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!!
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!! 


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. 


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 nonabelian). So you can take $g=E_{1,2}(1), h=E_{2,3}(1)$. 

