Symmetric groups are normal-comparable

Statement
The symmetric group on any set is a normal-comparable group: any two normal subgroups of the group can be compared, i.e., one of them must lie inside the other.

Facts used

 * 1) uses::Baer-Schreier-Ulam theorem

Proof
We consider various cases:


 * For an infinite set, the result follows from fact (1) (the Baer-Schreier-Ulam theorem) that classifies all normal subgroups of the symmetric group.
 * For a finite set of size $$n \ge 5$$ or $$n = 3$$: The alternating group is a simple normal subgroup, and is also of index two. Thus, any other proper nontrivial normal subgroup must intersect it trivially, and hence must be of order two. However, a normal subgroup of order two contains a non-identity central element, and we know that the group is centerless. Thus, the alternating group is the only proper nontrivial normal subgroup, and thus, any two normal subgroups are comparable.
 * For a finite set of size $$n = 2$$: In this case, there are no proper nontrivial subgroups.
 * For a finite set of size $$n = 4$$: In this case, there are two normal subgroups: a subgroup of order four comprising the double transpositions and the identity map, and the alternating group, which is of order twelve. The subgroup of order four lies inside the subgroup of order twelve.