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?

  • 5
    $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    Commented Apr 6, 2021 at 11:53
  • $\begingroup$ 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 $G/[G,G]$. $\endgroup$
    – anon
    Commented Apr 6, 2021 at 19:57

1 Answer 1


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: $|C_G(g)|=\underset{\chi \in Irr(G)}\sum |\chi(g)|^2$.


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