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?
Best Answer
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 $i \in \lbrace 0,1,\ldots,p-1\rbrace$.