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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?
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3 Answers 3

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

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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... –  Nick Gill May 12 '12 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? –  Nick Gill May 14 '12 at 12:46
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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. –  Mark Wildon May 14 '12 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 $\frac12|G|$. –  Nick Gill May 15 '12 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(???) –  Nick Gill May 15 '12 at 9:23
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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.

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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. –  Marty Isaacs May 18 '12 at 23:56
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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.

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thanks natalie, that is a promising start... –  Nick Gill May 11 '12 at 21:18
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