User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:59:14Z http://mathoverflow.net/feeds/user/5023 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20089/conjugacy-classes-insersecting-subgroups-of-finite-groups Conjugacy classes insersecting subgroups of finite groups profilesbtilly 2010-04-01T17:53:13Z 2011-04-29T13:00:27Z <p>I figured out the first part of this years ago, but completely forget how I did it. I looked at the second, but don't think I figured it out.</p> <p>This I am sure is true, but don't remember why. Suppose that G is a finite group of size $n$, and H is a normal subgroup with |G/H| = $k$. Then at least $\frac{1}{k}$ of the conjugacy classes of G are within H.</p> <p>This I don't know the answer to. If H is allowed to be an arbitrary subgroup, must H intersect at least $\frac{1}{k}$ of the conjugacy classes?</p> <p>An example of a simple consequence of the first statement is that if we look at $S_n$ and $A_n$ is that at least half the partitions of $n$ have an even number of even parts.</p> http://mathoverflow.net/questions/19840/how-do-we-recognize-an-integer-inside-the-rationals/19848#19848 Answer by profilesbtilly for How do we recognize an integer inside the rationals? profilesbtilly 2010-03-30T16:09:01Z 2010-03-30T16:09:01Z <p>I'm very rusty on this stuff, but I think you need to be very careful in how to pose this question. The cautionary example to stay aware of is that while there is no complete set of first order axioms for the integers, the standard axioms for the real numbers are known to be both consistent and complete. The reason this can be is exactly because there is no way in first order logic to identify the integers as a subset of the reals.</p> <p>If you are defining the rationals as equivalence classes of ratios, then #1 seems to be the best approach. If you have some more abstract axiomatization of the rationals, it is not clear to me that there is an answer within first order logic.</p> http://mathoverflow.net/questions/20089/conjugacy-classes-insersecting-subgroups-of-finite-groups/20104#20104 Comment by 2010-04-01T21:38:03Z 2010-04-01T21:38:03Z D'oh. Of course. And thanks for answering the second half as well. http://mathoverflow.net/questions/20089/conjugacy-classes-insersecting-subgroups-of-finite-groups/20104#20104 Comment by 2010-04-01T20:38:27Z 2010-04-01T20:38:27Z I'm sure there is a simple reason why $|C_H(g)| \geq \frac{1}{k}|C(g)|$ but I'm missing it at the moment.