Show that $G_X\le{\rm Sym}(A)$ where $X\subseteq A$ and $G_X=\{g\in{\rm Sym}(A)\mid g(X) = X\}.$

Question

Show that $G_X \le{\rm Sym}(A)$ where $X \subseteq A$ and $G_X =\{g \in{\rm Sym}(A)\mid g(X) = X\}$

I'm a bit confused with this question and how to approach it. First I believe I need to show $G_X \subseteq{\rm Sym}(A)$. Then, we know $g$ permutes the elements of $X$ in $X$ but the image of $X$ remains in $X$. When you restrict $g$ to $X$ I feel like you get a bijection (I'm not sure how to show this). I don't know how to show that is a subgroup (that $xy^{-1} \in G_X$) and how it relates to the symmetric group of $A$.

Also because I believe it is a bijection, does that mean that if the order of $A$ is equal to $n$, the order of $G_X$ is equal to $n! (|S_n| = n!)$?

I'm really confused

Answer 1

You don’t have to show that $G_X\subseteq\operatorname{Sym}(A)$: that fact is part of the definition of $G_X$. And there is no reason (or need) to assume that $A$ is finite.

Let $f,g\in G_X$. By definition $g\in\operatorname{Sym}(A)$, so $g$ is a bijection from $A$ to $A$ and therefore has an inverse $g^{-1}$ in $\operatorname{Sym}(A)$. Since $g\in G_X$, $g[X]=X$, so $g^{-1}[X]\supseteq X$. And $g$ is a bijection, so in fact $g^{-1}[X]=X$: otherwise there would be $x_0,x_1\in X$ and $a\in A\setminus X$ such that $g(x_0)=x_1=g(a)$, and $g$ would not be injective. Thus, $g^{-1}\in G_X$, and $f\big[g^{-1}[X]\big]=f[X]=X$, so $fg^{-1}\in G_X$.

Saying that $G_X$ is a subgroup of $\operatorname{Sym}(A)$ just amounts to saying that every permutation of $A$ that leaves the set $X$ setwise fixed has an inverse that leaves $X$ setwise fixed, and the composition of permutations of $A$ that leave $X$ setwise fixed is a permutation of $A$ that leaves $X$ setwise fixed.

Added: As Chris Custer points out below, I really ought to say explicitly that $G_X$ is clearly non-empty, since the identity map on $A$ is in $G_X$.

Answer 2

Think about what $G_X$ is. It is the set of things in $\operatorname{Sym}(A)$ which permute the elements of $X$ among themselves. Isn't it obvious that if $g_1$ and $g_2$ permute the elements of $X$ among themselves, then so does $g_1g_2$?

Yes, if you restrict $g \in G_X$ to $X$, you get a bijection from $X$ to $X$. This is also obvious, because $g$, being an element of $\operatorname{Sym}(A)$, is already a bijection from $A$ to $A$, and so its restriction to $X$ is a bijection from $X$ onto the image of $X$, which is again $X$.

No, the order of $G_X$ is usually not going to be $n!$. Remember that $\operatorname{Sym}(A)$ is the group of bijections from $A$ to itself, and it has order $n!$. Since $G_X$ is a subgroup of $\operatorname{Sym}(A)$, it will generally have fewer elements. Maybe you are confused because you are thinking of $G_X$ as a subgroup of $\operatorname{Sym}(X)$? It is not, since the elements of $G_X$ are bijections from $A$ to $A$, not from $X$ to $X$. However, there is a surjective homomorphism

$$G_X \rightarrow \operatorname{Sym}(X)$$

$$g \mapsto g|_X$$

where you "forget" what $g$ does to all the elements of $A$ which are not in $X$.