Finding subgroups of $S_n$ using any subset of $S = \{1, ..., n \}$.

Question

I came across the following strategy that is supposed to help finding subgroups of $S_n$ (the symmetric group of degree $n$.)

• Let $A$ be any subset of $S = \{1, ..., n \}$.
• Let $G$ be the subset of the symmetric group $S_n$ that consists of all permutations that map each element of $A$ to another element of $A$.
• It can be shown that $G$ is a subgroup of $S_n$.

I went through the proof that $G$ is indeed a subgroup of $S_n$, and I get the technicalities of it. However I can't really picture in my head what is actually happening.

Can someone explain more intuitively why it makes sense to use a subset of $S = \{1, ..., n \}$ as a vehicle to find subgroups of $S_n$? I worked it out on paper for $S_3$ and of course it works, but I just can't get the intuition of why this makes sense.

Answer 1

In general, let $X$ be a set and $Y \subseteq X$. For a given bijection $u$ on $X \setminus Y$, any bijection $\alpha$ on $Y$ can be extended to a bijection $f_u(\alpha)$ on $X$ by:

\begin{alignat}{1} &f_u(\alpha)_{|Y}:=\alpha \\ &f_u(\alpha)_{|X \setminus Y}:=u \\ \tag 1 \end{alignat}

Note that $f_u(\alpha)=f_u(\beta) \Rightarrow f_u(\alpha)_{|Y}=f_u(\beta)_{|Y} \Rightarrow \alpha=\beta$, so that $f_u$ is injective for all $u \in \operatorname{Sym}(X \setminus Y)$. Moreover,

$$f_u(\alpha\beta)=f_u(\alpha)f_u(\beta) \iff u=\iota_{X \setminus Y} \tag 2$$

($\iota_B$ is the identical map on the set $B$: $\iota_B(b)=b$ for every $b \in B$.) Therefore, if we take $u=\iota_{X \setminus Y}$, we have $f_u : {\rm{Sym}}(Y) \hookrightarrow {\rm{Sym}}(X)$. Now, take $X$ your $S$ (and then $\operatorname{Sym}(X)=S_n$) and $Y$ your $A\subseteq S$, and you're done.

More straightforwardly (but driven by what above, actually), consider the set of all the permutations of $S_n$ which fix the elements of $S\setminus A$. This is a finite subset of $S_n$ closed under composition, and thence it is a subgroup of $S_n$.