# Symmetric group on finite or cofinite subset is conjugacy-closed

This article gives the statement, and proof, of a particular subgroup in a group being conjugacy-closed: in other words, any two elements of the subgroup that are conjugate in the whole group, are also conjugate in the subgroup
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This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup satisfying a particular subgroup property (namely, Conjugacy-closed subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).

## Statement

Suppose $S \subseteq T$ are sets. Then, the symmetric group $\operatorname{Sym}(S)$ embeds naturally as a subgroup of $\operatorname{Sym}(T)$: any permutation of $S$ extends to a permutation of $T$ as the identity map on $T \setminus S$.

With this embedding, if either $S$ or $T \setminus S$ is finite, $\operatorname{Sym}(S)$ is a conjugacy-closed subgroup in $\operatorname{Sym}(T)$. In other words, if two elements of $\operatorname{Sym}(S)$ are conjugate in $\operatorname{Sym}(T)$, they are also conjugate in $\operatorname{Sym}(S)$.

## Related facts

### Stronger formulation

Symmetric group on finite or cofinite subset is subset-conjugacy-closed: Not only can we perform conjugation of single elements, we can also perform conjugation of subsets inducing exactly the same map on each element of the subset.

### Breakdown for infinite coinfinite subsets

If both $S$ and $T \setminus S$ are infinite, then $\operatorname{Sym}(S)$ is not conjugacy-closed in $\operatorname{Sym}(T)$. Further information: Symmetric group on infinite coinfinite subset is not conjugacy-closed