# Are There Always More Conjugacy Classes in the Kernel of a Morphism to $\Bbb Z_2$ than Not?

Let $$G$$ be a finite group and let $$\phi:G\to\Bbb Z_2$$ be a homomorphism to the group with two elements. Is it always the case that there are at least as many conjugacy classes in the kernel of $$\phi$$ as 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 $$\Bbb Z_2$$ is replaced by another group $$H$$ (possibly required to be abelian).

• If $G$ is (finite) abelian (of even order), I think the number of conjugacy classes in the kernel is exactly equal to the number not in the kernel. May 28 at 0:36
• Please edit your question so that it says what you mean to say, and not something else. May 28 at 1:13
• Here are two very nice proofs for the partition identity: math.stackexchange.com/questions/92191/… May 28 at 2:43
• Simul-posted to MO, mathoverflow.net/questions/393934/… without notification to either site. May 28 at 6:30
• I posted a solution to the cyclic quotient case on the MO post, which follows the idea of the solution here. May 28 at 20:21

This is true!

Let $$\phi : G \to \mathbb{Z}/2$$ be a group homomorphism, where $$G$$ is finite. By composing with $$n \mapsto (-1)^n : \mathbb{Z}/2 \to \mathbb{C} \setminus \{0\}$$, we produce a $$1$$-dimensional character of $$G$$. All the entries in the corresponding row of the character table will be $$\pm 1$$, and we want to show that the sum of the entries in this row is non-negative. But the sum of the entries of any row of the character table of any finite group is always a non-negative integer! A proof follows.

Proof. Let $$\psi$$ be the permutation character of $$G$$ acting on itself by conjugation. By thinking about the permutation matrices, we see that $$\psi(g) = \lvert C(g) \rvert$$ (the order of the centralizer). Since $$\psi$$ is a character, $$\langle \chi, \psi \rangle$$ is a non-negative integer for all irreducible characters $$\chi$$. But

$$\langle \chi, \psi \rangle = \frac{1}{\lvert G \rvert} \sum_{g \in G} \chi(g) \overline{\psi(g)} = \frac{1}{\lvert G \rvert} \sum_{g \in G} \lvert C(g) \rvert \chi(g)$$

is precisely the sum of the entries of the $$\chi$$-row of the character table.

The following also holds

Claim Let $$f : G \to \mathbb{Z}/3$$ be a homomorphism. Then $$\ker f$$ contains at least $$1/3$$rd of the conjugacy classes of $$G$$.

Proof. By composing with $$n \mapsto \zeta^n : \mathbb{Z}/3 \to \mathbb{C} \setminus \{0\}$$, where $$\zeta = e^{2 \pi i / 3}$$, we get a $$1$$-dimensional character $$\chi$$ of $$G$$. The entries in the corresponding row of the character table are all $$1$$, $$\zeta$$, or $$\overline{\zeta}$$, and these sum to some $$m \in \mathbb{N}$$.

So, we have $$a + b \zeta + c \overline{\zeta} = m$$ for some $$a, b, c \in \mathbb{N}$$, where $$a$$ is the number of conjugacy classes of $$G$$ which are contained in $$\ker f$$ and $$a+b+c$$ is the total number of conjugacy classes of $$G$$. By equating imaginary parts, we get that $$b = c$$, so $$a - b = a + b(\zeta + \overline{\zeta}) = m \geq 0$$. This means that $$a \geq b$$, so $$a+b+c = a + 2b \leq 3a,$$ as desired.

• @ClarkLyons (the OP) gave a generalisation in the same spirit to maps to any cyclic group over at MathOverflow. May 28 at 20:31