TITLE: Are there always more conjugacy classes in the kernel of a morphism to $Z_2$ than not?
QUESTION [28 upvotes]: Let $G$ be a finite group and let $\phi:G\to Z_2$ be a homomorphism to the group with two elements. Is it always the case that there are more conjugacy classes in the kernel of $\phi$ than conjugacy classes not in the kernel of $\phi$?
I've tried a little bit of messing around algebraically and written down some exact sequences of $G$-modules to try to apply the methods of group cohomology, but I haven't gotten anything to work.
My inspiration here is the special case when $G$ is the symmetric group $S_n$ and $\phi$ is the sign homomorphism. In this case conjugacy classes of $G$ correspond to partitions, and the problem becomes about counting partitions of $n$ with an even number of even parts versus an odd number of even parts. I was able to prove (via generating functions and also bijectively) that the number of partitions of $n$ with an even number of even parts minus the number of partitions of $n$ with an odd number of even parts is equal to the number of partitions of $n$ with all parts odd and distinct. I could not find a reference for this fact after some googling, so I would be interested to know if this is a well-known partition identity.
I'm also interested in possible extensions of this problem where $Z_2$ is replaced by another group $H$ (possibly required to be abelian).
(I just posted this on stackexchange and then learned that mathoverflow might be more appropriate but I'm not sure.)

REPLY [10 votes]: This question was the motivation for two papers of mine with John R. Britnell. In Commuting conjugacy classes, an application of Hall's Marriage Theorem to group theory, J. Group. Th, 12 (2009) 795–802, we showed that there is a bijection between the non-split classes in the kernel and those not in the kernel, such that if $C \subseteq \ker \phi$ is paired with $D \subseteq G \setminus \ker \phi$ then $C$ and $D$ have representatives that commute. Here a class $C$ is non-split if its centralizer is not contained in $\ker \phi$; clearly this is a necessary condition for the claimed bijection. This immediately implies that there are at least as many conjugacy classes in $\ker \phi$ as conjugacy classes in $G \setminus \ker \phi$.
The proof is elementary, by counting the number of conjugacy classes that can be paired with each $C$, and showing that this number satisfies the conditions for Hall's Marriage Theorem.
In the case of the symmetric group this implies the identity counting partitions into an odd/even number of even parts mentioned in the question. (But while it says a bijection exists, it does not give one explicitly.) In Combinatorial proof of a theorem on partitions into an even or odd number of parts, J. Comb. Th. Ser A. 21 (1976) 100–103, Gupta gave an explicit bijective proof of this identity that happens to have the commuting property: see Section 3.1 of my joint paper.
In On the distribution of conjugacy classes between the cosets of a group in a cyclic extension, Bull. Lond. Math. Soc. 40 (2008) 897–906, we generalised the counting part of this result to arbitrary cyclic quotients.

REPLY [5 votes]: I will restrict to the case that $G$ is finite. There are many interesting results of this type already in the literature, some well-known, some not so well-known (some repeatedly re-discovered, so I will do my best with attributions). Of great relevance are Clifford's theorem, Frobenius reciprocity, and Brauer's permutation lemma, among other things.
Brauer's permutation lemma plays a role in changing this to a question about complex irreducible characters. Let $N$ be a normal subgroup of a finite group $G$. Then $G$ acts by conjugation on $N$, and there is a natural action of $G$ on (complex) irreducible characters of $N$ (in both cases, $N$ is acting trivially, and the action is really one of $G/N$).  Brauer's permutation lemma tells us that for any $x \in G$, the number of conjugacy classes of $N$ fixed by $x$ is the same as the number of irreducible characters fixed by $x$. The usual orbit counting formula then tells us that $G$ has the same number of orbits  on conjugacy classes of $N$ as it does on irreducible characters of $N$.
Hence the number of conjugacy classes of $G$ contained in $N$ is the number of $G$ orbits on irreducible characters of $N$.
Before I continue, I will fix some notation. Let $G$ be a finite group, $H$ be a (not necessarily normal) subgroup of $G$ and $N$ be a normal subgroup of $G$. We let $k(G)$ denote the number of conjugacy classes of $G$, and we let $k_{G}(N)$ denote the number of $G$-orbits on conjugacy classes of $N$, which is the same as the number of conjugacy classes of $G$ contained in $N$. Notice that it is not generally true that
$k(N) = k_{G}(N).$
Frobenius reciprocity is relevant because (even in the case that $H$ is not normal in $G$ ), we may note that for each irreducible character $\mu$ of $H$, ${\rm Ind}_{H}^{G}(\mu)$ has at most $[G:H]$ irreducible constituents (counting multiplicity), while each irreducible character $\chi$ of $G$ occurs as an irreducible constituent of some such induced character. It follows that $k(G) \leq [G:H]k(H)$.
Similarly, we find that if $\chi$ is an irreducible character of $G$, then ${\rm Res}^{G}_{H}(\chi)$ has at most $[G:H]$  irreducible constituents, so we obtain
$k(H) \leq [G:H]k(G).$
Hence we have $\frac{k(H)}{[G:H]} \leq k(G) \leq [G:H]k(H)$ whenever $H$ is a subgroup of the finite group $G$. These inequalities were proved by P.X. Gallagher around 1962. We will see that they are also relevant to the application of Clifford's theorem, as Gallagher himself knew.
Clifford's theorem is relevant for the following reason. If $\chi$ is an irreducible character of $G$, and $N \lhd G$, then the irreducible constituents of ${\rm Res}^{G}_{N}(\chi)$ lies in a single $N$-orbit (and all occur with equal multiplicity). Furthermore,
if $\mu$ is an irreducible character of $N$, then there is a bijection between irreducible characters $\chi$ of $G$ such that $\mu$ occurs with non-zero multiplicity in ${\rm Res}^{G}_{N}(\chi)$, and irreducible characters of the $G$-stabilizer of $\mu$, usually denoted $I_{G}(\mu)$, which restrict to $N$ as a multiple of $\mu$.
Using projective representations (in Schur's sense), it can be proved that the number of irreducible characters of $I_{G}(\mu)$ which restrict to a multiple of $\mu$ is at most $k(I_{G}(\mu)/N).$ The main idea is to reduce to the case that $N$ is central and $\mu$ is a linear character of $N$ by another part of Clifford's theorem(s).
Applying Gallagher's inequalities with $I_{G}(\mu)/N$ in the role of $H$, and $G/N$ in the role of $G$, , we obtain that
$k(I_{G}(\mu)/N) \leq [G:I_{G}(\mu)]k(G/N).$ Notice that $[G:I_{G}(\mu)]$ is the length of the $G$-orbit of $\mu$. Letting $\mu$ run through orbit representatives for the action of $G$ on irreducible characters of $N$, we obtain another inequality of P.X. Gallagher (also proved independently by H. Nagao around 1962): $k(G) \leq k(N)k(G/N).$
A variant of the above argument which is sometimes useful, and is relevant to the present question was proved by L.G. K'ovacs and myself in 1993: if $N$ is a normal subgroup of $G$, then we have $k(G) \leq m k_{G}(N)$, there $m$ is the maximum number of conjugacy classes of any subgroup of $G/N$. Notice that if $G/N$ is Abelian,
then $m = [G:N] = k(G/N).$ In particular, when $G/N$ is Abelian, we obtain a sharpening of the Gallagher-Nagao bound, for then we have $k(G) \leq k(G/N)k_{G}(N).$
Recalling that $k_{G}(N)$ is the number of $G$-conjugacy classes contained in $N$, we see that whenever $N$ is a normal subgroup of $G$ with $G/N$ Abelian, then we have
$k(G) \leq k(G/N)$ $\times$ (the number of conjugacy classes of $G$ contained in $N$).
This answers the question in the case $[G:N] =2$
Notice that when $N$ is a normal subgroup of $G$ with $G/N$ non-Abelian, it may not be straightforward to obtain the precise value of $m$, the maximum number of conjugacy classes of any subgroup of $G/N$ (though of course we always have $m < [G:N]$ given that $G/N$ is non-Abelian. Later edit: in fact, it is an easy exercise to check that a non-Abelian group $X$ has at most $|X|-3$ conjugacy classes, and if it has $|X|-3$ conjugacy classes, then $|X| \in \{6,8 \}).$