# Size of the abelianization of a finite group is always less than or equal to the number of conjugacy classes?

This question came to me when looking at irreducible representations of a finite group over the complex numbers.

We know by Artin-Wedderburn theory that the number of irreducible complex representations of a finite group equals the number of conjugacy classes. We also know that we have a $$1:1$$ correspondence between one dimensional representations of $$G$$ and one dimensional representations of $$G^{ab}$$. Then it must follow that $$|G^{ab}| \leq |\{\text{conjugacy classes of } G\}|$$? since every one dimensional representation is trivially irreducible. If this is in fact true, then possibly there is a much simpler way requiring less machinery?

• The elements in a set of of inverse images in $G$ of the set of elements of $G/[G,G]$ must lie in distinct conjugacy classes in $G$. Apr 6, 2021 at 11:53
• In other words, conjugate elements represent the same cosets of $[G,G]$ (since their "quotient" is a commutator), so there is a surjection from conjugacy classes of $G$ to elements of $G/[G,G]$.
– anon
Apr 6, 2021 at 19:57

Each coset $$gG'$$ of $$G'$$ is a normal subset: for every $$x \in G$$, one has $$x^{-1}gxG'=g[g,x]G'=gG'$$. So $$gG'$$ is a disjoint union of conjugacy classes, among them $$Cl_G(g)$$: again, $$x^{-1}gx=g[g,x] \in gG'$$ for every $$x \in G$$, whence $$Cl_G(g) \subseteq gG'$$.
Hence every coset $$gG'$$ contains the conjugacy class $$Cl_G(g)$$, which has a void intersection with the other cosets. So the number of conjugacy classes $$k(G)$$ of $$G$$, is at least the number of cosets of $$G'$$.
It also follows that $$|G:G'|=k(G)$$ if and only if $$G$$ is abelian (look at the coset of $$1$$). So if $$G$$ is non-abelian, $$k(G) \gt |G:G'|$$.
In a similar vein, one can show that $$\#Cl_G(g) \leq |G'|$$ (consider the embedding $$x^{-1}gx \mapsto [g,x]$$, from the class of $$g$$ to $$G'$$, which is injective). Hence $$|C_G(g)| \geq |G:G'|$$. And this can also be obtained via the Second Orthogonality Relation: $$|C_G(g)|=\underset{\chi \in Irr(G)}\sum |\chi(g)|^2$$.