By knit product (alias: Zappa-Szép product), I mean a product $ AB$ of subgroups for which $ A\cap B=1$ . In particular, note that neither subgroup is required to be normal, thus making this a generalization of the semidirect product.

**Synopsis of questions** (in order):

(1) Can someone provide subgroups $ A,B$ of $ S_{10}$ for which $ S_{10}=AB$ , $ A\cong S_6$ , and $ B\cong S_7$ ? (Note that by cardinality considerations, necessarily $ A\cap B=1$ if this happens, in which case $ S_{10}$ really is the knit product of the two.)

(2) Can it be proven, without a computer exhaust, that $ S_{10}$ does not have such a decomposition?

(3) How would one go about, with a computer exhaust, showing $ S_{10}$ does not have such a decomposition? This has as a subquestion: how would we know we captured all the weird ways each $ S_k$ with $ k=6,7$ embeds as a subgroup of $ S_{10}$ ?

For reference, this is similar to the question here, but even there it was pointed out there were additional ways the embeddings could occur.

*History:* Once upon a time (i.e., a number of years ago), I was contemplating ways one could factor a symmetric group $ S_n$ as a knit product of two symmetric subgroups $ A\cong S_a$ and $ B\cong S_b$ with positive integers $ a,b$ . Obviously, a necessary condition for this to happen is that $ n! = a!b!$ , so a natural question to ask is the corresponding number theory problem: when is it possible to write $ c!$ as a product $ a!b!$ ? Via computer runs, I quickly discovered two infinite families (breaking the symmetry between $ a$ and $ b$ , I’ll only write triples with $ a\leq b$ ), which are $ (a,b,c) = (1,n,n)$ over integers $ n\geq 1$ and $ (a,b,c)=(n,n!-1,n!)$ over integers $ n\geq 3$ , and an outlier example $ (a,b,c)=(6,7,10)$ .

Returning these examples to the motivating group theory question, the first family obviously corresponds to the (extremely trivial) product of $ 1=S_1$ and $ S_n$ . Meanwhile, a Frattini argument applied to the right regular action of $ S_n$ on itself can be used to show $ \mathrm{Sym}(S_n)$ is the knit product of $ \mathrm{Sym}(S_n\smallsetminus \langle1\rangle)$ with the group $ H$ which is the image of the Cayley embedding $ S_n\hookrightarrow\textrm{Sym}(S_n)$ . This then yields the second family of factorizations.

All of this leads to the question: is there a factorization of $ S_{10}$ as a product of $ S_6$ and $ S_7$ , thus providing group theoretic reason for the triple $ (6,7,10)$ ? I seem to recall, but cannot find the e-mail, that a friend of mine did a computer run to verify there is no copy-of-$ S_6$ , copy-of-$ S_7$ pair for which the product is $ S_{10}$ and which intersect trivially.

If I’m wrong in my recollection, and there does exist a decomposition of $ S_{10}$ as a knit product of a copy-of-$ S_6$ times a copy-of-$ S_7$ , I would appreciate enough details to be convinced it is true, including knowledge about which copy of each $ S_k$ is being considered (e.g., generating set of the $ S_k$ -copy, or a monomorphism $ S_k\rightarrow S_{10}$ ).

If I do recall correctly that there is no such factorization, then can someone provide a proof of that fact (directly or via reference)?

Barring the first being true and the second being fulfilled, my fallback position is that I would like to reproduce that computation for myself, except I don’t have a solid feel for how many different ways each $ S_k$ , $ k=6,7$ can embed into $ S_{10}$ . Therefore, a necessary step in an algorithmic process is coming up with a full list of copies-of-$ S_k$ .

Likely the best way to gather that information would be to provide a representative for each conjugacy class. (If there is a better way to perform the computation, I am all ears.)

As to the conjugacy classes of which I am aware:

$ \bullet$ The symmetric groups that move exactly $ k$ letters among the $ 10$ letters are the conjugates of the usual subgroup interpretation of $ S_k$ .

$ \bullet$ There is, generally speaking, an embedding $ S_k$ into $ A_{k+2}$ given by mapping members of $ A_k$ to themselves and mapping $ \sigma(1\;2)$ in the coset $ A_k(1\;2)$ to $ \sigma(1\;2)(k+1\;k+2)$ . This yields the conjugacy class representative $ A_k\cup \bigl(A_k(1\;2)(k+1\;k+2)\bigr)$ .

As an aside for anyone who might be interested, while I have been given reason to believe $ 10! = 6!7!$ does not come up as a symmetric group factorization (via the aforementioned, now lost e-mail), it does come up as a permutation group factorization. Via a Frattini argument applied to the sharply $ 3$ -transitive action of the Mathieu group $ M_{10}$ on $ 10$ letters, the symmetric group $ S_{10}$ is the knit product of $ S_7$ and $ M_{10}$ , and $ |M_{10}|=720=6!$ . This makes me think that the sporadic example really is sporadic, in that it (likely) arises through similar “happy accidents” of small numbers that allows $ A_6$ to have nontrivial outer automorphisms. I am very curious if the two families and this sporadic example really do represent the only solutions $ (a,b,c)$ to $ c!=a!b!$ , but even if true a proof of that fact is not likely to materialize any time soon.