Since pairs of transpositions generate for by construction, we need only show that any pair of transpositions can be written as a product of 3-cycles. Noting that , the following list exhausts any pair of transpositions:

proving ^S1.

Now for can in fact be generated only from 3-cycles of the form with fixed in , since every 3-cycle can be expressed as such:

Now assume contains a 3-cycle, say . By normality it follows for any

Hence contains all 3-cycles of the form and hence all 3-cycles, thus by ^S1 it is , proving ^S2.

Now let for be nontrivial. Then one of the following holds:

Case 1: Suppose there exists with a cycle of length , so without loss of generality (by relabelling) where and are disjoint. Since , it follows

so contains a 3-cycle.

Case 2a: Suppose there exists a which contains two disjoint 3-cycles (and nothing longer). Without loss of generality, for disjoint , , and . Since , it follows

which falls under case 1.

Case 2b: Suppose there exists a containing exactly one 3-cycle and otherwise only transpositions. Without loss of generality where and are disjoint, and . Then , so contains a 3-cycle.

Case 2c: If contains a 3-cycle we are already done.

Case 3: The only remaining possibility is that there exists a which is a product of disjoint transpositions, and an even number thereof since . Without loss of generality with , , disjoint and . Then

whence

thus contains a 3-cycle.

This proves ^S3 and therewith the simplicity of for .