## Does the symmetric group $S_{10}$ factor as a knit product of symmetric subgroups $S_6$ and $S_7$?

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.

## Determine whether the following elements of $S_6$ are of the form $\sigma^2$.

The Problem:

Let $$S_6$$ be the symmetric group on six letters. Determine whether the following elements of $$S_6$$ are squares (i.e., of the form $$\sigma^2$$).

(a) $$(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$$.

(b) $$(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$$.

(c) $$(1 \hspace{1mm} 2 \hspace{1mm} 3)(4 \hspace{1mm} 5)$$.

My Progress:

(a) is clear. $$\sigma^2$$ must be an even permutation, and $$(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$$ is an odd permutation. Therefore there can be no $$\sigma \in S_6$$ such that $$\sigma^2 = (1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$$.

(b) is not so clear to me. $$(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$$ is, indeed, an even permutation, so the parity doesn’t help me here. I’ve pretty much exhausted all of my tools. I’m basically at the trial-and-error point now (e.g., trying to write $$(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$$ as the square of different permutations) which is getting me nowhere fast. Basically, I’m not sure what else can be said about an element of the form $$\sigma^2$$ other than the fact that it’s even.

(c) is giving me the same issues as (b).