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 $B \setminus A$.

Examples

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.