Is there a finite group of order greater than $1$, where all the elements are commutators?
I thought about this question once, but I couldn't think of a single example. Obviously, such a group must be perfect.
Perhaps the following fact can be used somehow:
$g \in G$ is commutator iff $$\sum\limits_{\chi \in Irr(G)} \frac{\chi(g)}{\chi(1)} \neq 0$$
Let $G$ be a group, let $H$ be the commutator subgroup, then $H$ is normal in $G$, and the quotient group $G/H$ is abelian. If $G$ is a nonabelian simple group, then it has no normal subgroups other than itself and the one-element subgroup, so its commutator subgroup must be the whole group. So any nonabelian simple group should do, e.g., the alternating group $A_n$ for $n\ge5$.
Now, the question is whether that means all elements are commutators. There are finite groups where there are products of commutators that aren't themselves commutators, but the smallest example is a group of order $96$, so for $A_5$ it must be the case that all the elements are commutators. The result about groups of order $96$ has been mentioned on previous questions about commutators on this website, e.g., Derived subgroup where not every element is a commutator
I can certainly recommend reading the overview paper On Commutators in Groups by L.C. Kappe and R.F. Morse, which appeared in the second volume of Groups St. Andrews 2005, p. 531-558. It contains a lot of very interesting information and 76 relevant references.
