most recent 30 from http://mathoverflow.net 2013-05-18T12:53:04Z http://mathoverflow.net/feeds/user/4058 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16026/the-finite-subgroups-of-sl2-c/16116#16116 Ben Wieland 2010-02-23T02:18:57Z 2010-02-23T02:18:57Z

<p>Don't forget Euclid book 13.</p>

http://mathoverflow.net/questions/15511/does-a-group-have-a-unique-pro-p-topology/15602#15602 Ben Wieland 2010-02-17T18:31:25Z 2010-02-17T18:31:25Z

<p>no.</p> <p>Let G(n,p) be the kernel of GL_n(Z_p) -> GL_n(Z/p). This is a pro-p group.</p> <p>Enough random matrices in G(n,p) generate a free group which has positive probability of being dense. In other words, the induced topology on the free group has completion G(n,p). If we take the same number of matrices in G(2,p) and G(3,p) this gives two topologies on the same free group with different completions, neither of which is the full pro-p topology. Without the randomness, the kernel of GL_2(Z) -> GL_2(Z/p) is free, but the topology coming from G(2,p) is not the full pro-p topology.</p> <p>In a more positive direction, the congruence subgroup property says that every finite index subgroup of GL_3(Z) contains the kernel of reduction mod m for some m. I think that this implies that the only p-topology on the kernel of reduction mod p is the one with completion G(3,p).</p>

http://mathoverflow.net/questions/15511/does-a-group-have-a-unique-pro-p-topology/15602#15602 Ben Wieland 2010-02-19T01:44:59Z 2010-02-19T01:44:59Z

G(2,p) has abelianization (Z/p)^4, while the free pro-p group on 4 generators has abelianization (Z_p)^4. (and the kernel of reduction mod p has is free on more generators than 4)