You can view a pdf of this entry here.

Let be a finite group that acts on a finite set, . Given elements and , we introduce the cycle notation, to denote that , but for all . We say that is a -cycle in . Conceptually, this is a natural construction: the action of on induces a map, , of into the symmetric group of . This allows us to decompose the image of any element into disjoint cycles.

Let the expression denote the number of -cycles of . Let be the set of all -cycles contained in elements of . In particular, for , because -cycles are fixed points of elements of , and fixes every element of . Note that if is nonempty, then acts on by conjugation: if , then is a cycle in , so . Parker’s lemma is the following:

**Lemma 1 (Parker’s Lemma)** *The number of orbits of on is*

At first glance, this lemma appears to be a straightforward application of Burnside’s orbit counting lemma:

In fact, the case is Burnside’s lemma: the number, , is the number of -cycles of , i.e. . However, for , if is nonempty, then . Thus, we cannot simply equate terms: , in general. The difficulties described above suggest that Parker’s lemma is a stronger statement than Burnside’s lemma. In fact, I am unaware of any proof in the literature that deduces Parker’s lemma as a consequence of Burnside’s Lemma. The aim of this post is to establish Parker’s lemma as a simple consequence of Burnside’s lemma.

**Proof: **We assume that is nonempty, otherwise the result is trivial. We note that by Burnside’s lemma

Define subsets by

For each , choose such that . Let be the pointwise stabilizer of . Then is the set of elements of that contain , and is the stabilizer of . Note that for , with . Indeed, if , where , then is disjoint from , i.e. , which is a contradiction. Thus, because , we see that is a partition of . Therefore, , and hence

because . Now, we count the order of and by counting points along the fiber of the canonical projection

Indeed,

Thus,

as desired.

### Like this:

Like Loading...

*Related*