# Order of subgroup of symmetric group

Let $X$ be a finite set, i.e. that $|X| = n$, and let $G = \operatorname{Sym}(X)$ be the symmetric group on $X$. Let $Y \subseteq X$ be a subset of $X$ and define the subset $G_Y \subseteq G$ to be the set $G_Y = \{g \in G\mid \forall y \in Y:\ g(y) \in Y \$and$\ g^{-1}(y) \in Y\ \}$, which is a subgroup of $G$.

I want to find the order of $G_Y$ as a function of $|X|=n$ and $|Y|$.

I have something but am not sure at all about it: We are looking for the number of bijections fixing $y \in Y$, thus, $|G_Y| = (|X|-1)!\cdot|Y|$.

Is this getting anywhere? Any help would be appreciated!

• In your definition of $G_Y$ it seems that elements of $Y$ should be sent to other elements of $Y$, not necessarily fixing elements of $Y$ as you state in the second to last paragraph. Which is it? Commented May 9, 2016 at 20:49

Every permutation of $X$ that leaves $Y$ invariant as a set is really composed of two distinct, independent and non-overlapping permutations: one is a permutation of $Y$, and the other is a permutation of $X \setminus Y$. Indeed, if a permutation of $X$ maps all the elements of $Y$ back into $Y$, then the restriction of the original permutation (on $X$) to the subset $Y$ is bijective (injective and going from $Y$ to itself means bijective if $Y$ is finite), and then this means $X\setminus Y$ must be mapped to itself.
As you may know or prove, a permutation of $Y$ and one of $X \setminus Y$ commute. It should now be easy to prove that $G_Y$ is the direct product of the symmetric group of $Y$ and the symmetric group of $X\setminus Y$. So if $X$ has $n$ elements and $Y$ has $m$ elements, $G_Y$ has $\frac{n!}{m!(n-m)!}$ elements.