What kind of group product is this? 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?
 A: 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 $\mathbb Z_n$.
A: 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.
