Let’s call a group $G$ $\omega$-soluble if $\bigcap_{i = 1}^{\infty} G^{(i)}$ is trivial. Here $\{G^{(i)}\}_{i = 1}^\infty$ stands for the derived series of the group.
Note, that not all $\omega$-soluble groups are soluble
My question is:
Is the class of all $\omega$-soluble groups a variety? And if it is, then by what laws can it be defined?
What have I thought on that topic:
To find out whether a class of group is a variety, one have to check, whether it is closed under subgroups, quotients and direct products. This one seems to be… Thus, if I am not mistaken, it is a variety.
The only two questions that remain are:
Am I mistaken?
If I am not, then what laws is this variety defined with?
No. As it was said in the comments $\omega$-soluble groups are not closed under quotients. If they were, it would have implied, that all groups are $\omega$-soluble, because free groups are, however this clearly is not the case (non-abelian simple groups are hit most "simple" counterexample).