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?

 A: 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.
A: 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: 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 $G.$
A: 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 $z.$
A: 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 $i \in \lbrace 0,1,\ldots,p-1\rbrace$.  
