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

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This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup satisfying a particular subgroup property (namely, Subgroup of double coset index two (?)) in a particular group or type of group (namely, 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 $G$.

Facts used

1. Product formula

Proof

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.