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## Conjugacy classes of finite index subgroups

Previously we learned how to count the number of finite index subgroups of a finitely generated group $G$. But for various purposes we might instead want to count conjugacy classes of finite index subgroups, e.g. if we wanted to count isomorphism classes of connected covers of a connected space with fundamental group $Gi$.

There is also a generating function we can write down that addresses this question, although it gives the answer less directly. It can be derived starting from the following construction. If $X$ is a groupoid, then $LX = [S^1, LX]$, the free loop space or inertia groupoid of $X$, is the groupoid of maps $S^1 \to X$, where $S^1$ is the groupoid $B\mathbb{Z}$ with one object and automorphism group $\mathbb{Z}$. Explicitly, this groupoid has

• objects given by automorphisms $f : x \to x$ of the objects $x \in X$, and
• morphisms $(f_1 : x_1 \to x_1) \to (f_2 : x_2 \to x_2)$ given by morphisms $g : x_1 \to x_2$ in $X$ such that

$x_1 \xrightarrow{f_1} x_1 \xrightarrow{g} x_2 = x_1 \xrightarrow{g} x_2 \xrightarrow{f_2} x_2$.

It’s not hard to see that $L(X \coprod Y) \cong LX \coprod LY$, so to understand this construction for arbitrary groupoids it’s enough to understand it for connected groupoids, or (up to equivalence) for groupoids $X = BG$ with a single object and automorphism group $G$. In this case, $LBG$ is the groupoid with objects the elements of $G$ and morphisms given by conjugation by elements of $G$; equivalently, it is the homotopy quotient or action groupoid of the action of $G$ on itself by conjugation.

In particular, when $G$ is finite, this quotient always has groupoid cardinality $1$. Hence:

Observation: If $X$ is an essentially finite groupoid (equivalent to a groupoid with finitely many objects and morphisms), then the groupoid cardinality of $LX$ is the number of isomorphism classes of objects in $X$.

I promise this is relevant to counting subgroups!

Where are those subgroups?

Now let $X$ be the groupoid $[B G, BS_n]$ of actions of a finitely generated group $G$ on $n$-element sets. The number of isomorphism classes of objects in this groupoid is the number of isomorphism classes of $G$-sets with $n$ elements, and so this number can also be identified with the groupoid cardinality of the free loop space $LX$. But this is just

$[B\mathbb{Z}, [BG, BS_n]] \cong [B(\mathbb{Z} \times G), BS_n]$

which can in turn be identified with the groupoid of $\mathbb{Z} \times G$-sets with $n$ elements. In other words, the number of isomorphism classes of $G$-sets with $n$ elements is

$\displaystyle \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!}$.

Now, the collection of all isomorphism classes of finite $G$-sets is a free commutative monoid (under disjoint union) on the isomorphism classes of finite transitive $G$-sets. In other words, such an isomorphism class is described by describing the multiplicity with which each finite transitive $G$-set occurs within it. This gives us the following count.

Theorem: Let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$ in $G$. Then

$\displaystyle \sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}$.

Incidentally, this result and the previous result about subgroups of index $n$ are both exercises in Stanley’s Enumerative Combinatorics: Volume II (more precisely, Exercise 5.13a and c).

It would be nice to write this in a form that lets us more clearly extract the coefficients $c_n(G)$ from the LHS. If $s_n(\mathbb{Z} \times G)$ denotes the number of subgroups of $\mathbb{Z} \times G$ of index $n$, then taking logarithms gives

$\displaystyle \sum_{n \ge 1} s_n(\mathbb{Z} \times G) \frac{z^n}{n} = \sum_{n \ge 1} \log \frac{1}{(1 - z^n)^{c_n(G)}} = \sum_{n \ge 1} c_n(G) \sum_{d \ge 1} \frac{z^{nd}}{d}$.

Extracting the coefficient of $\frac{z^n}{n}$ from both sides gives

$\displaystyle s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$

and hence Möbius inversion gives the following.

Theorem: With $s_n$ and $c_n$ as above, we have

$\displaystyle c_n(G) = \frac{1}{n} \sum_{d \mid n} \mu \left( \frac{n}{d} \right) s_d(\mathbb{Z} \times G)$.

Example. Let $G = \mathbb{Z}$. Then $c_n(\mathbb{Z}) = 1$ for all $n$. This recovers

$\displaystyle s_n(\mathbb{Z}^2) = \sum_{d \mid n} d = \sigma_1(n)$

as expected.

Example. Now let $G = \mathbb{Z}^2$. For abelian groups counting conjugacy classes of subgroups is the same as counting subgroups, so $c_n(\mathbb{Z}^2) = s_n(\mathbb{Z}^2) = \sigma_1(n)$ and $s_n(\mathbb{Z}^3)$ turns out to be

$\displaystyle s_n(\mathbb{Z}^3) = \sum_{d \mid n} \sigma_1(d) = \sum_{d_1 \mid d_2 \mid n} d_1 d_2 = \sum_{d_1 d_2 d_3 = n} d_1^2 d_2$.

In the same way that $\sigma_1(n)$ has Dirichlet series $\zeta(s) \zeta(s - 1)$, this function has Dirichlet series $\zeta(s) \zeta(s - 1) \zeta(s - 2)$. By induction, we find that the number of subgroups of $\mathbb{Z}^k$ of index $n$ is

$\displaystyle s_n(\mathbb{Z}^k) = \sum_{d_1 d_2 \dots d_k = n} d_1^{k-1} d_2^{k-2} \dots d_{k-1}$

which has Dirichlet series $\zeta(s) \zeta(s - 1) \dots \zeta(s - (k-1))$. We leave it as an entertaining exercise for the reader to give a direct proof of this.

A slower discussion

The generating function given above can be interpreted as the weighted groupoid cardinality of the groupoid of $\mathbb{Z} \times G$-sets, and the Möbius inversion formula gives some sort of relationship between transitive $\mathbb{Z} \times G$-sets and transitive $G$-sets. What exactly is this relationship, and can we use it to give a more direct proof of the Möbius inversion formula?

For starters, a $\mathbb{Z} \times G$-set is the same thing (in the sense that we have an equivalence of categories) as a pair $(X, f)$ consisting of a $G$-set $X$ and an automorphism $f : X \to X$ of $X$. (We already used this fact when we passed from the free loop space description to talking about $\mathbb{Z} \times G$ above.) The $\mathbb{Z} \times G$-set is transitive if the combination of the action of $G$ and the automorphism $f$ is transitive. And we can identify subgroups of $\mathbb{Z} \times G$ with pointed transitive $\mathbb{Z} \times G$-sets. So what do these look like?

If $(X, f)$ is a finite transitive $\mathbb{Z} \times G$-set, then its decomposition as a $G$-set consist of a number $\frac{n}{d}$ of copies of the same finite transitive $G$-set $Y$ of size $d$, which are cyclically permuted by the automorphism $f$. If $(X, f)$ is pointed, then one of these copies has a basepoint $y \in Y$ in it, so $Y$ can canonically be identified with $G/H$ where $H$ is the stabilizer of $y$. The automorphism $f$ can be used to identify all of the other copies of $G/H$ with the copy containing the basepoint, so the only remaining data in this automorphism is the induced automorphism $g = f^{n/d}$ of $G/H$.

The automorphism group of $G/H$ is $N_G(H)/H$, where $N_G(H)$ denotes the normalizer

$N_G(H) = \{ g \in G \mid gHg^{-1} = H \}$

of $H$ in $G$, and $g \in N_G(H)$ acts by left multiplication.

Altogether we’ve described a bijection between

• subgroups of $\mathbb{Z} \times G$ of index $n$ and
• triples $(d, H, g)$ of a divisor $d$ of $n$, a subgroup $H$ of $G$ of index $d$, and an element $g \in N_G(H)/H$.

This is close to, but not quite, the count we wanted, which was in terms of conjugacy classes of subgroups $H$ of $G$ of index $d$. To get this count we need to know how many conjugates a given subgroup $H$ of index $n$ has. Every conjugate appears as the stabilizer of a point in $G/H$, but two points that are related by the action of the automorphism group $N_G(H)/H$ will have the same stabilizer, and conversely two points with the same stabilizer are related by the action of the automorphism group. The automorphism group itself acts freely, so every orbit has size $|N_G(H)/H|$. Altogether we find that $H$ has

$\displaystyle \frac{n}{|N_G(H)/H|}$

conjugates, so after grouping all of the conjugates of $H$ together in the above bijection we find that the contribution of conjugacy classes of subgroups of $G$ of index $d$ to the count of subgroups of $\mathbb{Z} \times G$ of index $n$ is $d c_d(G)$ as desired.