Timeline for Generating a finite group from elements in each conjugacy class
Current License: CC BY-SA 3.0
2 events
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| Jan 31, 2012 at 19:15 | comment | added | Marty Isaacs | Actually, more is true. Let H be a proper subgroup of a finite group G. Not only is it true that some element of G lies in no conjugate of H, but in fact, there must be at least |H| such elements. This can be proved by a variation on the argument given in the comment by Harden. | |
| Nov 30, 2011 at 8:04 | history | answered | DavidLHarden | CC BY-SA 3.0 |