Commutator subgroup of $S_n$ generated by commutators of transpositions Is it true that the commutator group $[S_n, S_n]$ is generated by elements of the form $[\tau_1, \tau_2]$ for $\tau_1, \tau_2$ transpositions?
I know that in general, when a group $G$ is generated by a set $S$, then one has to take the normal closure of the group generated by commutators of elements in $S$ to obtain $[G,G]$.
 A: Every element of $S_n$ is a product of transpositions.
I use the definition $[x,y]=x^{-1}y^{-1}xy$; if you use the other convention, the identities below need to be tweaked, but they are essentially the same.
Now notice that
$$\begin{align*}
[xy,z] &= [x,z]^y[y,z] = [x^y,z^y][y,z]\\
[x,yz] &= [x,z][x,y]^z = [x,z][x^z,y^z].
\end{align*}$$
Since the conjugate of transposition is a transposition, this allows you to decompose any commutator into a product of commutators of transpositions.
Note that this does not require you to know that the commutator subgroup is $A_n$, or how $A_n$ can be generated, just that every element of $S_n$ is a product of transpositions, and that the conjugate of a transposition is a transposition.
A: The commutator of $(1,3)$ and $(2,3)$ is $(1,2)(2,3)=(1,2,3)$, so any $3$ cycle is a commutator of two transpositions.
For $S_3$ we are done.
Now for $n\geq 4$, it is know that every prouct of an even number of transposition (an alternate permutation) is a product of 3 cycle.
Indeed if the two transpositions have disjoint support $(12)(34)= (123)(234)$, and if not $(12)(23)=(123)$
Now it is known that the derived group of $S_n, n\geq 4$  (the alternate group) is generated by the 3-cycles.
So any product of an even number of transposition is a product of 3 cycles, and a product of commutator of transposition. But the derived group of $S_n$ is the kernel of the signature, so the set of permutation which are the product of an even number of transposition..
