Given a group $(G, \cdot)$ is there a way to find its derived subgroup other than calculating it by hand element by element?
For instance, if a group is abelian you know that its derived subgroup is $\{1_G \}$. This is one simple example but is there a similar trick to use to make it easier to calculate $G'$?
Thanks in advance
Another trivial case is when you know that $G$ is simple and non-abelian, so $G = G'$, because there are no other choices, as $G'$ is a normal subgroup.
Moreover, the derived subgroup is the smallest normal subgroup $N$ of $G$ such that $G/N$ is abelian. So a common way is to find a candidate normal subgroup $H$ such that $G/H$ is abelian and prove that $H$ is minimal with this property. Therefore $G' = H$.
Example: $G = S_3$, $H = \langle (123) \rangle$. $H$ has index $2$, so it is normal and $G/H$ is abelian. There are no proper non-trivial subgroups in $H$ and $G$ is not abelian, so $G' = H$.
There are several characterizations of the derived subgroup. See Commutator Subgroup Characterization or https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/characterization-of-the-commutator-subgroup-of-a-group/D103777CE40256FB37E2F7B2085F2788 or https://www.tandfonline.com/doi/pdf/10.1080/00927879208824542
