Conjugacy classes in group extensions

Question

Consider a normal subgroup $N$ of a finite group $G$.

How are the conjugacy classes of $G$ related to $G/N$ and $N$?

Answer 1

Let $k(G)$ denote the number of conjugacy classes of $G.$ It is known that the inequality $k(G) \leq k(N)k(G/N)$ holds. This was proved ( I assume independently) by P. X. Gallagher and H.Nagao. Equality can hold under some circumstances, even when $N$ is not a direct factor of $G,$ but the precise conditions for equality are somewhat elaborate.

Answer 2

Geoff Robinson gave the great upper bound, so I thought I'd mention a few obvious things on lower bounds that happen to be sharp.

Proposition: If G is a finite group with normal subgroup N, then $k(G) \geq k(G/N)$ with equality iff $N=1$. In other words, a finite group has more conjugacy classes than its proper quotient groups.

Proof: If x and y are conjugate in G, say by g, then xN and yN are conjugate in $G/N$, for instance by gN. Hence the map taking $x^G$ to $xN^{G/N}$ is well-defined and obviously surjective from the conjugacy classes of G to the conjugacy classes of $G/N$. If $x \in N \setminus 1$, then $x^G \mapsto 1N^{G/N}$ but $x^G \neq 1^G$, so we get strict inequality if N is nontrivial.

Proposition: There is a family of finite groups $G_n$ with normal subgroups $N_n$ such that $k(G_n) = o(k(N_n))$. In other words, a finite group can have radically fewer conjugacy classes than one of its normal subgroups.

Proof: The idea of course is to take $N$ to be elementary abelian, and $G/N$ to be a fairly swirly group of automorphisms with few conjugacy classes. One can take $G_n$ to be the hyperoctahedral group (the group of invertible $n\times n$ matrices with only one nonzero entry in each row and column, and that nonzero entry must be $\pm1$). The normal subgroup is the group $N_n$ of diagonal matrices. $k(G_n)$ is OEIS:A000712 and $$k(G_n) \sim \tfrac{\sqrt[4]{3}}{12}n^{(-5/4)} \exp\left(\tfrac{2 \pi}{\sqrt{3}} \sqrt{n}\right) \quad \ll\quad k(N_n) = 2^n$$ For what it is worth, another such sequence is $G_n=\operatorname{A\Gamma L}(1,2^n)$ with $k(G_n)$ given by OEIS:A178752.

The first proposition is sharp: Consider the Sylow p-normalizer G in the symmetric group on p points, and N its Sylow p-subgroup. Then $k(G) = p$, $k(G/N)=p-1$, $k(N) = p$, so G only has about the square root as many conjugacy classes as the product $G/N \times N$, and in fact $k(G) = k(G/N) + 1$ is the least integer strictly greater than $k(G/N)$.

Obviously there is no upper bound for $k(G)$ in terms of just $k(G/N)$ or just $k(N)$ as the case of $G = G/N \times N$ shows (where $k(G) = k(G/N) \times k(N)$ as indicated in Geoff's answer).

Answer 3

I'm not sure if this is what you are after, but if $x$ and $y$ are conjugate in $G$, then $xN$ and $yN$ are conjugate in $G/N$. This implies that under the canonical map from $G$ to $G/N$, the preimage of a conjugacy class in $G/N$ is a union of conjugacy classes in $G$. And this in turn comes in handy when you are trying to find the character table of $G$; you may be able to find characters of $G/N$ more easily (since, if we're dealing with finite groups, $G/N$ is smaller than $G$), and this relation between conjugacy classes allows you to "induce" an irreducible character on $G$ from each irreducible character on $G/N$.