# Cosets and conjugacy classes

I'm interested in the following situation:

• $G$ is a finite group;
• $C$ is a conjugacy class in $G$;
• $H$ is the centralizer of an element $h$ of $C$.

I want information on $|C\cap Hg|$ as $g$ varies across $G$. In particular I'd like to prove that there exists $k<1$ such that for all $g\in G$ we have $$|C\cap Hg| \leq k|H|.$$

Unfortunately for me such a bound does not exist in complete generality: consider $C_p\rtimes C_{p-1}$ for a prime $p$ (semidirect product of two cyclic groups). Let $C$ be the conjugacy class of elements of order $p$, all of which have the same centralizer $H$. Then $C$ is a subset of $H$ and we have $$|C\cap H| = (p-1/p)|H|.$$ So as $p$ goes to infinity we have $(p-1/p)\to 1$.

So we can only prove a bound of the given form for particular cases. With this in mind here are some questions:

• Is it true that $|C\cap Hg|\leq |C\cap H|$? Edit: No it is not true. Mark Wildon has provided counter-examples in his answer below. If we assume that $G$ is simple does a bound of the given form with $k<1$ exist?
• Does anyone know if this problem appears in the literature in an alternative formulation? I'm interested even in particular cases, e.g. taking G to be a particular family of simple groups and C a particular family of conjugacy classes.
• Edit: As discussed in comments below, the case when $|C\cap Hg|=1$ for all $g\in G$ corresponds precisely to the situation $G=HC$. An example of this phenomenon is given below when $G=C_p\rtimes C_{p-1}$, a Frobenius group. Does this ever happen for $G$ simple? Has the problem of decomposing a group $G$ into the product of a centralizer and conjugacy class been studied in the literature?

It is not always the case that $|C \cap Hg| \le |C \cap H|$, even if G is simple. Here are two examples in small degree permutation groups, found by a brute-force search.

(1) Let $G$ be the symmetric group of degree $6$, and let $C$ be the conjugacy class of all $6$-cycles. Then $h = (1,2,3,4,5,6) \in C$ and $\mathrm{Cent}_G(h) = \left< h \right>$ contains exactly two $6$-cycles, namely $h$ and $h^{-1}$. If $g = (1,3)(2,6)$ then $\mathrm{Cent}_G(h)g$ has three $6$-cycles, namely $hg$, $h^{-1}g$ and $h^3 g$.

(2) Let $G$ be the alternating group of degree $7$, and let $C$ be the conjugacy class of elements of cycle type $(4,2,1)$. Then $h = (1,2,3,4)(5,6) \in C$ and $\mathrm{Cent}_G(h) = \left< h \right>$ contains exactly two elements of $C$, namely $h$ and $h^{-1}$. If $g = (1,5,6,7,3)$ then
$$\mathrm{Cent}_G(h)g = \lbrace (1,2)(3,4,5,7), (1,5,6,7,3), (1,4)(2,5,7,3), (2,4)(3,5,6,7)\rbrace$$ has three elements in $C$.

One small remark (related to your example): it is possible that each coset of $H$ contains a unique element of $C$. Let $G$ be a Frobenius group with cyclic kernel $K = \left< k \right>$ of prime order $p$ and complement $H = \left< h \right>$ of order dividing $p-1$. Then the conjugacy class of $h$ is $hK$. The centralizer of $h$ is $H$, so the distinct intersections in your problem are $hK \cap Hg^i = \lbrace hg^i \rbrace$, for $i \in \lbrace 0,1,\ldots,p-1\rbrace$.

• Mark, this is very interesting - thank you! Your examples close one possible line of inquiry. I won't accept this as an answer (yet) though as I'm still interested in other ways of attacking the general problem of an upper bound for simple groups... May 12, 2012 at 19:16
• Mark, I've just realised that your "small remark" has an interesting interpretation. Define the map $\phi: H\times C \to G, (h,c)\mapsto hc.$ Now consider $\phi^{-1}(g)$, the pre-image of an element $g\in G$. One can see that $$|\phi^{-1}(g)| = |C\cap Hg|.$$ Now, keeping the orbit-stabilizer theorem in mind, we see that each coset of $H$ contains a unique element of $C$ if and only if $\phi$ is surjective. So another way of approaching this problem.... When is $\phi$ surjective? Can it happen for $G$ a simple group? May 14, 2012 at 12:46
• Hi Nick. That's an interesting question! Just to check I understand: $\phi$ is surjective (and so bijective) if and only if $G=\mathrm{Cent}_G(h)h^G$? This feels like a very strong condition. In particular: (1) the only conjugate of $h$ in $\left< h \right>$ is $h$ itself, and (2) if $h^G \cup \lbrace 1 \rbrace$ contains a subgroup $K$ then $\mathrm{Cent}_G(h) \cap K = \lbrace 1 \rbrace$. It think these rule out any example where G is a non-abelian alternating group, since (1) shows that $h$ must be an involution, and then (2) rules out involutions. May 14, 2012 at 20:56
• Yes, that condition is correct. And your reasoning for the alternating group rules them out. In $PSL_n(q)$ there might be more hope for this to happen - I'm thinking about Singer cycles. Some of these are real - and so violate your condition (1) - but for appropriate $n$ and $q$ it is easy enough to find Singer cycles for which (1) holds. Of course one is still a long way off showing that $\phi$ is surjective. I tried Singer cycles for $PSL(3,2)$, $PSL(3,3)$ and $PSL(4,2)$ and none worked, although in each case the image of $\phi$ was large - much more than $\frac12|G|$. May 15, 2012 at 9:21
• One more comment: the condition that $\phi$ is surjective, i.e. $G={\mathrm Cent}_G(h) h^G$ bears a vague resemblance to both Szep's conjecture on centralizers and Thompson's conjecture on conjugacy classes. So there's a good chance that this might have been studied before(???) May 15, 2012 at 9:23

Let's consider further the question of when it happens that every right coset of $H$ contains a unique element of the class $C$, or in other words, $C$ is a right transversl of $H$ in $G$. Nick Gill expressed an interest in these questions in the case where $G$ is simple. It appears likely that for nonabelian simple groups, it never happens that a class $C$ is a right transversal for $H$, where $H$ is the centralizer of $h \in C$. At least, I can prove that in the special case where $h$ has prime order. In fact, more is true: if $G$ is simple and $h$ has prime order, then $|C \cap H| > 1$.

Suppose $|C \cap H| = 1$. Then in the conjugation action of $h$ on $C$, there is exactly one fixed point, namely $h$. If the prder of $h$ is a power of a prime $p$, it follows that $|C| \equiv 1$ mod $p$, and thus $|G:H| = |C|$ is not divisible by $p$, and hence $H$ ccontains a Sylow $p$-subgroup $P$ of $G$, and necessarily, $h \in P$. Also, no element of $P$ other than $h$ is conjugate to $h$ in $G$. But if $h$ has prime order, this is impossible in a simple group. If $p = 2$, this follows by Glauberman's Z* theorem, and if $p> 2$, it is a consequence of a result of Artemovich (1988). [Thanks to Nick Gill for telling me about the Artemovich result.]

One could ask how much weaker is the condition $|C \cap H| = 1$ than the original contition, that $C$ is a transversal for $H$ in $G$. Perhaps it is not weaker at all. A few Magma experiments turned up no examples where $|C \cap H| = 1$, but $C$ is not a transversal.

• I have finally found a counterexample to the question I asked in the last paragraph of my answer, above. The group is of order 168 = 2^3 3 7, constructed as a semidirect product of a nonabelian group of order 21 acting faithfully on an elementary abelian group of order 8. May 18, 2012 at 23:56

This is not a solution for your questions but a remark which might help you: The number of elements in one conjugacy class, which lie in a coset is constant over all cosets which lie in a fixed double coset. By this I mean the following:

Let $Hx_1$ and $Hx_2$ be cosets which both lie in the same double coset $HgH$, let $C$ be a conjugacy class and fix $g_0\in C\cap Hx_1$. We like to show that $|Hx_1\cap C|=|Hx_2\cap C|$.

By assumption there are elements $h_l,h_l^{\prime}$ for $l = 1,2$ such that $x_l = h^{\prime}_lgh_l$. Thus, $g_0∈C∩Hx_1 =C∩Hgh_1$, so that $(h_1^{-1}h_2)^{−1}g_0(h_1^{-1}h_2)\in C∩Hgh_2 =C∩Hx_2$.

So at least for cosets which lie in the same double coset you get an answer for the first question.

If you define for each representative $g$ of the double cosets of $H$ in $G$ a valency to be the number $k_g=|H|^{-1}|D_g|$, then by the above, we have that

$|C\cap D_g|/k_g$ is a natural number for all conjugacy classes $C$. Maybe this helps.

A. Stein [J. Algebra 239 (2001), no. 1, 365–390] proved that if a conjugacy class $C$ of a finite group $G$ is a transversal to a subgroup $H$ of $G,$ then $\langle C\rangle,$ the subgroup generated by $C,$ is solvable. Therefore, if $G$ is simple (or has a trivial solvable radical), then no nontrivial conjugacy class of $G$ can be a transveral to any subgroup of $G.$

Here are a couple of character theoretic observations which do not require CFSG. In the situation of the question (where $C$ denotes the conjugacy class of $g$), we have $|C \cap Hx| = 1$ for every $x \in G$ if and only if $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle = 0$ whenever $\chi \in {\rm Irr(G)}$ is a non-trivial character with $\chi(g) \neq 0.$ More generally, $\sum_{ t \in T} |C \cap Ht|^{2} = [G:H] \sum_{\chi \in {\rm Irr}(G)} \frac{|\chi(g)|^{2}}{\chi(1)} \langle {\rm Res}^{G}_{H}(\chi,1 \rangle,$ where $T$ is a transversal to $H$ in $G.$

This formula is derived by considering the product (in the group algebra $\mathbb{Z}G$), of class sums ${\tilde C}{\tilde C^{-1}},$ where $C^{-1}$ denotes the class of $g^{-1}$ and we use ${\tilde C}$ to denote the class sum of the class of $g.$ The coefficient of $x \in G$ in this product is well-known to be $\frac{|G|}{|H|^{2}} \sum_{ \chi \in {\rm Irr}(G)} \frac{|\chi(g)|^{2} \chi(x^{-1})}{\chi(1)}.$ This is always non-negative, and if we sum these quantities over $x \in H,$ the claims follow easily ( in the first case, the trivial character already contributes $[G:H]$ to the RHS, and all other terms on the RHS are non-negative. In the second case, the conjugates of $g$ in $Ht$ contribute $|C \cap Ht|^{2}$ elements of $H$ to the given product of class sums, including multiplicities).

Note that we easily obtain $\sum_{t \in T}(|C \cap Ht|-1)^{2} \leq d(|G|-[G:H]),$ where $d$ is the maximum value over non-trivial irreducible characters $\chi$ with $\chi(g) \neq 0$ of $\frac{\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle}{\chi(1)}$.

Later edit: In fact, this problem is quite closely related to an earlier one on MO about doubly transitive action on a conjugacy class: if $p$ is a prime, and $G$ is a (putative) doubly transitive permutation group whose point stabilizer $H$ has a non-trivial center (and with $F(G) =1$), then it can be shown that for an element $z \in Z(H)$ of prime order, there is one conjugate of $z$ in each coset of $H,$ ie the conjugates of $z$ form a transversal to $H.$

Also, I mention (without proof, but in case it is useful to anyone else), the following facts which may be proved using block theory: if $G$ is a finite group and $z \neq 1$ is an element of order a power of a prime $p$ whose conjugates form a transversal to $H = C_{G}(z),$ the following hold:

Whenever $y$ is a $p$-regular element of $H,$ we have ${\tilde C}_{zy} {\tilde H} = [H:C_{H}(y)]{\tilde G}$ (where, as before, for $S$ a subset of $G,$ we let ${\tilde S}$ denote the sum of the elements of $S$ in the group algebra $\mathbb{Z}G,$ and where $C_{u}$ denotes the conjugacy class of $u.$

Whenever $x$ is a $p$ -element (possibly the identity element)of $H,$ we have $|S_{p}^{G}(x)| = [G:H]|S_{p}^{H}(x)|$ and ${\tilde S}_{p}^{G}(x){\tilde H} = |S_{p}^{H}(x)|{\tilde G},$ where $S_{p}^{H}(x)$ denotes the $p$-section of $x$ in $G$ ( that is, the set of elements of $G$ whose $p$-part is conjugate to $x$).

By way of explanation, these last facts follow because the trivial character is the only constituent in the principal $p$-block of the character ${\rm Ind}_{H}^{G}(1),$ and central characters associated to irreducible characters outside the principal $p$-block annihilate $p$-section sums by Brauer's Second Main Theorem. For the first, we also have the more precise fact that the trivial character is the only constituent of ${\rm Ind}_{H}^{G}(1)$ in a $p$-block with a defect group containing $z,$ so all non-trivial constituents of ${\rm Ind}_{H}^{G}(1)$ vanish on the $p$-section of $z.$

• Nice argument! In the first paragraph, $\chi$ must be a non-trivial character. Also I suggest to add that $C$ is the conjugacy class of $g$. Oct 27, 2014 at 11:30
• Thanks for this Geoff. I will need some to digest what you've written. I originally asked this question 2 years ago because I had an application in mind for a paper that is now written - it will take me some time to see if I can make use of your comments in that context now.... Oct 28, 2014 at 15:58
• OK, I can also go into more detail by email if you like. In fact, I originally started working on the odd Z^{*}-theorem as my thesis problem in 1976 ( pre CFSG days), and have maintained an interest in proving it directly by character-theoretic it ever since (the condition that conjugates form a coset to the centralizer is strictly stronger than the isolated $p$-element condition in general, though they are equivalent for involutions. Oct 28, 2014 at 16:10