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I am quite familiar with the Burnside/Cauchy-Frobenius Lemma, which states that for a group $G$ acting on a set $X$, where $O$ is the number of orbits of $X$ under the action of $G$, we have:

$$O=\frac{1}{|G|}\sum\limits_{g\in G}|\{x\in X: xg=x\}|,$$ that is, the number of orbits is found by finding the number of fixed points of the action for each group element and adding them up.

I'm trying to write this lemma down in notation more friendly to the specific situation where we have a symmetric design $D$ and its automorphism group $\text{Aut}(D)$, but I am not entirely comfortable with what I came up with, and thought I would check with the community.

I have already proven (and it is well-known) that any automorphism on a symmetric design $D$ fixes the same number of blocks and points, further blurring things. I believe $D$ is playing the role of $X$ above, and it makes sense to consider both blocks and points fixed by any element $\sigma\in\text{Aut}(D)$. I also think it makes sense that $\text{Aut}(D)$ is playing the role of $G$ above.

So I arrive at the following statement for Burnside's Lemma in the context of a symmetric design and its automorphism group:

$$O=\frac{1}{|\text{Aut}(D)|}\sum\limits_{\sigma\in\text{Aut}(D)}|\{p\in D:\sigma p=p\}|,$$

where $p$ represents a point.

Does this look good to you?

Main concern:

$D$ isn't really a set, in the traditional sense - it is an incidence structure, so writing "$p\in D$" feels wrong/notationally abusive, but writing it out for blocks $b$ has the same problem. I suppose I could further muddy things by saying $X$ is the set of points on which our design lives, but then writing $p\in X$ seems to destroy the relation between my group ($\text{Aut}(D)$) and my set ($D$?... $X$?...)

The questions: Does my statement of Burnside's Lemma for this situation look right? Is everything I'm doing consistent with a good understanding of how a design's automorphism group interacts with the design itself? Is my set in the summand reasonable and not too notationally weird?

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    This seems to be perfectly fine. As you said you just have to take the set of points. Obviously $Aut(D)$ acts on it and it can be easily seen as a subgroup of bijections of set $X$ (via the forgetful functor). The relation between $X$ and $Aut(D)$ is partially encoded in $Aut(D)$ itself since not every bijection of $X$ can be turned into an autmorphism of $D$. But you are right, you lose some information. But whether that's good or bad totally depends on what you are trying to achieve. – 2017-02-20
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    @freakish, thanks very much! I feel better about this. if you want, please write this as an answer so I can upvote and accept it! – 2017-02-20

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It's true that $D$ isn't really a set, so that is the part that seems awkward to me. I feel that you should say that you are viewing $\mathrm{Aut}(D)$ as a subset of $\mathrm{Sym}(\mathcal{P})$, $\mathrm{Sym}(\mathcal{B})$, or $\mathrm{Sym}(\mathcal{P} \cup \mathcal{B})$, to clarify whether you are viewing an automorphism as being defined by a permutation of points, blocks, or both.

After double checking, in the thesis of Aschbacher he refers to the automorphism group of a design $D$ as acting on $D$, the implied meaning being that the set $X = \mathcal{P} \cup \mathcal{B}$, so I guess this aligns directly with your notation (note that this means $O$ consists of both point and block orbits).

The version you have of Burnside's lemma will apply in any of these cases.

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    Thanks very much! This is very helpful! – 2017-02-21