TITLE: Center-commutator duality
QUESTION [62 upvotes]: I'm reading this article by Keith Conrad, on subgroup series. I'm having trouble with a statement he does at page 6:

Any subgroup of $G$ which contains $[G,G]$ is normal in $G$.

He says this as evidence that commutator and center play dual roles, since any subgroup of $G$ contained in $Z(G)$ is normal in $G$. Now, this I'm sure I understand, but I don't see how the quoted line holds.
What I have read is that $[G,G]$ is the least normal subgroup of $G$ such that the quotient is abelian, which seems related.
Also, while we're at it: are the center and the commutator "really" (as in, categorically) dual constructions? I'm quite a novice in category theory, so please excuse me if this question is trivial.

REPLY [3 votes]: Here is a sort of "duality" (though very weak) between the center and commutator of a finite group, which occurred to me recently:
Let $\rm{Irr}(G)$ be the set of irreducible complex characters of the finite group $G$. Then the commutator subgroup is equal to $$\bigcap_{\chi \in \rm{Irr}(G),\ \chi(1)=1}\rm{ker}(\chi)$$
But if we look at something which in a certain sense is the complement of this, namely $$H = \bigcap_{\chi\in\rm{Irr}(G),\ \chi(1)\neq 1}\rm{ker}(\chi)$$ then this subgroup is contained in the center of $G$.
Proof (only of the last part): We know that the intersection of the kernels of all irreducible characters is trivial, so we get that $H\cap G' = \{ 1\}$. But it is a general fact that any normal subgroup that intersects the commutator subgroup trivially is central. To see this, let $N$ be such a subgroup, $x \in N$ and $g\in G$. Then the element $xgx^{-1}g^{-1}$ is in $N$ since $N$ is normal, and it is in $G'$ by definition, so it is $1$, and hence $N$ is central.
The above is of course not anywhere close to a proper duality, as we really look at a sort of complement instead, and we only get something contained in the center, rather than the entire center.
Remark: The above is a special case of a more general result, which states that if $m$ is the degree of some irreducible character of $G$ then $$\bigcap_{\chi\in\rm{Irr}(G),\ \chi(1)\neq m}\rm{ker}(\chi)$$ is abelian, and in fact, one can go further and leave out $2$ degrees and get something of derived length at most $2$, and even leave out $3$ and get something of derived length at most $3$ (though for this last result, one needs to assume that $G$ is solvable to begin with). These results are proved in the paper "Irreducible character degrees and normal subgroups" By Isaacs and Knutson.