Cosets and conjugacy classes - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:14:34Z http://mathoverflow.net/feeds/question/96652 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96652/cosets-and-conjugacy-classes Cosets and conjugacy classes Nick Gill 2012-05-11T08:32:40Z 2012-05-16T23:52:50Z <p>I'm interested in the following situation:</p> <ul> <li>$G$ is a finite group;</li> <li>$C$ is a conjugacy class in $G$;</li> <li>$H$ is the centralizer of an element $h$ of $C$.</li> </ul> <p>I want information on $|C\cap Hg|$ as $g$ varies across $G$. In particular I'd like to prove that there exists $k&lt;1$ such that for all $g\in G$ we have $$|C\cap Hg| \leq k|H|.$$</p> <p>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 $(p-1/p)\to 1$.</p> <p>So we can only prove a bound of the given form for particular cases. With this in mind here are some questions:</p> <ul> <li>Is it true that $|C\cap Hg|\leq |C\cap H|$? <strong>Edit:</strong> 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 $k&lt;1$ exist?</li> <li>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.</li> <li><strong>Edit:</strong> 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?</li> </ul> http://mathoverflow.net/questions/96652/cosets-and-conjugacy-classes/96706#96706 Answer by Natalie for Cosets and conjugacy classes Natalie 2012-05-11T20:36:29Z 2012-05-11T20:36:29Z <p>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:</p> <p>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 $|Hx_1\cap C|=|Hx_2\cap C|$.</p> <p>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 $(h_1^{-1}h_2)^{−1}g_0(h_1^{-1}h_2)\in C∩Hgh_2 =C∩Hx_2$.</p> <p>So at least for cosets which lie in the same double coset you get an answer for the first question.</p> <p>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 </p> <p>$|C\cap D_g|/k_g$ is a natural number for all conjugacy classes $C$. Maybe this helps.</p> http://mathoverflow.net/questions/96652/cosets-and-conjugacy-classes/96721#96721 Answer by Mark Wildon for Cosets and conjugacy classes Mark Wildon 2012-05-12T00:34:51Z 2012-05-12T00:34:51Z <p>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.</p> <p>(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 $\mathrm{Cent}_G(h) = \left&lt; h \right>$ 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 $h^3 g$. </p> <p>(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 $\mathrm{Cent}_G(h) = \left&lt; h \right>$ contains exactly two elements of $C$, namely $h$ and $h^{-1}$. If $g = (1,5,6,7,3)$ then<br> $$\mathrm{Cent}_G(h)g = \lbrace (1,2)(3,4,5,7), (1,5,6,7,3), (1,4)(2,5,7,3), (2,4)(3,5,6,7)\rbrace$$ has three elements in $C$.</p> <p>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 $K = \left&lt; k \right>$ of prime order $p$ and complement $H = \left&lt; h \right>$ 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 $i \in \lbrace 0,1,\ldots,p-1\rbrace$. </p> http://mathoverflow.net/questions/96652/cosets-and-conjugacy-classes/97167#97167 Answer by Marty Isaacs for Cosets and conjugacy classes Marty Isaacs 2012-05-16T23:52:50Z 2012-05-16T23:52:50Z <p>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 $|C \cap H| > 1$.</p> <p>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.]</p> <p>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.</p>