Let $G$ be a group with subgroup $H$ of finite index. Let $X$ be a $G$-set, (ie. G acts on $X$), then we can define the tensor $G$-set $$G \otimes_H X := (G \times X) / \simeq $$ where the equivalence relation is defined as $(gh,x) \simeq (g,hx)$ for all $g \in G, h \in H, x \in X$. There is a $G$-action on this set: $$ g' (g, x) := (g'g, x)$$
I am trying to verify the following statement: if $X$ has finite $G$-stabilizers, then so does $G\otimes_H X$. Here is what I've done so far:
Suppose for contradiction that $(g, x) \in G \otimes_H X$ has infinitely many $G$-stabilizers. Since $H \subset G$ is of finite index, we can choose some finite transversal $\left \{ s_1, s_2, … , s_n \right \} \subset G$, so that for any $g \in G$, there exists some $s_k$ and $h \in H$ such that $g = s_k h$. Since the cosets partition $G$, then each $G$-stabilizer of $(g, x)$ lies in some coset $s_j H$. Furthermore, since there are finitely many cosets and infinitely many $G$-stabilizers, then there must be some $j \leq n$ such that there are infinitely many $G$-stabilizers in the coset $s_jH$. So we have $g_1, g_2, g_3, … \in s_jH$ such that $(g_i g, x) \simeq (g, x) $. Furthermore. we can express $g = s_k h$ so that these equivalences become: $ (g_i s_k, hx) \simeq (g_i s_k h, x) \simeq (s_k h, x) \simeq (s_k, hx) $, so replacing $x$ by $hx$, we can make the following statement:
There exists $s_j$ and $s_k$ in the transversal such that there exists $g_1, g_2, g_3, … \in s_j H$ and $(g_i s_k, x) \simeq (s_k, x) \hspace{4pt} \forall i \in \mathbb{N}$
Or equivalently:
there exists $s_j$ and $s_k$ in the transversal such that there exists a sequence $h_1, h_2, h_3, … \in H$ with $$(s_j h_1 s_k, x) \simeq (s_j h_2 s_k, x) \simeq (s_j h_3 s_k, x) \simeq … \simeq (s_k, x)$$
Because $s_k$ can only be expressed as a product of elements in $G$ and $H$ in the trivial way (ie. $s_k = s_k * 1$), it seems as though these elements $ (s_k, x)$ are in some sense $\textit{irreducible}$, and this seems like it could offer up some sort of contradiction, but I'm not really sure how to conclude. Does this look like the right approach in the first place? Any help would be appreciated.
Best Answer
You need to prove that every pair in the tensor product has a finite stabilizer. You have established that every pair up to equivalence has the form $(s_i, x)$. Now suppose that this point has infinite stabilizer $g_1,g_2,...$ Since there are only finitely many cosets we can assume that all $g_k$ are in the same coset $s_jH$, $g_k=s_jh_k$, $k=1,2,...$. then $g_k(s_i, x)=(s_j h_ks_i, x)=(s_is_m, h_k'x)=(s_i,hh_k'x)$ where $h_ks_i=s_mh_k'$ (again we can assume the same $m$ for all $k$) and $s_js_m=s_ih$ for some $h\in H$. But this means that $hh_k'$ stabilizes $x$ in $X$ for all $k$. Since the stabilizer of $x$ is finite, we have many equalities $h_k'=h_l'$, a contradiction.