Derived set VS derived subgroup In Topology:

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*the $\textbf{derived set}$ of a subset S of a topological space is the set S' of all accumulation points of S.

*If S'=S, then S is said $\textbf{perfect}$.

In Group Theory:

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*the $\textbf{derived subgroup}$ of a group G is the subgroup G'  is the subgroup generated by all the commutators of the group.

*If G'=G, then G is said $\textbf{perfect}$.

I ask: there is a correlation between this terminologies  (for example with opportune topology in a group)?
 A: This question kept coming back and forth to me, so I kept thinking about it and found some elements towards the idea that both notions have the same name coincidentally.
Given $\mathbb{R}$ as an additive abelian topological group (whose topology is endowed by $|\,\cdot\,|$), let $\mathbb{R}$ being itself one of its subgroups. On one hand, in terms of topology, one has $\mathbb{R}'=\mathbb{R}$. On the other hand, in terms of groups, $D(\mathbb{R})=\lbrace0\rbrace$, since $\mathbb{R}$ is abelian. This means that $\mathbb{R}$ is perfect as a topological subspace, but not as a (sub)group.
Another example of a non-abelian group is the case of $\mathfrak{S}_5$, endowed with the trivial topology $\lbrace\varnothing,\mathfrak{S}_5\rbrace$. Consider the subgroup $\mathfrak{A}_5$. Now $D(\mathfrak{A}_5)=\mathfrak{A}_5$, meaning that it is perfect in terms of groups, but $\mathfrak{A}_5'=\mathfrak{S}_5$, since the only neighborhood of any $\sigma\in\mathfrak{S}_5$ is $\mathfrak{S}_5$ itself, which contains other points in $\mathfrak{A}_5$, meaning $\mathfrak{A}_5$ is not perfect as a topological subset.

I think the issue for both notions to agree comes from the fact that in terms of groups, the derived object is a notion that makes sense for the whole group, whereas in terms of topology, the derived object makes sense for
