Determine the order of the subgroup H of $S_n$ for $n \geq 3$

Question

From the chapter on Permutation Groups in Gallian's 'Abstract Algebra' 9th Ed., we are asked to prove that for $n \geq 3,$ $$H=\{\beta \in S_{n}|\beta (1) \in \{1,2\} \land \beta (2)\in\{1,2\}\}$$ is a subgroup of $S_{n}$ and then determine (not prove) what its order is. To the extent that it matters, a sketch of my proof of the first part is...

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"Note $H \neq \emptyset$ since clearly the identity permutation $\epsilon \in H$. Now take $\alpha, \gamma \in H$. Consider $\alpha\gamma(1)=\alpha(\gamma(1)) \in S_{n}$. Note $\gamma(1)\in\{1,2\}$ by definition of $\gamma \in H$. It follows that $\alpha\gamma(1) \in \{1,2\}$ by definition of $\alpha \in H$. A similar proof shows $\alpha\gamma(2)\in\{1,2\}$ and thus $\alpha\gamma \in H$, proving $H \leq G$ by the 'Finite Subgroup Test.' QED."

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The last part of the question stuck me. My limited combinatorics talents are rusty so I played around with examples for $S_{3}, S_{4}, S_{5}, S_{6}$ before throwing in the towel since I could find no apparent patterns that relied only on $n\geq3$. My examples did give me the correct orders of the $H$ in each case - that is, $2, 4, 12,$ and $48$ - and I did this by the tried and true method of listing the lengths of the disjoint cycles of all elements in $S_{n}$, noting which disjoint cycles would allow $\beta(1)\in\{1,2\}$ and $\beta(2)\in\{1,2\}$, and then meticulously counting them.

tldr; Upon giving up, I found that $|H|=2|S_{n-2}|=2(n-2)!$ on the internet/ solutions manual without any proof as to why. This is interesting and would like to know why for my own edification. So, my question ultimately is, what is a [hopefully] elementary algebraic/ combinatorial proof as to why $|H|=2|S_{n-2}|=2(n-2)!$? Any help or insight would be greatly appreciated!

Answer 1

If $\beta\in H$ then, as both $\beta(1)$ and $\beta(2)$ are in $\{1,2\}$, and $\beta$ is a permutation, we conclude that either

1. $\beta(1)=1$ and $\beta(2)=2$

or

2. $\beta(1)=2$ and $\beta(2)=1$

For the rest of the numbers in $\{3,\ldots,n\}$, note that $\beta$ has to permute these numbers (since the action on $1$ and $2$ is already constrained). But there are no other constraints on what $\beta$ does to $\{3,\ldots,n\}$.

In other words an element of $H$ breaks down into a permutation of $\{1,2\}$ and a permutation of $\{3,\ldots,n\}$. So we can uniquely specify such a $\beta$ by choosing one of the two options above and then choosing a permutation of $\{3,\ldots,n\}$.

Since there are $(n-2)!$ permutations of $\{3,\ldots,n\}$, this means $2(n-2)!$ total choices.

Another way to look at it is to let $S$ be the group of permutations of $\{3,\ldots,n\}$. Clearly $S\cong S_{n-2}$. Then the above description induces a natural ismorphism from $H$ to $S_{2}\times S\cong S_{2}\times S_{n-2}$.