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. As requested in the comments, we note that $10!=6!\cdot 7!$, so that the necessary cardinality considerations have been satisfied.

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.

Edited (3 Apr 2019) for motivational background:

Now that I remember, I thought I'd include the reason I was starting to look at such knit products in the first place. I had a need for a formal construction of an outer automorphism of $S_6$ that was guaranteed to have order $2$, and so started to follow standard constructions of such outer automorphisms, but with an eye to making choices so that the construction yielded one with this order. My route was to use a transitive copy $H$ of $S_5$ inside of $S_6$ and have $S_6$ act on the right cosets of $H$. Having a transversal stand in for the full set of cosets made sense computationally, so I looked for a transversal $T$ for $H$ in $S_6$ that would be ''easy to spot'' (really easy to find the (unique) $\tau\in T$ so that $H\tau =H\sigma$ for any given $\sigma\in S_6$). Using the ''natural ordering'' provided by my coding, I discovered that $S_3$ would serve as a transversal.

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    $\begingroup$ I'd start with the observation $10!=7!6!$ at the first line. $\endgroup$
    – YCor
    Mar 1, 2019 at 22:36
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    $\begingroup$ A possible better way than going for each conjugacy class is to use the fact that $S_n$ can be generated by $2$ element: an $n$-cycle and a transposition. $\endgroup$
    – user44191
    Mar 1, 2019 at 22:48
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    $\begingroup$ Subgroups of $S_{10}$ isomorphic to $S_7$ are easy to classify, using that $S_7$ has only three (up to conjugation) subgroups of index $\le 10$, namely of index 1,2,7. Hence there are only 2 conjugacy classes of subgroups $\simeq S_7$ in $S_{10}$: $S_7$ itself, and another subgroup of index 2 in $S_7\times S_2\subset S_{10}$. $\endgroup$
    – YCor
    Mar 1, 2019 at 22:53
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    $\begingroup$ Subgroups of index $\le 10$ in $S_6$ are a bit more complicated. Besides the three obvious ones (up to conjugacy) of index 1,2,6, there is one more of index 6, and one of index 10. If I'm correct, this produces a number of conjugacy classes of subgroups of $S_{10}$ isomorphic to $S_6$: the standard $S_6$, with 0,1, or 2 transpositions, and the transitive $S_6$: 4 conjugacy classes. The next step is to determine if such a subgroup can have trivial intersection with one of the subgroups isomorphic to $S_7$. $\endgroup$
    – YCor
    Mar 1, 2019 at 23:00
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    $\begingroup$ @YCor: $S_6$ acts transitively on the 10-element set of partitions into two 3-element sets. Pitifully, any set of 3 partitions is stabilized by a non-trivial permutation. $\endgroup$ Mar 2, 2019 at 0:52

2 Answers 2


Here are a few comments and a slightly different approach, though we take advantage of some of the earlier comments. We first note the well-known (at least to people who work with factorizations) fact that if the finite group $G$ has a factorization of the form $G = AB$ with $A \cap B = 1 $ and $A,B$ subgroups, then we have $A \cap B^{g} = 1$ for all $g \in G$- for we also have $G = BA,$ and if we write $g = ba$ for some $b \in B, a \in A,$ then we have $A \cap B^{g} = A \cap B^{a} = (A \cap B)^{a} = 1.$

This means that if $G = S_{10}$ has a factorization of the form $G = AB$ with $A \cong S_{6}$ and $B \cong S_{7}$ (so that $A \cap B = 1$ as noted in the body of the question), then no non-identity element of $A$ can have the same disjoint cycle structure (in the given embedding) as any non-identity element of $B.$

Now $S_{6}$ contains commuting (and conjugate) distinct involutions which are odd permutations (in the natural representation). Hence the subgroup $A$ above contains an involution which is an even permutation in the embedding in $S_{10}.$ This is either a product of two disjoint transpositions or product of four disjoint transpositions.

It is noted in comments that $B$ must be of the form $(S_{7} \times C_{2}) \cap A_{10},$ where the $S_{7}$ is a "natural" $S_{7}$ inside $S_{10},$ ie fixing three points. It follows that $B$ contains both involutions which are products of two disjoint transpositions, and involutions which are product of four disjoint transpositions, contrary to the remarks above.

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    $\begingroup$ Yes I like that! Even in terms of computation it reduces the calculation by a factor of $360$. $\endgroup$
    – Derek Holt
    Mar 2, 2019 at 13:14
  • $\begingroup$ @DerekHolt I agree. While I greatly appreciate your code (and see a personal use for it), I have to go with Geoff's answer. Thank, everybody! $\endgroup$
    – John McVey
    Mar 4, 2019 at 1:20
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    $\begingroup$ @Geoff Robinson: While I've used variants of that first paragraph, I've never seen one this succinct nor this broad before. Thanks especially for that. $\endgroup$
    – John McVey
    Mar 4, 2019 at 1:23

As has been pointed out in comments, the only subgroups $G$ and $H$ of $S_{10}$ isomorphic to $S_7$ and $S_6$ that could possibly have trivial intersections are the copy of $S_7$ that lies in $A_{10}$ and the copy of $S_6$ that is transitive on the $10$ points.

Note that $S_6 \cong {\rm P \Sigma L}(2,9)$, the extension of ${\rm PSL}(2,9)$ by a field automorphism, and the transitive $S_6$ in $S_{10}$ can be described that way.

There are only $360$ conjugates of $G \cong S_7$ in $S_{10}$, so it is an easy computer calcualtion to check that they all have intersection of order $2$ with ${\rm P \Sigma L}(2,9)$: see Magma calcualtion below. So there is no such factorization.

> G := sub<Sym(10) | (1,2,3), (1,2,3,4,5,6,7), (1,2)(3,4)(5,6)(8,9)>;
> isiso := IsIsomorphic(G,Sym(7)); isiso;                            
> H := PSigmaL(2,9);
> isiso := IsIsomorphic(H,Sym(6)); isiso;
> N := Normaliser(Sym(10),G);
> Index(Sym(10),N);
> T := Transversal(Sym(10),N);
> forall{t: t in T | #(G^t meet H) eq 2};
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    $\begingroup$ I think this can be seen directly, ie without computer. $\endgroup$ Mar 2, 2019 at 8:30
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    $\begingroup$ @GeoffRobinson Yes I am sure it can! One gets lazy with advancing age! $\endgroup$
    – Derek Holt
    Mar 2, 2019 at 8:40
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    $\begingroup$ @GeoffRobinson thanks: indeed your well-known remark answers the question in my previous comment: indeed since the transitive copy of $A_6$ contains quadruple transpositions (as images of double transpositions), and so does the copy of $S_7$ in $A_{10}$, we deduce that such subgroups will not work. So $A_{10}$ is not a knit product of a copy of $S_7$ and a copy of $A_6$. $\endgroup$
    – YCor
    Mar 2, 2019 at 13:11
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    $\begingroup$ On the other hand, $A_{10}$ is a knit product of a (standard) copy of $A_7\times S_2$ and the transitive $A_6$. Indeed, the only conjugacy classes in $S_{10}$ of elements of the transitive $A_6$ have type $1^{10}$, $1^13^3$, $5^2$, and $1^22^4$. This includes no element conjugate into $A_7$, and no odd element, so the intersection with the standard $A_7\times S_2$ is trivial. $\endgroup$
    – YCor
    Mar 2, 2019 at 13:15
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    $\begingroup$ Late comment, but if $G = AB$, then $G = A^x B^y$ for all $x,y \in G$. (Proof: write $xy^{-1} = ab$, then $G = Axy^{-1}B$ so $G = x^{-1}Axy^{-1}By$.) So I think you just have to compute one intersection. $\endgroup$
    – spin
    Dec 4, 2021 at 1:02

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