$A_n$ Can be Generated by 3-Cycles with Consecutive Elements

group-theorypermutations

Let $ A_n $ be the alternating group on $n$ elements. I have already proven that $A_n$ can be generated by the set of 3-cycles with distinct elements. However, I am not sure how to further prove that $A_n$ can be generated by the set of 3-cycles with consecutive elements i.e. $ A_n = \langle (k,k+1,k+2) : 1 \leq k \leq n-2 \rangle$.

The natural thing to do is to show that an arbitrary 3-cycle with distinct elements can be generated by 3-cycles with consecutive elements, but I don't know how to do this.

Best Answer

Induction on $n$ plus transitivity of $A_n$, $n\ge3$, settle this.

Base Case.

The claim is clear when $n=3$.

Inductive Step.

I will use the natural identification of elements of $A_n$ with those elements of $A_{n+1}$ that map the number $n+1$ to itself.

Assume that $n\ge3$ and that the claim holds for $A_n$. We show that the claim holds for $A_{n+1}$ as well.

Let $\sigma\in A_{n+1}$ be arbitrary. Let $d=\sigma(n+1)$. If $d=n+1$, then $\sigma\in A_n$. And by the induction hypothesis, it is a product of 3-cycles of consecutive elements involving $\{1,2,\ldots,n\}$ only.

Otherwise $d<n+1$. Because $A_n$ is transitive, there is an element $\alpha\in A_n$ such that $\alpha(d)=n$. It follows that $$ \beta:=(n-1;n;n+1)\alpha\sigma\in A_n $$ because it maps the number $n+1$ to itseld.

Solving for $\sigma$ from this gives $$ \sigma=\alpha^{-1}(n-1;n;n+1)^2\beta. $$ The induction hypothesis says that as elements of $A_n$ both $\beta$ and $\alpha^{-1}$ can be written in terms of the prescribed generators. Hence so can $\sigma$. As $\sigma$ was an arbitrary element of $A_{n+1}$, the inductive step is valid.

An alternative, possibly closer to what the OP had in mind, could be the following:

  1. Brute force it for $A_4$.
  2. In the inductive step we can then assume that $n\ge4$. Thus we know that $A_n$ is doubly transitive.
  3. In the inductive step it suffice to show that all 3-cycles of the form $(a;b;n+1)$ can be generated. But, by double transitivity there exists an element $\alpha\in A_n$ such that $\alpha(n-1)=a$ and $\alpha(n)=b$. The usual conjugation rule of cycles says that $$ \alpha (n-1;n;n+1) \alpha^{-1}=(a;b;n+1). $$ Here both $\alpha$ and its inverse can be written in terms of the prescribed generators, so $(a;b;n+1)$ is included, and we are done.
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