How can a set of fixed points be viewed as a $N_{G}(H)/H$ - set?

Question

Okay, so let $G$ be a group and $H$ - its subgroup. Moreover, let $X$ be a $G$-set. Let's consider a set of fixed points of $H$ on $X$ and let's denote it by $X^{H}$. The author of the paper I'm reading says that it can be viewed as a $N_{G}(H)/H$-set, where $N_{G}(H)$ is a normalizer of $H$ in $G$, but does not provide any proof or explanation. Is this statement even true, and if it is, how to prove or explain that? I don't have the faintest idea how to even approach this.

Answer 1

If $g\in N_G(H)$, $h\in H$, and $x\in X^H$, then you know that $g^{-1}hg(x) = x$ (since $g^{-1}hg\in H$). That means that $h(gx) = gx$, and hence that $gx$ is also fixed by $h$; as $h$ is arbitrary, this shows that $gx\in X^H$. This holds for all $g\in N_G(H)$, so $N_G(H)$ sends $X^H$ to itself. The action of $H$ on that set is trivial, so the action of $N_G(H)$ on $X^H$ factors through $N_G(H)/H$ (which makes sense since $H\triangleleft N_G(H)$. The actions is the action of $G$, restricted to $X^H$.

Answer 2

Suppose $g\in N_G(H)$, $h\in H$, and $x\in X^H$. Then since $g$ normalizes $H$, $hg=gh'$ for some $h'\in H$. We thus have $hgx=gh'x=gx$. So, since $h\in H$ was arbitrary, $gx\in X^H$.

Thus every element of $N_G(H)$ actually maps $X^H$ to itself, and so $N_G(H)$ acts on $X^H$. But every element of $H$ fixes $X^H$, so $H$ is contained in the kernel of this action, and so it induces an action of the quotient group $N_G(H)/H$.

Answer 3

Since two people have already provided concrete answers, an abstract way to see this is to observe that the functor $X \mapsto X^H$ is represented by $G/H$, meaning that $\text{Hom}_G(G/H, X) \cong X^H$; in other words, $G/H$ is the free $G$-set on an $H$-fixed point. This means $X^H$ canonically acquires an action of the automorphism group $\text{Aut}_G(G/H)$ (in fact this group is precisely the group of automorphisms of the $H$-fixed points functor, by the Yoneda lemma).

An automorphism $\varphi : G/H \to G/H$ is in particular a map satisfying

$$\forall g \in G : \varphi(gxH) = g \varphi(xH).$$

Since $G$ acts transitively on $G/H$ such a map is determined by where it sends the coset $eH$ of the identity, which (by the universal property of $G/H$) must be another coset $yH$ fixed by the action of $H$, meaning that

$$\forall h \in H : hyH = yH \Leftrightarrow \forall h \in H : y^{-1} hy \in H$$

which says exactly that $y \in N_G(H)$. Moreover $y$ is only well-defined up to right multiplication by elements of $H$ which says exactly that $yH \in N_G(H)/H$. And conversely every element of $N_G(H)/H$ gives an automorphism of $G/H$, so this is exactly the automorphism group of $G/H$.

The same argument in the linear setting gives that if $V$ is a $G$-representation over a field $k$ then the functor $V \mapsto V^H$ canonically acquires an action of the endomorphism algebra $\text{End}_G(k[G/H])$, where $k[G/H]$ is the free $G$-representation on an $H$-fixed point, and again by the Yoneda lemma this is precisely the endomorphism algebra of the fixed point functor. This algebra is the Hecke algebra associated to the pair $H \hookrightarrow G$, and it has a distinguished basis given by double cosets; see, for example, this MO answer.