Subset containment gives inclusion of symmetric groups

Statement
Suppose $$A$$ is a subset of $$B$$. Then, the symmetric group on $$A$$ can be identified with a subgroup of the symmetric group on $$B$$ as follows:

For any permutation $$\sigma$$ of $$A$$, we treat $$\sigma$$ as a permutation of $$B$$ by setting it as the identity map on all elements in $$B \setminus A$$.

In particular, the symmetric group on the subset $$A$$ is the subset of the symmetric group on $$B$$ comprising those permutations that fix every element in $$B \setminus A$$.

An ordinary example
Suppose $$A = \{ 1,2,3 \}$$ and $$B = \{ 1,2,3,4,5 \}$$. Suppose $$\sigma$$ is the permutation of $$A$$ given by:

$$\sigma(1) = 3, \sigma(2) = 1, \sigma(3) = 2$$.

Then, we can treat $$\sigma$$ as a permutation of $$B$$ simply by defining it as the identity map on the elements $$4$$ and $$5$$:

$$\sigma(1) = 3, \sigma(2) = 1, \sigma(3) = 2, \sigma(4) = 4, \sigma(5) = 5$$.

Extreme examples

 * The empty subset is a subset of every set. The symmetric group on the empty set, which is the trivial group, is thus naturally a subgroup of the symmetric group on every set.
 * Every set is a subset of itself. The corresponding subgroup of the symmetric group is the whole group.
 * The symmetric group on a subset that is the complement of one element is precisely the stabilizer of that element. For instance, in the set $$\{ 1,2,3,4 \}$$, the symmetric group on the subset $$\{ 1,3,4 \}$$ is precisely the stabilizer of $$2$$ in the whole symmetric group.