6
$\begingroup$

Homomorphisms $B_n \to B_{2n}$ and $B_n \to S_{2n}$ have been classified in Chen–Kordek–Margalit - Homomorphisms between braid groups and Lin - Braids and permutations respectively. I am interested in the corresponding question for symmetric groups: what are the homomorphisms $S_n \to S_{n+k}$ (up to conjugation)?

I know from Explicit description of all morphisms between symmetric groups. that such a classification is difficult in full generality—I am willing to restrict my attention to suitably small $k$. The answer to the linked question discusses the classification of maximal subgroups of $S_m$ isomorphic to $S_n$, but I do not see how this is directly useful, as the image of a map $S_n \to S_{n+k}$ needn't be maximal.

When $k = 1$ and $n + 1 \neq 6$, the only index $n+1$ subgroups of $S_{n+1}$ are point stabilizers. Thus, any non-cyclic map $S_n \to S_{n+1}$ is conjugate to the obvious inclusion.

When $k > 1$, we may combine* the identity map $S_n \to S_n$ with a sign representation $S_n \to \mathbb{Z}_2 \to S_{k}$ to produce a map $S_n \to S_{n+k}$ not conjugate to an inclusion. Following Chen–Kordek–Margalit, we might hope that these are all such homomorphisms (for small $k$).

Is anything known about these homomorphisms?

*In case it wasn't clear: this combination is found by letting $S_n$ act on $[1,\ldots, n]$ and $S_k$ act on $[n+1,\ldots,n+k]$

$\endgroup$
6
  • $\begingroup$ Related: mathoverflow.net/questions/302185/… $\endgroup$
    – YCor
    Apr 26, 2022 at 12:47
  • $\begingroup$ Small comment: I think the exception for $k = 1$ is when $n+1=6$, not $n=6$. $\endgroup$ Apr 26, 2022 at 12:56
  • $\begingroup$ What is a cyclic map $S_n \to S_{n + 1}$? \\ Your composition was originally $S_n \to \mathbb Z_2 \to S_k$, but, since you wanted the result to be a map $S_n \to S_{n + k}$, I figured the final target should be $S_{n + k}$, and edited accordingly. I hope that this was all right. $\endgroup$
    – LSpice
    Apr 26, 2022 at 13:14
  • 1
    $\begingroup$ @LSpice I am using Lin's terminology--a cyclic map is a map with cyclic image. Here, this just means "doesn't kill $A_n$". I have reverted your edit and elaborated on what I mean by "combine the identity map..." $\endgroup$ Apr 26, 2022 at 13:36
  • $\begingroup$ Ah, I see. I apologise for my wrong edit. $\endgroup$
    – LSpice
    Apr 26, 2022 at 14:03

1 Answer 1

7
$\begingroup$

As spin observed in a comment below, if you know all (conjugacy classes of) subgroups of $S_n$ of index up to $m$, then you can determine the equivalence classes of homomorphisms $\phi:S_n \to S_m$, because the subgroups tell you the possible actions on the orbits of the image of $\phi$. The subgroups of index up to $m=n^2$ are listed explicitly in Mikko Korhonen's answer to the the MSE post Large subgroups of Symmetric Group.

As an illustration, I will answer the question on the assumption that $k<n$.

For $n \ge 5$, a homomorphism from $S_n$ that is not injective has image of order $1$ or $2$, so we can restrict attention to injective maps $\phi:S_n \to S_{n+k}$.

From the MSE post referred to above we find that, for $n>6$, the only subgroups of index less than $2n$ are $S_n$, $A_n$, and the point stabilizers, which are isomorphic to $S_{n-1}$.

So if $S_n$ is acting faithfully on the set $\Omega := \{1,2,\dotsc,n+k\}$ with $k<n$, then there must be a single orbit $\Delta$ of length $n$ on which $S_n$ acts faithfully.

Since we are interested in classifying maps up to conjugation, we can assume that $\Delta = \{1,2,\dotsc,n\}$ and that the image of $\phi(g)$ restricted to $\Delta$ is $g$ for all $g \in S_n$.

Furthermore, the image of $\phi$ restricted to $\Omega \setminus \Delta$ has order $1$ or $2$.

So, $\phi(g)_{\Omega \setminus \Delta} = 1$ for $g \in A_n$, and for $g \in S_n \setminus A_n$, we have, up to conjugation, $\phi(g)_{\Omega \setminus \Delta}$ can be $1$, or $(n+1,n+2)$, or $(n+1,n+2)(n+3,n+4)$, etc., which gives a total of $\lfloor \frac{k+2}{2} \rfloor$ equivalence classes of injective homomorphisms $\phi$.

$\endgroup$
5
  • $\begingroup$ Does the subscript in $\phi(g)_{\Omega \setminus \Delta}$ indicate restriction? $\endgroup$
    – LSpice
    Apr 26, 2022 at 18:49
  • 1
    $\begingroup$ Yes that's right, the restriction of $\phi(g)$ to $\Omega \setminus \Delta$. $\endgroup$
    – Derek Holt
    Apr 26, 2022 at 19:05
  • $\begingroup$ I guess the basic principle is that if you know all transitive actions of $S_n$ on $\leq k+n$ points, then you know all homomorphisms $S_n \rightarrow S_{n+k}$. Each homomorphism corresponds to a partition $\{1,\ldots,n+k\} = X_1 \cup \cdots \cup X_t$ where $X_i$ is a transitive $S_n$-set. Then two homomorphisms are equivalent if and only if up to isomorphism of $G$-sets, then the set of $X_i$ that occur and the number of times they occur are the same. $\endgroup$
    – spin
    Apr 27, 2022 at 1:07
  • $\begingroup$ And finding all transitive actions of $S_n$ on $\leq k+n$ points up to isomorphism of $G$-sets amounts to finding subgroups $H < S_n$ with $[S_n : H] \leq k+n$ up to conjugacy. For $k = n^2-n$ the list is in the link MSE answer, I guess $k = n^2$ will not need much more work. $\endgroup$
    – spin
    Apr 27, 2022 at 1:09
  • $\begingroup$ @spin Yes that's right - I've edited my answer to clarify that point. $\endgroup$
    – Derek Holt
    Apr 27, 2022 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.