The alternating group is generated by three-cycles

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.

What do you think ?

Solution 1:

A permutation is an element of $A_n$ if and only if it is a product of an even number of transpositions.

We first note that the three cycles do not generate more than $A_n$ since for distinct $i,j,$ and $k$, we have $(ij)(ik)=(ijk)$.

For the other inclusion, we note that $(ijk)=(ij)(ik)$ for distinct $i,j,$ and $k$. And for distinct $i,j,k,$ and $l$, we have that $(ij)(kl)=(ijk)(jkl)$. Thus the three cycles generate all products of an even number of transpositions.

Solution 2:

The idea of the proof is that we will show that any product of two transpositions is a product of 3-cycles ; and ,if so, then since every member of the A_n is a product of even number of transpositions, we are done.

To begin with lets us suppose that we have a product of transpositions a¹ and a² where both have a common number a which belongs to the set {1,2,.....}- say they both move a. Then we have a¹ of the form (a b) and a² of the form (a c) . In this case , we have

a¹ a² = (a b)(a,c)= (a,c,b) and we are done. Now suppose that a¹ and a² move different numbers .

a¹ = (a b) a² = (c d) Then

a¹a² = (a,b)(c,d) = ( d a c ) ( a b d ). And we are done.