Isotropy of a point has double coset index two in finitary symmetric group

Statement
Suppose $$S$$ is a set of size at least two. Let $$G = \operatorname{FSym}(S)$$ be the finitary symmetric group on $$S$$. Let $$x \in S$$ and $$H$$ denote the subgroup of $$G$$ comprising those permutations that fix $$x$$; in other words, $$H$$ is the isotropy subgroup of $$x$$ and equivalently $$H$$ is the finitary symmetric group on $$S \setminus \{ x \}$$.

Then, $$H$$ has double coset index two in $$G$$.

Related facts

 * Isotropy of a point has double coset index two in symmetric group: The analogous result holds even when we are considering arbitrary permutations, i.e., permutations that are not necessarily finitary.
 * Isotropy of a point has double coset index two in finitary alternating group
 * Isotropy of finite subset has finite double coset index in symmetric group: If a subset has size $$n$$, then the double coset index of its isotropy subgroup is bounded by a function of $$n$$. In fact, if the whole set has size at least $$2n$$, the double coset index is precisely equal to that function of $$n$$.
 * Isotropy of finite subset has finite double coset index in finitary symmetric group: If a subset has size $$n$$, then the double coset index of its isotropy subgroup is bounded by a function of $$n$$. In fact, if the whole set has size at least $$2n$$, the double coset index is precisely equal to that function of $$n$$.
 * Isotropy of finite subset has finite double coset index in finitary alternating group: If a subset has size $$n$$, then the double coset index of its isotropy subgroup is bounded by a function of $$n$$. In fact, if the whole set has size at least $$2n$$, the double coset index is precisely equal to that function of $$n$$.

Facts used

 * 1) uses::Product formula

Direct proof
Given: $$S$$ is a set of size at least two, $$G = \operatorname{FSym}(S)$$, $$H$$ is the isotropy subgroup of $$x$$.

To prove: $$H$$ has double coset index two in $$G$$.

Proof:


 * 1) The transpositions $$(x,y)$$ (along with the identity element) form a system of left coset representatives for $$H$$ in $$G$$: Indeed, given any finitary permutation $$\sigma \notin H$$, we have $$\sigma(x) \ne x$$. Define $$\tau = (x,\sigma(x))$$. The permutation $$\tau^{-1}\sigma$$ fixes $$x$$, so $$\sigma \in \tau H$$. Thus, every element of $$G$$ outside $$H$$ can be expressed in the form $$(x,y)H$$ for some $$y$$, so every left coset contains a transposition of the form $$(x,y)$$. Further, for $$y \ne z$$, we have that $$(x,y)^{-1}(x,z)$$ is not in $$H$$, so distinct transpositions correspond to different left cosets of $$H$$ in $$G$$. (Note that the left coset representative for $$H$$ itself is the identity element).
 * 2) The transpositions $$(x,y)$$ and $$(x,z)$$ are conjugate via $$(y,z)$$. In particular, they are conjugate via $$H$$: In other words, $$(y,z)(x,y)(y,z)^{-1} = (x,z)$$.
 * 3) Pick any $$y \ne x$$. We claim that if $$\alpha = (x,y)$$, then $$G = H \cup H\alpha H$$: Pick any $$\sigma \notin H$$. By step (1), $$\sigma \in \tau H$$ for some $$\tau = (x,z)$$. If $$y = x$$, $$\sigma \in \alpha H$$ and we are done. Else, by step (2), we have $$\tau = (y,z)(x,y)(y,z)^{-1} \in H(x,y)H$$. Thus, $$\sigma \in \tau H \subseteq H \alpha H$$, completing the proof.

Proof by size computations
(This proof works only for finite sets).

Given: A finite set $$S$$ of size $$n \ge 2$$. $$G$$ is the symmetric group on $$S$$ and $$H$$ is the subgroup comprising permutations that fix a particular point $$x \in S$$.

To prove: $$H$$ has double coset index two.

Proof: We need to prove that for $$\alpha \in G \setminus H$$, we have $$|H \alpha H| = |G \setminus H|$$. We know that:

$$|H \alpha H| = |H \alpha H \alpha^{-1}| = |H (\alpha H \alpha^{-1})| = \frac{|H||\alpha H \alpha^{-1}|}{|H \cap \alpha H \alpha^{-1}|}$$.

(The last step is from the product formula, fact (1).

Now, note that for any $$\alpha \notin H$$, $$\alpha H \alpha^{-1}$$ is the isotropy subgroup of $$\alpha(x)$$. Hence:


 * $$G$$ has size $$n!$$.
 * Both $$H$$ and $$\alpha H \alpha^{-1}$$ are subgroups of size $$(n-1)!$$: They are the symmetric groups on the sets $$S \setminus \{ x \}$$ and $$S \setminus \{ \alpha(x) \}$$ respectively.
 * The intersection is a subgroup of size $$(n-2)!$$: It is the symmetric group on the set $$S \setminus \{ x, \alpha(x)\}$$.

Plugging in the formula yields:

$$|H\alpha H| = \frac{(n-1)!^2}{(n-2)!}$$.

This simplifies to:

$$|H\alpha H| = (n-1)(n-1)! = n! - (n-1)! = |G \setminus H|$$.

Cardinality considerations thus yield:

$$H \alpha H = G \setminus H$$.

This completes the proof.