Timeline for Question about some element in a group commutes with its all conjuagates.
Current License: CC BY-SA 3.0
6 events
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Jun 16, 2011 at 8:38 | comment | added | Tom De Medts | @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 $1 + 2 + 3 + 3 + 6 = 15$. | |
Jun 15, 2011 at 10:54 | comment | added | Zuriel | 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? | |
Jun 15, 2011 at 10:52 | vote | accept | Zuriel | ||
Jun 15, 2011 at 10:52 | vote | accept | Zuriel | ||
Jun 15, 2011 at 10:52 | |||||
Jun 14, 2011 at 16:29 | comment | added | Jack Schmidt | 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). | |
Jun 14, 2011 at 13:40 | history | answered | Tom De Medts | CC BY-SA 3.0 |