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

From Groupprops
Jump to: navigation, search
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.

Related facts

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.