. Apply this to $X' = G'/U'N$ where $U'$ is a small open subgroup of $G'$. Then we see that $\mathop{\mathrm{Ker}}(h) \subset U'N$ for all $U'$. Since $N$ is closed this implies $\mathop{\mathrm{Ker}}(h) \subset N$, i.e., (1) holds.

Suppose that (1) holds. This means that $N = \mathop{\mathrm{Ker}}(h)$. Let $X'$ be a finite $G'$-set such that $G''$ acts trivially. This means that $\mathop{\mathrm{Ker}}(G' \to \text{Aut}(X'))$ is a closed normal subgroup containing $\mathop{\mathrm{Im}}(h')$. Hence $N = \mathop{\mathrm{Ker}}(h)$ is contained in it and the $G'$-action on $X'$ factors through $G$, i.e., (2) holds.

Suppose that (3) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, there is a surjection of $G'$-sets $X \to X'$ where $X$ is a $G$-set. Clearly this means the action of $G'$ on $X'$ factors through $G$, i.e., (2) holds.

The implication (2) $\Rightarrow $ (3) is immediate. This finishes the proof. $\square$

Proof. We may replace $H'$ by the corresponding functor between the categories of finite $G'$-sets and finite $G''$-sets.

Assume $h' : G'' \to G'$ is injective. Let $H'' \subset G''$ be an open subgroup. Since the topology on $G''$ is the induced topology from $G'$ there exists an open subgroup $H' \subset G'$ such that $(h')^{-1}(H') \subset H''$. Then the desired diagram is

Conversely, assume (2) holds for the functor $\textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$. Let $g'' \in \mathop{\mathrm{Ker}}(h')$. Pick any open subgroup $H'' \subset G''$. By assumption there exists a finite $G'$-set $X'$ and a diagram

of $G''$-sets with the left arrow surjective and the right arrow injective. Since $g''$ is in the kernel of $h'$ we see that $g''$ acts trivially on $X'$. Hence $g''$ acts trivially on $Y''$ and hence trivially on $G''/H''$. Thus $g'' \in H''$. As this holds for all open subgroups we conclude that $g''$ is the identity element as desired. $\square$

Proof. This translates into the following statement for the continuous group homomorphism $h' : G'' \to G'$: the image of $h'$ is normal if and only if every open subgroup $U' \subset G'$ which contains $h'(G'')$ also contains every conjugate of $h'(G'')$. The result follows easily from this; some details omitted. $\square$