**Main idea shortly:** As we discussed recently [MO272045](https://mathoverflow.net/q/272045/10446), there is beautiful fomula which
counts index-n subgroups in terms of homomorphisms to $S\_n$.
Let me give ["field with one element"](https://en.wikipedia.org/wiki/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](https://mathoverflow.net/q/272045/10446), Qiaochu Yuan's [blog1(proof)](https://qchu.wordpress.com/2015/11/04/the-categorical-exponential-formula/),
[blog2](https://qchu.wordpress.com/2015/11/15/finite-index-subgroups-of-the-modular-group/), [blog3](https://qchu.wordpress.com/2015/11/25/conjugacy-classes-of-finite-index-subgroups/), 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](https://en.wikipedia.org/wiki/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](https://mathoverflow.net/q/272498/10446))
* 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](https://en.wikipedia.org/wiki/Exponential_formula)",
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](http://www.neverendingbooks.org/DATA/KapranovSmirnov.pdf) (see also [MO272698](https://mathoverflow.net/questions/272698/what-are-the-points-of-simple-algebraic-groups-over-extensions-of-mathbbf-1)) 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](https://en.wikipedia.org/wiki/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 ?



Field with one element look at counting index-$n$ subgroups in terms of Homs to $S_n$, generalization to $F_{1^k}$?
Yes this is known. The paper ["On Wohlfahrt series and wreath products"](http://www.sciencedirect.com/science/article/pii/S0001870806001642) by Y. Takegahara contains slightly more general results and builds on previous work (Muller's paper). Corollary 2.7 says:


$$\sum\_{n=0}^{\infty}\frac{|\operatorname{Hom}(A,G\wr S\_n)|}{|G|^nn!}x^n=\exp\left(\sum\_{B\subset A}\frac{|\operatorname{Hom}(B,G)|}{|G||A:B|}x^{|A:B|}\right)$$
where $A$ is any group that has finitely many subgroups of each index, $G$ is finite and $|A:B|$ is the index of $B$ as a subgroup of $A$.


