Welcome to MathOverflow

Question

3

2

Let G be a group and g and h be elements in G. If g commutes with all conjugates of g and h commutes with all conjugates of h, can one conclude that gh commutes with all conjugates of gh?

Thanks!!

flag	asked Jun 14 2011 at 12:56 Zuriel 290●1●6

Tom De Medts · Accepted Answer · 2011-06-14 13:40:22Z UTC

9

No; the smallest counterexamples are given by the groups SmallGroup(54,5) and SmallGroup(54,6) (in GAP's SmallGroups library); these are groups of the form $$G_1 = ((3 \times 3) : 3) : 2 \text{ and } G_2 = (9 : 3) : 2.$$ In both cases, there are 15 elements with the given property (i.e. that they commute with all their conjugates), so they can't possibly form a subgroup.

answered Jun 14 2011 at 13:40

Tom De Medts
2,886●7●17

2

These are similar to Mark Sapir's example, except the extra element of order 2 makes the normal closure of gh a little bigger, and so non-abelian. In his example, too many of the normal subgroups are abelian, and your extra automorphism removes some normal subgroups (of the Fitting). – Jack Schmidt Jun 14 2011 at 16:29

Thank you so much for your answer! May I know how $G_1$ and $G_2$ are defined and how do you know that there are exactly 15 elements in each group with the given property? – Zuriel Jun 15 2011 at 10:54

@Zuriel: I wrote a tiny computer program in Sage to find these examples. But Jack's comment indicates how to construct $G_1$ explicitly as a subgroup of $PSL(3,3)$: it is the subgroup generated by the Sylow $3$-subgroup consisting of upper-triangular matrices with $1$'s on the diagonal, together with the involution $\operatorname{diag}(1, -1, -1)$. You can now explicitly compute the conjugacy classes by hand, if you wish, and you will see that $5$ of these classes are commutative, with sizes $1 + 2 + 3 + 3 + 6 = 15$. – Tom De Medts Jun 16 2011 at 8:38

Mark Sapir · Answer 2 · 2011-06-14 14:56:36Z UTC

You have that the normal subgroups $U, V$ generated by $g$ and by $h$ are Abelian. This only implies that $UV$ is normal and solvable of class 2. Take the group of unitriangular 3 by 3 matrices $H_3$ over ${\mathbb Z}_p$.

It has two normal Abelian subgroups $U,V$ containing the center (= the derived subgroup) generated by the elementary matrix $E_{1,3}(1)$: $U$ is generated as a subgroup by the center and $E_{1,2}(1)$, $V$ is generated by the center and $E_{2,3}(1)$. Both subgroups are normal because they contain the derived subgroup. The product of these two subgroups is the whole $H_3$ (hence non-abelian). So you can take $g=E_{1,2}(1), h=E_{2,3}(1)$.

Question about some element in a group commutes with its all conjuagates.

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.