(a) Let $A$ be a finite set, and $B\subseteq A$. Let $G$ be the subset of $S_A$ (permutations of $A$) consisting of all the permutations $f$ of $A$ such that $f(x)\in B$ for every $x\in B$. Prove that $G$ is a subgroup of $S_A$.
(b) Is the conclusion of part (a) still true if $A$ is an infinite set?
To check whether $G$ is a subgroup of $S_A$, we need to check closure and inverse.
Closure: Suppose $f,g$ are permutations such that $f(x)\in B$ and $g(x)\in B$ for every $x\in B$. Then $fg$ is a permutation such that $f(g(x))\in B$ for $x\in B$. This holds whether $A$ is finite or infinite.
Inverse: Suppose $f$ is a permutation such that $f(x)\in B$ for every $x\in B$. If $A$ is finite, $B$ is also finite, so $f^{-1}(x)\in B$ for every $x\in B$. So $f^{-1}\in G$. But what if $A$ is infinite?
Best Answer
Consider the permutations of $\mathbb{Z},$ and consider the set $B=\mathbb{N},$ and $f$ a permutation defined by $f(x) = x+1.$ This maps natural numbers to natural numbers, but what about its inverse?