4
$\begingroup$

An involution $a$ of a group $G$ is called central if there exists a sylow $2$-subgroup $H$ of $G$ such that $a \in C_G(H)$, or equivalently if the centralizer of $a$ contains a sylow $2$-subgroup.

Clearly if an involution is central then its every conjugate is also central.

Under what conditions on $G$ is the following statement true:

Product of two distinct elements $a$, $b$ in a conjugacy class $C$ of central involutions belongs to $C$ if and only if $a$ and $b$ commute.

Clearly if $G$ has a unique conjugacy class of involutions then the result is true. What if $G$ has more than one conjugacy classes of involutions but precisely one which contains a central involution? Is the product of two distinct central involutions which commute again a central involution? Are there any characterisations of $G$ for which the above statement is always true?

As a follow up, can we classify all finite groups that contain a single conjugacy class of central involutions?

EDIT1: Derek Holt has given an example in $A_8$ showing that having a unique conjugacy class of central involutions is not sufficient.

EDIT2: Central involutions are the same thing as involutions that lie in the center of some sylow $2$-subgroup. Moreover, the number of conjugacy classes of central involutions is at most the number of involutions in the center of a sylow $2$-subgroup.

Improve this question
$\endgroup$
8
  • $\begingroup$ The condition is certainly necessary, since the product of two involutions is an involution if and only if they are distinct and commute. $\endgroup$
    – Colin Reid
    Aug 6, 2014 at 23:49
  • 1
    $\begingroup$ I don't know if you really said what you mean, but if there is a unique conjugacy class of involutions in $G$, the statement is true. By the way, I think you need to specify that $a$ and $b$ are distinct involutions. $\endgroup$
    – Geoff Robinson
    Aug 6, 2014 at 23:49
  • $\begingroup$ Yes, I didn't notice that it is quite straightforward to show that the result is true if G has a unique conjugacy class of involutions. I am slightly rephrasing the question now. @ColinReid: I was talking about conditions on G and not the part of the statement following "if and only if" $\endgroup$
    – Anurag
    Aug 7, 2014 at 1:43
  • 1
    $\begingroup$ The question reduces to fusion in a Sylow $2$-subgroup $S$ of $G$. We want to know, given $a \in S, b \in Z(S)$ distinct involutions that are in a given conjugacy class $C$ of $G$, does $ab$ lie in $C$? In other words, is $C'$ a union of cosets of $C' \cap Z(S)$, where $C' = (C \cup \{1\}) \cap S$? (We certainly need $C' \cap Z(S)$ to be a non-trivial group.) $\endgroup$
    – Colin Reid
    Aug 7, 2014 at 2:50
  • 1
    $\begingroup$ In fact, the number of classes of central involutions is the number of $N_{G}(S)$ orbits by conjugation on $\Omega_{1}(Z(S)) \backslash \{1_{G}\},$ where $S$ is a Sylow $2$-subgroup and $\Omega_{1}(Z(S))$ is the subgroup of $Z(S)$ generated by its involutions. $\endgroup$
    – Geoff Robinson
    Aug 11, 2014 at 23:18

1 Answer 1

Reset to default
7
$\begingroup$

$A_8$ has two classes of involutions with representatives $x=(1,2)(3,4)$, $y=(1,2)(3,4)(5,6)(7,8)$, and $y$ is central but $x$ is not.

However $y$ commutes with $y'=(1,2)(3,4)(5,7)(6,8)$ but $yy' = (5,8)(6,7)$ is not central.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Interesting. I wonder how I missed that when I was checking this for alternating groups. So, clearly commutativity is not enough to deduce that product of two central involutions is also central. Can you think so some other conditions that will suffice? Some famous examples of groups where this sort of closure is true and used are $J_2$ and $Suz$. $\endgroup$
    – Anurag
    Aug 7, 2014 at 9:19
  • $\begingroup$ But I suppose those groups are too special and this result won't be true in general. $\endgroup$
    – Anurag
    Aug 7, 2014 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.