This question is inspired by a [recent question](https://math.stackexchange.com/questions/59816/why-is-the-set-of-commutators-not-a-subgroup) by user Jaymes about the set of commutators not being a group.

Following a link to MO, Gerry Myerson posted an example from Carmichael. I quote

Let $G$ be a subgroup of $S\_{16}$ generated by the following eight elements:
$$
\eqalign{(ac)(bd);&(eg)(fh);\cr(ik)(jl);&(mo)(np);\cr(ac)(eg)(ik);&(ab)(cd)(mo);\cr(ef)(gh)(mn)(op);&(ij)(kl).\cr}
$$

Then the commutator subgroup is generated by the first four elements above, and is of order $16$. Moreover,
$$
\alpha=(ik)(jl)(mo)(np)
$$
is in the commutator subgroup, but is not a commutator. 

However, no proof is given. Can somebody offer a proof of this? 

Regards,