# Is there a classification of homomorphisms $S_n \to S_{n+k}$ for small $k$?

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

• – YCor
Apr 26, 2022 at 12:47
• Small comment: I think the exception for $k = 1$ is when $n+1=6$, not $n=6$. Apr 26, 2022 at 12:56
• 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. Apr 26, 2022 at 13:14
• @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..." Apr 26, 2022 at 13:36
• Ah, I see. I apologise for my wrong edit. Apr 26, 2022 at 14:03

## 1 Answer

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.

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 $$k, then there must be a single orbit $$\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 $$\phi$$.

• Does the subscript in $\phi(g)_{\Omega \setminus \Delta}$ indicate restriction? Apr 26, 2022 at 18:49
• Yes that's right, the restriction of $\phi(g)$ to $\Omega \setminus \Delta$. Apr 26, 2022 at 19:05
• 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.
– spin
Apr 27, 2022 at 1:07
• 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.
– spin
Apr 27, 2022 at 1:09
• @spin Yes that's right - I've edited my answer to clarify that point. Apr 27, 2022 at 7:54