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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.

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.

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 $\tau H=\sigma H$ 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.

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.

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 $\tau H=\sigma H$ 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.