**Main idea shortly:** As we discussed recently MO272045, there is beautiful fomula which
counts index-n subgroups in terms of homomorphisms to $S_n$.
Let me give "field with one element" interpretation of that formula,
and propose a formula which can be seen as
as $F_{1^k}$-generalization ($S_n$ changed to $\mu_k \wr{S}_n=GL(n,F_{1^k})$, subgroups enriched by pairs: subgroup + character).

*My question (see details below)*: whether such generalization is true / known ?

**Details.**

**1. The formula:**
Let $G$ be any finitely generated group,
the following is true (see MO272045, Qiaochu Yuan's blog1(proof),
blog2, blog3, R. Stanley EC2 Exercise 5.13):

$$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,S_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~\text{subgroups of}~ G|}nz^n \right) . $$

Here $| \mathrm{Hom}(G,S_n) | $ is the number of homomorphisms from $G$ to the symmetric group $S_n$ (pay attention that we count homomorphisms themselves - not up to conjugation). $|\mathrm{Index}~n~\text{subgroups of}~ G|$ is the number of index $n$ subgroups in $G$ (pay attention subgroups themselves, not up to conjugation).

**2. Field with one element interpretation:**
Although field with one element is quite mysterious, but there
are simple and widely agreed heuristics about it:

- $S_n$ is $GL(n,F_1)$ (see e.g. MO272498)
- Respectively $Hom(G,S_n)$ or an action on n-points set is n-dimensional representation of $G$
- Respectively
*transitive*action on finite set is*IRREDUCIBLE*representation, i.e. irreducible representations comes from index n-subgroups of $G$ via action on $G/H$

So we come to :

$$ \sum_{n \ge 0} \frac{| n\text{-dim}~F_1 ~ \text{representations of}~ G ~\text{with choice of basis} | }{|Automorphisms| } z^n = \exp\left( \sum_{n \ge 1} \frac{|n\text{-dim}~F_1~\text{IRREDUCIBLE representations of}~ G|}{|Automorphisms|} z^n \right) . $$

That interpretation seems clarifying for me, it puts the formula into general framework of "exponential formulas", counting all representations as exponential of irreducible ones.

Analogies above hopefully explain the interpretation. Let me comment why $n$ in the denominator at the RHS is interpreted as |Automorphism|. Indeed for n-point set $M$ with transitive action of $G$, we can send any point "x" to any other point "y" and extend it to map $ \phi:M->M$ by $\phi(gx)=g\phi(x)$ in a unique way, thus getting automophism of $M$ which commutes with action of $G$. So we have $n$ such automorphisms.

**3. Question on generalization to $F_{1^k}$**
Kapranov and Smirnov 1995 (see also MO272698) propose
to consider algebraic extension $F_{1^k}$ of the $F_1$ for each $k$
and $GL(n,F_{1^k})$ to be generalized symmetric group $\mu_k\wr {S}_n$ (consisting of generalized permutation matrices whose nonzero entries are in the cyclic group $\mu_k$ of $k$-th roots of unity)

So, proposal:

$$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mu_k \wr{S}_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~G\text{-subgroups and their characters to k-th roots of unity }|}nz^n \right) . $$

The idea is that $Ind^G_H V_{trivial} = G/H$ as representation, so we just consider $Ind^G_H V_{character}$ - it gives representation like $G/H$ but twisted by roots of unity. That seems the way to get all irreducible representation of $G$ over $F_{1^k}$.

So we arrive to the formula motivated by $F_1$, but its formulation is completely independent of anything related to mysteries of $F_1$.

**QUESTION:** Is the formula true/known ?