What can we say about the kernel of $\phi: F_n \rightarrow S_k$

Question

Let $F_n$ denote the free group on $n$ generators and let $S_k$ denote the symmetric group on the integers $\{1,\dots, k\}$, and the action of homomorphism $\phi$ (as given in the title) on the generators of $F_n$ be known. Let $K_{\phi}$ denote the kernel.

My question is two-fold.

1. I've had GAP spit out a couple dozen random examples of such homomorphisms/kernels and in every case so far $K_{\phi}$ appears to be of infinite generation. Is this always the case? I can't imagine how it couldn't be if $\phi$ is chosen to be nontrivial (my reasoning is that, even if $\phi$ kills off all but one generator, so that $K_{\phi}$ is a subset of an infinite cyclic subgroup of $F_n$, $K_{\phi}$ is still not finitely generated, and if $\phi$ kills off fewer-than-all-but-one generators the complexity only increases) but I also can't see how I would begin a proof, unless my reasoning in the previous parenthetical is the proof strategy. (Supposing, though, that there is a counterexample, is there a way to tell from the presentation of $\phi$ whether or not $K_{\phi}$ is finitely-presented?)

2. Can anything be said about the "combinatorics" of the embedding $\iota: K_{\phi} \hookrightarrow F_n$? (To clarify what I mean by this: Of course, if all the groups that arise as kernels in this way are on infinite generators then any two are isomorphic by Nielsen-Schrier, but even if two such arise from the same $F_n$, they're clearly not necessarily the same subgroup.)

Edit: I seem to have forgotten or never learned something important--As James points out in a comment below, $K_{\phi}$ is a finite-index subgroup of a finitely-generated group, so it's necessarily finitely-generated by what I just learned is called Schreier's lemma. With that in mind, I think it's sensible to revise the first part of my question to "What can we say about a generating set of $K_{\phi}$ given what we know about $\phi$?" The second part of the question remains a question for the time being.

I've given this the reference request tag in case there's a standard source on these topics that I should be familiar with. If anything is unclear, please post a comment and I'll try to fix it.

Answer 1

By the standard theorems on covering spaces, conjugacy classes of maps $\varphi : F_n \to S_k$ are naturally in bijection with $k$-fold covers $Y$ (not necessarily connected) of a wedge $X = \bigvee_{i=1}^n S^1$ of $n$ circles. The cycle type of the image of a generator of $F_n$ tells you what the preimage of the corresponding copy of $S^1$ should be in this cover. Covering spaces of graphs are always graphs, so it is possible to be quite explicit about what these covers look like. You can draw lots of pretty pictures.

Let me now assume that $\text{im}(\varphi)$ is a transitive subgroup of $S_k$. Then the corresponding $k$-fold cover is connected, and its fundamental group $H$ is the stabilizer of a point in $\{ 1, 2, ..., k \}$. Moreover, $\text{ker}(\varphi)$ is the kernel of the action of $G = F_n$ on the cosets $G/H \cong \{ 1, 2, ..., \}$, which can be identified with the intersection

$$\bigcap_{g \in G} gHg^{-1}$$

of the conjugates of $H$ in $G$. It is again possible to be quite explicit about this; not only is the fundamental group of any connected graph a free group, but if you choose a spanning tree of the graph then the set of edges not in the spanning tree forms a free generating set. Moreover, we can predict in advance how many generators we'll get as follows. First, the above shows that if $X = (V, E)$ is a connected graph then its fundamental group is free on $|E| - |V| + 1$ generators. Second, the Euler characteristic $\chi(X) = |E| - |V|$ is multiplicative under covers, and since $X$ has Euler characteristic $n - 1$, it follows that the corresponding cover $Y$ has Euler characteristic $k(n - 1)$, hence that the fundamental group $H$ of $Y$ is free on $k(n - 1) + 1$ generators. This doesn't get you $\text{ker}(\varphi)$ directly; you can try to get $\text{ker}(\varphi)$ directly by constructing the Galois closure of the cover above, but I actually don't know how to do this in full generality. Applying the above argument to the Galois closure shows that $\text{ker}(\varphi)$ is free on $|\text{im}(\varphi)|(n - 1) + 1$ generators.

Edit: Here is a simple example. The first graph below is a $2$-fold cover of the second graph, with the colors of the edges matching up.

The corresponding homomorphism $\varphi : F_2 \to S_2$ sends the red generator $r$ to the identity and sends the blue generator $b$ to the transposition. Either of the blue edges is a spanning tree of the covering graph, and it follows that the fundamental group of the cover is free on generators $r, b^2, brb$. This subgroup is already normal, hence is $\text{ker}(\varphi)$ as desired.

Answer 2

As the following shows, GAP can work out generators of the kernel. Perhaps you were put off by the fact that it said "no generators known". That just means that it has not calculated them yet.

The other possibility is that the example you gave it was too large. If you tried mapping $F_{20}$ to $S_{20}$ for example, then $K$ would have far too many generators for GAP to want to start calculating and storing them. That is the main reason why it does not try to calculate the generators before you ask it to.

 gap> F:=FreeGroup(2);;
 gap> S:=SymmetricGroup(5);;
 gap> f := GroupHomomorphismByImages(F, S, [F.1,F.2], [(1,2),(1,2,3,4,5)]);;
 gap> K:=Kernel(f);
   Group(<free, no generators known>)
 gap> GK:=GeneratorsOfGroup(K);;
 gap> Length(GK);
  121