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.)



Are there always more conjugacy classes in the kernel of a morphism to $Z_2$ than not?
$\DeclareMathOperator\tr{tr}\DeclareMathOperator\Int{Int}\DeclareMathOperator\Cent{Cent}$The OP (@ClarkLyons) gave a lovely [generalisation](https://mathoverflow.net/a/394013) of an [answer](https://math.stackexchange.com/a/4153721) of @diracdeltafunk over on MSE. I believe that the technique can be pushed still further to handle maps to *any* group $H$, as in the original question. The argument has no new ideas from me.


Suppose that $\phi : G \to H$ is any homomorphism. For $g \in G$, write $G\cdot g$ for the $G$-conjugacy class through $g$; and similarly $H\cdot h$ for $h \in H$. Write $f : H \to \mathbb Z\_{\ge 0}$ for the function given by
$$
f(h) = (\#H\cdot h)^{-1}\sum\_{g \in \phi^{-1}(H\cdot h)} (\#G\cdot g)^{-1}
$$
for all $h \in H$.


The Fourier transform $\hat f : \rho \mapsto (\#H)^{-1}\sum\_{h \in H} f(h)\tr \rho(h)$ satisfies
$$
(\#H)\hat f(\rho)
= \sum\_{h \in H} (\#H\cdot h)^{-1}\sum\_{g \in \phi^{-1}(H\cdot h)} (\#G\cdot g)^{-1}\tr \rho(h)
= \sum\_{g \in G}
 (\#G\cdot g)^{-1}\tr (\rho \circ \phi)(g),
$$
which is the inner product of the character of $\rho \circ \phi$ with the character $g \mapsto \#\Cent\_G(g)$ of the permutation representation of $G$ acting on itself by conjugation, so that $\hat f(\rho)$ is non-negative, for all irreducible complex representations $\rho$ of $H$.


By column orthogonality of the character table, we have that
$$
f(h) = \sum\_{\rho \in \hat H} \hat f(\rho)\overline{\tr \rho(h)}
$$
for all $h \in H$, so that
$$
f(h) = \lvert f(h)\rvert \le \sum\_{\rho \in \hat H} \lvert\hat f(\rho)\rvert\cdot\lvert\tr \rho(h)\rvert \le \sum\_{\rho \in \hat H} \hat f(\rho)\tr \rho(1) = f(1)
$$
for all $h \in H$.


In words, upon multiplying both sides of the inequality $f(h) \le f(1)$ by $\#H\cdot h$, we obtain that the number of $G$-conjugacy classes (in $G$) that map into the $H$-conjugacy class of $h$ is bounded by the product of the size of the $H$-conjugacy class of $h$ and the number of $G$-conjugacy classes in the kernel of $\phi$.


