When is $G \subset S_{2n}$ isomorphic to $S_m \times \Bbb Z/2\Bbb Z$? Consider the subgroup $G \subset S_{2n}$ consisting of permutations $\sigma\in S_{2n}$ such that $\sigma (i) + \sigma (2n+1-i) = 2n+1$.
I want to know if it is true that $G \subset S_{2n}$ is isomorphic to $S_m \times \Bbb Z/2\Bbb Z$ when $ m = n$? If not, what is the smallest possible $m$? And if there is a general solution for this, or we must find individual $m$ for every $n$ like $2,3,4,... $.
I'm thinking that since the value of $\{\sigma(i):i \leq n \}$ uniquely determines the $\sigma \in G$, then one may devise some map that maps $S_n$ to $G$ in this manner. But for each $i$, we can choose to map it to either $\sigma(i)$ or $2n+1-\sigma(i)$ and fix the value of $2n + 1 - i $ accordingly; yet it seems in this way the more suitable option is $S_n \times\Bbb Z/2^n\Bbb Z$.
 A: Proving that $G$ is a Group
Closure : Let $\tau,\sigma \in G$. Then, we have:
$$x+(2n+1-x)=2n+1 \implies \sigma(x)+\sigma(2n+1-x)=2n+1$$
$$\implies \tau(\sigma(x))+\tau(\sigma(2n+1-x))=2n+1 \implies \tau \circ \sigma \in G$$
which proves closure of $G$.
Associativity : This directly follows from the fact that $G \subseteq S_{2n}$.
Identity : The identity element of $G$ is the same as the identity element of $S_{2n}$.
Inverse : Let $\sigma \in G$. Let $\sigma^{-1}$ be the inverse of $\sigma$ in $S_{2n}$. We have:
$$\sigma(\sigma^{-1}(x))+\sigma(\sigma^{-1}(2n+1-x))=2n+1$$
$$\implies \sigma^{-1}(x)+\sigma^{-1}(2n+1-x)=2n+1 \implies \sigma^{-1} \in G$$
and thus, $\sigma^{-1}$ serves as the inverse of $\sigma$ in $G$.
Thus, $G$ is a group.

Isomorphism Type of $G$
We claim that $G \cong S_n \times (\mathbb{Z}/2\mathbb{Z})^n$. Consider the action of $G$ on the elements of $\{1,2,\ldots,2n\}$. We can see that for $S=\{\{1,2n\},\{2,2n-1\},\ldots,\{n,n+1\}\}$, $\sigma(S)=S$ if and only if $\sigma \in G$. We see that $\sigma$ permutes all the $n$ elements in $S$. Moreover, if $\sigma(\{x,2n+1-x\})=\{y,2n+1-y\}$, we have the choice of taking $\sigma(x)=y$ or $\sigma(x)=2n+1-y$. This gives $2^n$ choices. Hence, $|G|=n! \cdot 2^n$.
Consider the following bijection from $G \to S_n \times (\mathbb{Z}/2\mathbb{Z})^n$:

*

*Number the elements of $S$ as $\{a_1, a_2, a_3, \ldots , a_n\}$.

*Let the permutation induced by $\sigma \in G$ on $S$ be isomorphic to $\tau_\sigma \in S_n$.

*Denote the elements of $(\mathbb{Z}/2\mathbb{Z})^n$ by $(b_1, b_2, \ldots, b_n)$ with $b_i \in \{0,1\}$.

*Given $\sigma \in G$, define $\alpha_\sigma \in (\mathbb{Z}/2\mathbb{Z})^n$, such that $\alpha_\sigma = (b_1, b_2, \ldots, b_n)$ where:
$$b_i=\begin{cases}
0 && \text{if $\sigma(i)<\sigma(2n+1-i)$} \\
1 && \text{if $\sigma(i)>\sigma(2n+1-i)$}
\end{cases}$$
The bijection we consider is $G \to S_n \times (\mathbb{Z}/2\mathbb{Z})^n$, which performs the map $\sigma \to (\tau_\sigma,\alpha_\sigma)$. We prove that this is an isomorphism. We must prove that $\tau_{xy} = \tau_x\tau_y$ and $\alpha_{xy}=\alpha_x\alpha_y$. Both of these are direct. This proves the required.
