Computation of the commutator of permutations of $J_n$ I've been slowly going through some of the material in Serge Lang's Alegbra, and I've just stumbled upon some computations that's puzzling me at the moment. It's a specific step in the proof of a theorem, namely:

Theorem: If $n\geq 5$ then $S_n$ is not solvable.

The step that's troubling me is the following one :

Let $i,j,k,r,s$ be five distincts integers in $J_n=\{1,2,\ldots,n\}$ and let $\sigma=[ijk]$ and $\tau=[krs]$. Then direct computation gives their commutator : $$\sigma\tau\sigma^{-1}\tau^{-1}=[rki]$$

I don't think I really understant how the computation works here, I'm definitely unable to get to the result. So if anyone could explain it, I'd appreciate it thanks.
 A: Hint
Compute the image of all $l \in J_n$ under $\omega = \sigma\tau\sigma^{-1}\tau^{-1}$
For $l \notin \{i,j,k,r,s\}$, $\omega(l) = l$ is clear.
Now for example
$$\omega(i)=\sigma\tau\sigma^{-1}\tau^{-1}(i)=\sigma\tau\sigma^{-1}(\tau^{-1}(i)) = \sigma\tau\sigma^{-1}(i)=\sigma\tau(\sigma^{-1}(i))=\sigma\tau(k)=\sigma(r)=r=[rki](i).$$
Do the same for $j, k, r,s$.
A shorter proof can use the fact that for any cycle $\tau = [i_1 \ i_2 \ \dots \ i_k]$ and any element $\sigma \in \mathcal S_n$, you have
$$\sigma \tau \sigma^{-1} = [\sigma(i_1) \ \sigma(i_2) \ \dots \ \sigma(i_k)].$$
Applying that to the commutator $\sigma\tau\sigma^{-1}\tau^{-1}$, you get
$$\sigma\tau\sigma^{-1}\tau^{-1} = [i \ r \ s ] \tau^{-1} = [i \ r \ s ][k \ s \ r ] = [r \ k \ i]$$
A: Recall that the product of two permutations written in cycle notation is defined as the composition, and therefore should be computed right-to-left.
As an example, to compute $(12)(23)$ you should write $(1$, then compute
$$([12][23]).1 = [12].([23].1) = [12].1 = 2$$
and hence write $(12$. Then
$$([12][23]).2 = [12].([23].2) = [12].3 = 3$$
and hence write $(123$. Then
$$([12][23]).3 = [12].([23].3) = [12].2 = 1$$
and so you close the bracket, and get $(123)$.
Now, just use the associative property to do one product at a time, as described above. You get:
$$[ijk][krs][ikj][ksr] = [ijk][krs][iksrj] = [ijk][irj] = [irk] = [rki]$$
The standard notation is to put as leftmost element of a cycle the smallest one, but when letters are involved you can freely shuffle them around. What I mean is that $[irk]=[rki]=[kir]$.
