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

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 (?)).

## Contents

## Statement

Suppose is a set of size at least two. Let be the finitary symmetric group on . Let and denote the subgroup of comprising those permutations that fix ; in other words, is the isotropy subgroup of and equivalently is the finitary symmetric group on .

Then, has double coset index two in .

## 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 , then the double coset index of its isotropy subgroup is bounded by a function of . In fact, if the whole set has size at least , the double coset index is precisely
*equal*to that function of . - Isotropy of finite subset has finite double coset index in finitary symmetric group: If a subset has size , then the double coset index of its isotropy subgroup is bounded by a function of . In fact, if the whole set has size at least , the double coset index is precisely
*equal*to that function of . - Isotropy of finite subset has finite double coset index in finitary alternating group: If a subset has size , then the double coset index of its isotropy subgroup is bounded by a function of . In fact, if the whole set has size at least , the double coset index is precisely
*equal*to that function of .

## Facts used

## Proof

### Direct proof

**Given**: is a set of size at least two, , is the isotropy subgroup of .

**To prove**: has double coset index two in .

**Proof**:

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

### Proof by size computations

(This proof works only for finite sets).

**Given**: A finite set of size . is the symmetric group on and is the subgroup comprising permutations that fix a particular point .

**To prove**: has double coset index two.

**Proof**: We need to prove that for , we have . We know that:

.

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

Now, note that for any , is the isotropy subgroup of . Hence:

- has size .
- Both and are subgroups of size : They are the symmetric groups on the sets and respectively.
- The intersection is a subgroup of size : It is the symmetric group on the set .

Plugging in the formula yields:

.

This simplifies to:

.

Cardinality considerations thus yield:

.

This completes the proof.