How can I tell if a permutation can be expressed as the commutator of another 2 permutations?

Question

Given a permutation, how could I tell if it can be decomposed into the commutator of 2 permuations? I have known the formula of commutators and the following fact $$[(a,c),(b,c)]=(a,b,c)$$ But for longer permutation, I don't know how to do. For example, can $(1,5,7,10)(2,8,3)(4,6,9)$ be decomposed in this way? Can the formula above be applied?

Answer 1

For the question in the post (now edited):

In any group other than the trivial group and the group of two elements, every element can be written as the product of two nontrivial elements. In particular, for every $n\geq 3$, if $\sigma$ is any permutation, and $\rho$ is a given nontrivial permutation, then there is a permutation $\pi$ such that $\sigma=\rho\pi$, namely $\pi=\rho^{-1}\sigma$. If $\rho\neq\sigma$ (which is always possible) then neither $\rho$ nor $\pi$ are trivial.

So every permutation in $S_n$, $n\geq 3$, is a product of two nontrivial permutations.

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For the question in the title:

An element of $S_n$ is a commutator if and only if it lies in $A_n$, i.e., if and only if it is an even permutation.

That commutators are even follows from their definition, as regardless of the parity of $\sigma$ and $\rho$, we have $[\sigma,\rho]=\sigma^{-1}\rho^{-1}\sigma\rho$ is even.

The converse is also true, proven by Ore in his paper Some remarks on commutators, Proc. Amer. Math. Soc. 2 (1950), pp. 307-314, Theorem 1.

I use the commutator convention that $[x,y]=x^{-1}y^{-1}xy$, and my permutations compose right-to-left.

Lemma. An element $\rho\in S_n$ is a commutator if and only if it can be expressed in the form $\rho=\tau\tau'$, where $\tau$ and $\tau'$ have the same cycle structure.

Proof. A commutator has the form $[\sigma,\pi] = (\sigma^{-1}\pi^{-1}\sigma)\pi$. Now, $\pi$ and $\pi^{-1}$ have the same cycle structure, and $\sigma^{-1}\pi^{-1}\sigma$ has the same cycle structure as $\pi^{-1}$, so every commutator has that form.

Conversely, if $\tau$ and $\tau'$ have the same cycle structure then they are conjugate, as are $\tau^{-1}$ and $\tau'$, so there exists $\sigma$ such that $\sigma\tau^{-1}\sigma^{-1}=\tau'$. Then $\rho=\tau\tau'=\tau\sigma\tau^{-1}\sigma^{-1} = [\tau^{-1},\sigma^{-1}]$. $\Box$

Theorem. Every element in the alternating group $A_n$ is a commutator in $S_n$.

Proof. Every element of $A_n$ has a disjoint cycle decomposition that consists of cycles of odd length, and an even number of cycles of even length. So the result will follow if one can show cycles of odd length and products of two disjoint cycles of even length are commutators of elements whose support is contained in the support of the cycles.

For cycles of odd length, $(a_1,\ldots,a_{2k+1})$, you can write $$(a_1,\ldots,a_{2k+1}) = (a_1,\ldots,a_{k+1})(a_{k+1},\ldots,a_{2k+1}),$$ a product of two cycles of length $k+1$, hence a commutator by the Lemma.

For disjoint products of two cycles of even length, $$(a_1,\ldots,a_{2i})(a_{2i+1},\ldots,a_{2i+2j}),\quad j\geq i,$$ we can express it as $$(a_1,\ldots,a_{2i},a_{2i+1},\ldots,a_{i+j+1})(a_{2i},a_{i+j+1},a_{i+j+2},\ldots,a_{2i+2j})$$ which is a product of two cycles of length $i+j+1$, hence a commutator. $\Box$

The paper inspired the Ore Conjecture (now a theorem), that every element in a finite nonabelian simple group is a commutator. See this math.overflow post discussing how a "conjecture" becomes a Conjecture, with some discussion of Ore's conjecture.

Answer 2

In addition to the excellent answer of Arturo, I would like to add the following. This stems from Ferdinand Georg Frobenius and is based on characters of a group.

Theorem Let $G$ be a (finite) group and let $Irr(G)$ be the set of irreducible characters. Then $g=[x,y]$ for some $x,y \in G$ iff

$$\sum_{\chi \in \text{Irr}(G)} \frac{\chi(g)}{\chi(1)} \neq 0.$$ See here for a proof.