Prove that, for $n \geq 3$, the three-cycles generate the alternation group $A_n$
Proof: We multiply on the left by 3-cycles to "reduce" an even permutation $p$ to the identity, using induction on the number of indices fixed by a permutation. How the indices are numbered is irrelevant. If $p$ contains a $k$-cycle with $k \geq 3$, we may assume that it has the form $p=(123\dots k)\dots$ Multiplying on the left by $(321)$ gives $$p'= (321)(123 \dots k)\dots=(1)(2)(3\dots k)\dots$$ More fixed indices.
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