# Are there always more conjugacy classes in the kernel of a morphism to $Z_2$ than not?

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.)

• I have seen that partition identity before, though unfortunately not in a publicly available source. I suspect it is "well known" but I'm not sure off the top of my head the best place to look. May 28 at 1:49
• The stackexchange post is math.stackexchange.com/questions/4153508/… The general rule here is you wait for several days before posting to a second site, and you link the two posts to spare people from duplication of effort. May 28 at 6:29
• Regarding the partition identity, see math.stackexchange.com/questions/92191 . May 28 at 12:32
• The equivalence of this identity with the one in the post linked by David is pretty easy, but it appears in exactly this form as chapter 9 exercise 7 of these notes. My guess is that this one doesn't appear in standard sources because it's straightforwardly equivalent to that better-known identity (attributed to Euler in Andrews's book) even though this formulation is in some ways more natural. May 29 at 1:21
• My answer below includes a reference to this paper of Gupta sciencedirect.com/science/article/pii/0097316576900510 that gives a bijective proof of the partition identity. May 29 at 11:20

$$\DeclareMathOperator\tr{tr}\DeclareMathOperator\Int{Int}\DeclareMathOperator\Cent{Cent}$$The OP (@ClarkLyons) gave a lovely generalisation of an answer 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 $$\phi$$.

• This is great! I was thinking that I could get it to work for H abelian, but I did not consider taking inverse images of conjugacy classes instead of conjugacy classes in inverse images. Very nice. May 29 at 4:06
• But I think you should modify the definition of $f$ so that it only counts each conjugacy class from $G$ once. So $f(h)$ should be the number of conjugacy classes of $G$ which $\phi$ sends to the conjugacy class of $h$. May 29 at 4:36
• @ClarkLyons, you are right. I edited accordingly. May 29 at 12:08

$$\def\Irrep{\mathrm{Irrep}}\def\Conj{\mathrm{Conj}}\def\Out{\mathrm{Out}}\def\CC{\mathbb{C}}\def\Hom{\mathrm{Hom}}$$Choose an element $$g$$ of $$G$$ not in $$H$$. Conjugation by $$g$$ is an automorphism of $$H$$. If we consider this automorphism as an element $$\sigma$$ of $$\Out(H)$$, then $$\sigma$$ is independent of the choice of $$g$$ and $$\sigma^2=1$$.

The outer automorphism group $$\Out(H)$$ acts on both the set $$\Conj(H)$$ of conjugacy classes of $$H$$ and the set $$\Irrep(H)$$ of (complex) irreps of $$H$$. Evaluation of characters gives a perfect pairing between $$\CC^{\Conj(H)}$$ and $$\CC^{\Irrep(H)}$$ which respects the $$\Out(H)$$ action, so $$\CC^{\Conj(H)}$$ and $$\CC^{\Irrep(H)}$$ are dual (and also isomorphic, since they are permutation representations) as representations of $$\Out(H)$$. Restricting our attention to the action of $$\sigma$$, we see that the number of orbits of orders $$1$$ and $$2$$ for $$\sigma$$ acting on $$\Conj(H)$$ and $$\Irrep(H)$$ must coincide. Let $$a$$ be the number of orbits of size $$1$$ and let $$b$$ be the number of orbits of size $$2$$. So $$\#\Conj(H) = \# \Irrep(H) = a+2b$$, and the number of $$G$$-conjugacy classes in $$H$$ is $$a+b$$.

Let $$\epsilon : G \to \pm 1$$ be the map with kernel $$H$$. Tensoring with $$\epsilon$$ is an involution of $$\Irrep(G)$$. It is "well known" that, if $$V \cong V \otimes \epsilon$$, then $$V|_H$$ decomposes as a direct sum of two irreps, forming an orbit for the $$\sigma$$ action and, if $$V \not\cong V \otimes \epsilon$$, then $$V|_H$$ is irreducible and fixed by $$\sigma$$. So $$\#\Irrep(G) = 2a+b$$. Of course, $$\#\Conj(G) = \#\Irrep(G)$$.

So the number of $$G$$-conjugacy classes in $$H$$ is $$a+b$$ and the number of $$G$$-conjugacy classes not in $$H$$ is $$(2a+b) - (a+b) = a$$. We have $$a+b \geq a$$ and we are done.

Proof of the well known claim. Let $$\chi$$ be the character of $$V$$. Note that the character of $$V \otimes \epsilon$$ is $$\chi \epsilon$$.

First, suppose that $$V \cong V \otimes \epsilon$$. Then $$\chi(g)=\chi(g) \epsilon(g)$$, so $$\chi$$ vanishes for all $$g \not \in H$$. Then $$\sum_{g \in G} |\chi(g)|^2 = \sum_{h \in H} |\chi(h)|^2$$, so we deduce that $$\mathrm{Hom}_H(V,V)$$ is $$2$$ dimensional and $$V|_H$$ must be a sum of two non-isomorphic irreps.

Now, suppose that $$V \not \cong V \otimes \epsilon$$. Then $$\Hom_G(V, V \otimes \epsilon)=0$$, so $$\sum_{g \in G} \chi(g) \overline{\chi}(g) \epsilon(g)=0$$ and we deduce that $$\sum_{h \in H} |\chi(h)|^2 = \sum_{g \in G - H} |\chi(g)|^2$$. Thus, $$\sum_{h \in H} |\chi(h)|^2 = \tfrac{1}{2} \sum_{g \in G} |\chi(g)|^2$$ and we deduce that $$\Hom_H(V,V)$$ is $$1$$ dimensional, so $$V|_H$$ is an irrep.

• $H = \ker \phi$? (I guess this comes later, when you define $\epsilon$; so I guess you just want to start with $H$ any index-2 subgroup?) May 28 at 3:29
• $\DeclareMathOperator\Hom{Hom}$Proof of the well known fact: $\Hom_H(V_H, V_H) = \Hom_G(V_H^G, V) = \Hom_G(V \oplus (V \otimes \epsilon), V) = 1 + [V \otimes \epsilon \cong V]$. May 28 at 3:33
• I suppose that the natural question this argument raises is to find an action of $\mathbb{Z}/2 \mathbb{Z}$ on $\mathrm{Conj}(G)$ with $a$ orbits of size $2$ and $b$ orbits of size $1$. (Since we have actions of $\mathbb{Z}/2 \mathbb{Z}$ on $\mathrm{Conj}(H)$, $\mathrm{Irrep}(H)$ and $\mathrm{Irrep}(G)$.) May 28 at 12:35
• By the way, the reason I asked for clarification about $H$ is that the question also uses $H$, with a different meaning. Speaking of that, wouldn't the same argument, with any target $\mathbb Z/p\mathbb Z$ with $p$ prime, show that $(p - 1)\cdot\#\text{conj. classes in$H$} \ge \#\text{conj. classes not in$H$}$, thus answering one possible version of the generalised question? May 28 at 18:10
• @LSpice Yes, I agree with your generalization to other primes. More specifically, I think it shows that the number of $G$-conjugacy classes in $H$ which stay conjugacy classes in $G$ is equal to the number of $G$-conjugacy classes lying over any fixed nonzero $a \in \mathbb{Z}/p \mathbb{Z}$. May 28 at 18:54

I have a solution to the case of a general cyclic quotient which follows @‍diracdeltafunk's answer on MSE.

If $$A=Z_n$$ is cyclic then we can treat $$\phi:G\to Z_n\hookrightarrow\mathbb{C}$$ as a one dimensional representation of $$G$$. We can also form representations $$\phi^a$$ for $$a=0,...,n-1$$. These give rise to characters $$\chi_a$$ for $$a=0,...,n-1$$. We view these characters as functions from the set $$\mathcal{C}$$ of conjugacy classes of $$G$$ to the complex numbers.

First we recall an important fact from the theory of characters of finite groups. If $$\chi$$ is the character of some representation, then $$\sum_{g\in\mathcal{C}}\chi(g)$$ is a nonnegative integer. This is because $$\sum_{g\in\mathcal{C}}\chi(g)$$ is exactly the value of the inner product of $$\chi$$ with the character associated to the conjugation action of $$G$$ on $$\mathbb{C}[G]$$.

For $$b\in Z_n$$ let $$f(b)$$ be the number of conjugacy classes of $$G$$ on which $$\chi_1$$ takes the value $$\zeta_n^b$$. So the problem now is to show that each of the $$f(b)$$ is bounded by $$f(0)$$. Applying the previously stated fact about group characters to each of the characters $$\chi_a$$, we have for $$a=0,...,n-1$$ that $$g(a)=\sum_{b\in Z_n}f(b)e^{2\pi i ab/n}\geq 0.$$ Now Fourier inversion tells us that for $$b=0,...,n-1$$ we have $$f(b)=\frac{1}{n}\sum_{a\in Z_n}g(a)e^{-2\pi iab/n}.$$ Therefore by the triangle inequality, for any $$b$$ we have $$f(b)=|f(b)|\leq\frac{1}{n}\sum_{a\in Z_n}g(a)=f(0).$$

• Nice argument! This seems closely related to @diracdeltafunk's answer to your MSE question, perhaps to the extent that it might be appropriate to acknowledge that answer explicitly. May 28 at 20:26

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.

• So $C \leftrightarrow D$, with split classes mapping to themselves, is the involution suggested in @DavidE.Speyer's comment. That's lovely! You mention an explicit version of the involution in the symmetric case; does it exist in general? May 29 at 17:15

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 $$|X| \in \{6,8 \}).$$

• Is it correct that this is a more abstract and general version of the techniques in @DavidE.Speyer's answer? May 29 at 16:16
• I would say that this answer covers more general situations than David's answer, and points out other directions for generalization. I am not sure whether it is "more abstract", but David's answer is more focused on the specific case of an index 2 (normal) subgroup. May 29 at 16:33
• @LSpice : I think I also begin to address the question at the end about kernels of homomorphisms from $G$ to other groups $H$ (not necessarily Abelian). May 29 at 16:43

There is a trivial example in which the number of conjugacy classes in the kernel and not in the kernel are equal. This is just the identity map $$i:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}$$.

• Or more generally any surjective map from an abelian group. But can it actually be smaller? May 28 at 1:55
• This should have been a comment.
– YCor
May 29 at 15:50
• @YCor I fully admit that this is not a particularly interesting answer. However, it is absurd to claim that it is not actually an answer to the question (as appearing in the title or in the body text: "Is it always the case that there are more conjugacy classes in the kernel of $\phi$ than conjugacy classes not in the kernel of $\phi$?").
– Buzz
May 30 at 2:25
• @YCor didn't say that it wasn't an answer, only that (though technically an answer) it would have been better as a comment—which I took to be advice, not chastisement. Posting it as such would have allowed @‍ClarkLyons to change the question to what was or should have been meant, whereas the etiquette has it that changing a question in a way that invalidates answers should not be done. May 31 at 0:03