Permutations in $S_{A}$ fixing all the elements in a subset $B\subset A$ form a subgroup.

Question

Let $A$ be a finite set, and $B$ a subset of $A$. Let $G$ be the subset of $S_A$ consisting of all the permutations $f$ of $A$ such that $f(x) = x$ for every $ x \in B$. Prove that $G$ is a subgroup of $S_A$.

I am thinking of starting off by showing that $G$ is closed with respect to the inverse, but I am unsure if this is the correct way to go about it.

Any help?

Answer 1

Show that $ab\in G$ for all $a,b$ in G. That is enough.

Answer 2

$A$ is finite, so $S_A$ is finite.

In general a subset of a finite group that is closed under multiplication is a subgroup of it.

For a proof of this see the answer on this question.

So it is enought to prove that $f,g\in G\implies f\circ g\in G$ wich can be classified as trivial.