# Normalizer of group action

Take a group $$G$$ acting faithfully on a set $$X$$, and let $$H \leq G$$. It can be easily shown that the elements of $$N_{G}(H)$$ stabilize the collection of orbits of $$H$$ (as a set, ie orbits are mapped to orbits).

Is the converse true? That is, if we take $$\mathcal{O}$$ to be the collection of orbits of $$H$$, do we always have that $$\mathrm{Stab}_{G}(\mathcal{O}) = N_{G}(H)$$? (I'm happy if everything is assumed to be finite, but more general answers are also welcome.)

Edit: Since, as mentioned in the answer of runway44, this can be considered by looking at what happens with each of the $$G$$-orbits on $$X$$, I would like to know if this is true for $$G$$ acting transitively and faithfully on $$X$$.

Suppose $$g\in G$$ stabilizes the orbit space $$H\,\backslash X$$ (that is, the collection of orbits of $$H\curvearrowright X$$).

Since $$g$$ cannot mix the $$G$$-orbits, it must stabilize each orbit space $$H\,\backslash \mathcal{O}_1$$, $$\cdots$$, $$H\,\backslash \mathcal{O}_k$$ (where $$\mathcal{O}_i$$ are the orbits of the full action $$G\curvearrowright X$$). Any $$G$$-orbit is isomorphic (as a $$G$$-set) to $$G/K$$, where $$K$$ is the stabilizer of some point in the orbit, so we ought to examine the situation for $$G/K$$.

Note that the $$H$$-orbit space $$H\,\backslash(G/K)$$ is the set $$H\,\backslash G/K$$ of double cosets. Also left and right actions can be converted using inverses, and in particular there is an action $$H\times K\curvearrowright G$$ given by $$(h,k)g:=hgk^{-1}$$ and the orbits are precisely the double cosets; $$(H\times K)\,\backslash G=H\,\backslash G/K$$. In particular, the double cosets partition the group $$G$$, just as orbits partition any $$G$$-set.

For $$g$$ to stabilize $$H\,\backslash G/K$$, we must have for all $$a\in G$$ there exists a $$b\in G$$ such that $$gHaK=HbK$$. As every element of an orbit is a representative, and $$ga\in gHaK$$, we have $$gHaK=HgaK$$. This is equivalent to saying $$H({}^cK)=(H^g)({}^cK)$$ for all $$c\in G$$, where $$H^g=g^{-1}Hg$$ and $${}^cK=cKc^{-1}$$ and $$c=ga$$. Thus,

$$\mathrm{Stab}_G(H\,\backslash G/K)=\{g\mid H({}^cK)=(H^g)({}^cK)~\forall c\in G\}.$$

I am not sure if this simplifies any better. Then $$\mathrm{Stab}_G(H\,\backslash X)$$ will be the intersection of the individual stabilizers $$\mathrm{Stab}_G(H\,\backslash \mathcal{O}_i)$$. Two extreme situations:

• $$\mathrm{Stab}_G(H\,\backslash G/G)=G$$
• $$\mathrm{Stab}_G(H\,\backslash G/1)=N_G(H)$$.

In general, $$\mathrm{Stab}_G(H\,\backslash X)$$ will be between $$N_G(H)$$ and $$G$$.

I wanted to extend the analysis of runway44's answer to the case where $$G$$ acts transitively on $$X$$, and give an example that shows that even here, the answer is negative.

If we assume that $$G$$ acts faithfully and transitively on $$X$$, then we can put $$K = \mathrm{Stab}_{G}(x)$$ for some arbitrary $$x \in X$$; then there is a natural correspondence between $$X$$ and $$G/K$$, the cosets of $$K$$ (in contrast to runway44's answer, I am going to be working with right cosets, and a right group action). So each element of $$X$$ corresponds to a coset $$Kg$$ for some $$g \in G$$, and an $$H$$-orbit on $$X$$ corresponds to the double-coset $$KgH$$.

Now, if some $$n \in G$$ stabilizes the $$H$$-orbits on $$X$$, then for each $$g \in G$$ we have $$KgHn = Kg^{\prime}H$$ for some $$g^{\prime} \in G$$. Since $$gn \in KgHn$$, we have $$KgHn = KgnH$$. On the other hand, we can write $$KgHn = Kgn(n^{-1}Hn)$$ giving us $$KgnH = Kgn(n^{-1}Hn)$$; since $$gn$$ covers all the elements of $$G$$ as $$g$$ does, this means $$KgH = Kg(n^{-1}Hn)$$ for all $$g \in G$$, or equivalently, $$x^{H} = x^{n^{-1}Hn}$$ for all $$x \in X$$. In other words, the most we can conclude is that $$H$$ and $$n^{-1}Hn$$ have the same orbits on $$X$$.

From this, we can see that even when $$G$$ acts faithfully and transitively on $$X$$, it is possible that $$N_{G}(H) < \mathrm{Stab}_{G}(O(H))$$. For example, let $$G = S_{6}$$ acting naturally on $$\{1,2,3,4,5,6\}$$, and take $$H = \langle (1,2,3)(4,5,6),\ (1,2) \rangle$$. If $$n = (1,4)(2,5)(3,6)$$, then $$n$$ stabilizes the orbits of $$H$$. But $$n^{-1}Hn = \langle (1,2,3)(4,5,6),\ (4,5)\rangle \ne H.$$