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 $[n+1,\ldots,n+k]$

  • $\begingroup$ Related: mathoverflow.net/questions/302185/… $\endgroup$
    – YCor
    Commented Apr 26, 2022 at 12:47
  • $\begingroup$ Small comment: I think the exception for $k = 1$ is when $n+1=6$, not $n=6$. $\endgroup$ Commented 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
    Commented 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..." $\endgroup$ Commented Apr 26, 2022 at 13:36
  • $\begingroup$ Ah, I see. I apologise for my wrong edit. $\endgroup$
    – LSpice
    Commented Apr 26, 2022 at 14:03

1 Answer 1


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 $\phi$.

  • $\begingroup$ Does the subscript in $\phi(g)_{\Omega \setminus \Delta}$ indicate restriction? $\endgroup$
    – LSpice
    Commented Apr 26, 2022 at 18:49
  • 1
    $\begingroup$ Yes that's right, the restriction of $\phi(g)$ to $\Omega \setminus \Delta$. $\endgroup$
    – Derek Holt
    Commented 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
    Commented 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
    Commented Apr 27, 2022 at 1:09
  • $\begingroup$ @spin Yes that's right - I've edited my answer to clarify that point. $\endgroup$
    – Derek Holt
    Commented Apr 27, 2022 at 7:54

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