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!!
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
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
-
2$\begingroup$ 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). $\endgroup$ Commented Jun 14, 2011 at 16:29
-
$\begingroup$ 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? $\endgroup$– ZurielCommented Jun 15, 2011 at 10:54
-
$\begingroup$ @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 $1 + 2 + 3 + 3 + 6 = 15$. $\endgroup$ Commented Jun 16, 2011 at 8:38
$\begingroup$
$\endgroup$
3
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 $g=E_{1,2}(1), h=E_{2,3}(1)$.
-
1$\begingroup$ This answers a slightly different question. For your g,h, the normal subgroup generated by gh is also abelian, though the subgroup generated by {g,h} is not abelian. $\endgroup$ Commented Jun 14, 2011 at 16:27
-
1$\begingroup$ @Jack: You are right. I did not understand the question. An element $g$ such that $g^G$ is commutative is called Engel of degree 2: it satisfies $[[x,g],g]=1$ for every $x\in G$. So the question asks whether the product of two Engel elements of degree 2 is an Engel element of degree 2. There are classical papers by Baer and others about it. $\endgroup$– user6976Commented Jun 14, 2011 at 19:32
-
$\begingroup$ Thank you for your answer! May I know more details (title, etc.) of those classical papers by Baer and others? I would like to read them. $\endgroup$– ZurielCommented Jun 15, 2011 at 10:57