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.