**Context:** The number of conjugacy classes equals to the number of irreducuble representations (over C) for any finite group.
Moreover for the symmetric group there is well-known "natural bijection" between the two sets via Young diagrams.
The question whether such "good" bijection may exist for other finite groups seems to be non-trivial one and discussed at MO (see e.g. G. Robinson, MO153731 and links there in.)

The point of the present post is to suggest an analogy between local Langlands correspondence and that bijection and rise few questions: it is well-known that Weyl groups can be thought of as algebraic simple groups over the field with one element, the arguments presented below seems to me convincing that certain simplified version of the local Langlands correspodence over $F_{un}$ should be precisely equivalent to the bijection between conjugacy classes and irreducible repsentations of Weyl groups (as well as alternating group).

So the analogy motivates to ask several questions, the hope is that despite both Langlands correspondence and "field with one element" are somewhat difficult and not always well-defined concepts, the two difficulties cancel each other and as a result one may get facts or conjectures which "everyone can understand". So questions:

**Question 0:** Is there "standard"-"good" bijection between conjugacy classes irreducible representations for Weyl groups (and $Alt_n$) ?

I assume the positive answer to that question - it is remarked in several places , however I cannot google somewhat a detailed exposition.

**Question 1:** What are the "good" properties of such bijection for Weyl groups ? (respecting Out(G), respecting reality/rationality, respecting reduction modula "p" etc.. (see MO ) for further properties and MO for related question for $S_n$ ).

**Question 2:** Can one suggest further analogies between Langlands correspondence and finite group theory ?
It might be for any finite group there exist a "Langlands dual" finite group such that conjugacy in one are in natural bijection with the irreps of the other (s.t. Weyl groups are self-dual)? what should correspond to "Langlands functoriality" i.e. for a homomorphism $H^L->G^L$ there should be transfer
of repsentations from $H$ to $G$ ? Are there analogues of "autmorphic forms", "zeta-functions", "Eisenstein series" in finite group setup ?

**The analogy** between the bijection for Weyl groups and the local Langlands correspondence over field with one element goes as follows:

The local Langlands correspondence over field $F_q((t))$ is roughly speaking certain bijection between certain irreducible complex representations of $G(F_q((t)))$ ("automotphic side") and morphisms of the Galois group $Gal(F_q((t)))$ to the Langlands dual group $G^L$ ("Galois side"). Here $G$ is reductive group.

Let us look on the modified setup: instead of $F_q((t))$ consider just $F_q$, instead of $F_q$ consider $q=1$ i.e. "field with one element". So :

1) On "automorphic side" we would see: irreducible representations of Weyl group - precisely one side of the bijection. (Since well-known heuristics $G(F_q)$ for $q=1$ becomes Weyl group).

2) On "Galois side" we should have morhphisms from Galois group of $F_{un}$, to $G^L$. To come to the other side of the bijection we remark that:

2a) Langlands dual for B-series is from C-series and vice versa, but the Weyl group for B is the same as for C ! It actually resolves the main trouble, since in other cases $G^L$ is the same $G$ up to some central elements, e.g. $SL(n)^L = PSL(n)$, but if we think of $F_{un}$ there is no place for such differences, e.g. both $SL(n)$ and $PSL(n)$ reduces to alternating group.

2b) The Galois group of $F_q$ for all $q$ is equal to $\hat Z$ (profinite compltetion of $Z$), so it is natural to expect it holds for $q=1$. Moreover

in Langlands correspondence one should consider not the whole Galois group,
but precisely $Z$ sitting in $\hat Z$ (that is the so-called Weil group,
which is defined as preimage of Frobenius element for the map Galois(LocalField) to $Gal(F_q$)).

So, "Galois side" reduces to: morphisms of $Z$ into $G$ (up to natural equivalence). That is precisely the conjugacy classes in $G$.

That completes the analogy.

If it is discussed somewhere else - can one give a reference ? I was unable to google something like that.

Well, there was certain cheating in my story: on the Galois side one considers morphisms from the Galois (Weil) group to $G^L(C)$ (algebraic group over complex numbers), while I consider $G^L(F_q)$. Well ... analogy is not complete. Nevertheless I think it is quite suggestive. Actually there can be certain ways rounds - like controlling ramifications and actually image always belongs to certain subset of C which might be related to $F_q$ ... But I am not sure that sure ... And let me leave it as it is...

We might ask what in general should Langlands be over $F_{un}$, but that will be a separate question.