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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).

Gerry Myerson
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Clark Lyons
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    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. – Gerry Myerson May 28 '21 at 00:36
  • I should have said explicitly that I am really just concerned with $G$ finite and that I only expect a non-strict inequality. Of course when $G$ is abelian the conjugacy classes are all just singletons and either exactly half or all of them are in the kernel. – Clark Lyons May 28 '21 at 01:09
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    Please edit your question so that it says what you mean to say, and not something else. – Gerry Myerson May 28 '21 at 01:13
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    Here are two very nice proofs for the partition identity: https://math.stackexchange.com/questions/92191/identity-involving-partitions-of-even-and-odd-parts – Mike Earnest May 28 '21 at 02:43
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    Simul-posted to MO, https://mathoverflow.net/questions/393934/are-there-always-more-conjugacy-classes-in-the-kernel-of-a-morphism-to-z-2-tha without notification to either site. – Gerry Myerson May 28 '21 at 06:30
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    I posted a solution to the cyclic quotient case on the MO post, which follows the idea of the solution here. – Clark Lyons May 28 '21 at 20:21
  • @‍diracdeltafunk's solution here and @ClarkLyons's generalisation, referenced above. – LSpice May 28 '21 at 20:33

1 Answers1

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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.

diracdeltafunk
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