What kind of group product is this?

Question

I participated in an academy contest this year. My topic was symmetric groups.

Every permutation in $S_n$ can be decomposed into a rotation in $\mathbb{Z}_n$ and a permutation in $S_{n-1}$. I named this left form: $$ \begin{pmatrix} 4 & 3 & 2 & 1 & 0 \\ 2 & 0 & 1 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 2 & 0 & 1 & 3 \\ 2 & 0 & 1 & 3 & 4 \end{pmatrix} \begin{pmatrix} 3 & 2 & 1 & 0 \\ 2 & 0 & 1 & 3 \end{pmatrix} $$ To put it in inverse, the $n-1$ ($4$ here) gets into the first place, and the remaining elements are sorted. Note that the domain of the rotation depends on the permutation in $S_{n-1}$.

I proved that the dual form, right form exists, and has one-to-one correspondence to the left form:

$$ \begin{pmatrix} 4 & 3 & 2 & 1 & 0 \\ 2 & 0 & 1 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 1 & 0 \\ 2 & 0 & 1 & 3 \end{pmatrix} \begin{pmatrix} 4 & 3 & 2 & 1 & 0 \\ 3 & 2 & 1 & 0 & 4 \end{pmatrix} $$ Now the rotation is independent, and thus can be represented as an element in $\mathbb{Z}_n$.

What to pinpoint here is that I established a bijection from $S_n$ to the product of $\mathbb{Z}_n$ and $S_{n-1}$. But what kind of group product is this? It's clear that it's not the direct product. Is it a semidirect product?

Answer 1

When neither of the subgroups is normal, it is apparently called a Zappa–Szép product.

Let us identify $\mathbb Z_n$ with the subgroup $\langle(0\ 1\ 2 \ldots n-1)\rangle\le S_n$, and $S_{n-1}$ with the permutations in $S_n$ that fix $n-1$. Then you showed that $S_n = S_{n-1}\mathbb Z_n$ (i.e. the right form exists).

Since $S_{n-1}\cap \mathbb Z_n = \{\mathrm{id}\}$, then there is a theorem that says that the decomposition is actually unique, and then the decomposition is called an Zappa–Szép product. That means that we can also construct $S_n$ from scratch as an external Zappa–Szép product of $S_{n-1}$ and $\mathbb Z_n$.

Answer 2

Here $S_{n-1}\subset S_n$ is a subgroup of index $n$ realised as permutations fixing $n-1$. So it has exactly $n$ cosets. So we can find $n$ elements in $S_n$ picking one representative each from those cosets. So if $S$ is a complete set of representatives of distinct cosets of a subgroup $H\subset G$ any any element $g$ has a unique expression as $g=sh$ with $s\in S, h\in H$. This is always possible for every subgroup in a group.

What is noteworthy here in your case is that you can pick those $n$ representatives to form another subgroup.

Unfortunately neither of them is a normal subgroup. When normal we would have obtained a semidirect product.