User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:34:05Z http://mathoverflow.net/feeds/user/7188 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29950/countable-open-subgroup Countable open subgroup unknown (google) 2010-06-29T18:36:47Z 2013-04-12T02:45:00Z <p>In a Hausdorff topological group, how can I show that every infinite topological group has a countable open subgroup?</p> http://mathoverflow.net/questions/30092/discrete-cyclic-subgroup Discrete cyclic subgroup. unknown (google) 2010-06-30T18:46:55Z 2010-08-02T05:59:07Z <p>Let T is a hausdorff group topology and (G,T) is locally compact abelian group.If (G,T) has no open compact subgroups then can we say G has an infinite discrete cyclic subgroup?</p> http://mathoverflow.net/questions/31511/totally-bounded-topology-and-indiscrete-topology Totally bounded topology and indiscrete topology unknown (google) 2010-07-12T07:22:40Z 2010-07-12T07:22:40Z <p>Is this true? A refinement of a totally bounded topology and an indiscrete topolgy is an indiscrete topology.</p> http://mathoverflow.net/questions/30094/finitely-generated-group Finitely generated group unknown (google) 2010-06-30T18:56:18Z 2010-06-30T19:40:41Z <p>If G is finitely generated how can we say G/nG is finite for every natural number n?</p> http://mathoverflow.net/questions/30093/totally-bounded-subgroup Totally bounded subgroup. unknown (google) 2010-06-30T18:51:29Z 2010-06-30T18:51:29Z <p>Is it true to say every totally bounded subgroup of a complete group is finite?</p> http://mathoverflow.net/questions/30092/discrete-cyclic-subgroup Comment by 2010-07-01T06:23:15Z 2010-07-01T06:23:15Z here is the article <a href="http://www.4shared.com/document/ih8d9nMA/29-Complemented_Topologies_on_.html" rel="nofollow">4shared.com/document/ih8d9nMA/&hellip;</a> and please see theorem 3.2 http://mathoverflow.net/questions/29950/countable-open-subgroup Comment by 2010-06-30T18:34:58Z 2010-06-30T18:34:58Z Thanks to all for your attentions. here is the theorem.<a href="http://666kb.com/i/bkj3854twsrqna1hf.bmp" rel="nofollow">666kb.com/i/bkj3854twsrqna1hf.bmp</a> and also requirment definition:<a href="http://666kb.com/i/bkj3c6i3ushyjzchv.bmp" rel="nofollow">666kb.com/i/bkj3c6i3ushyjzchv.bmp</a> .And if you need Article here it is:<a href="http://www.4shared.com/document/ih8d9nMA/29-Complemented_Topologies_on_.html" rel="nofollow">4shared.com/document/ih8d9nMA/&hellip;</a> http://mathoverflow.net/questions/29950/countable-open-subgroup Comment by 2010-06-29T20:52:27Z 2010-06-29T20:52:27Z Some requirement Definition. A sequence (an) of elements of a group G is a T-sequence if there exists a topology on G making this sequence vanish. .A group G furnished with a maximal topology in which a given T-sequence (an) vanishes we say topology on a group G is determined by T-sequence (an). http://mathoverflow.net/questions/29950/countable-open-subgroup Comment by 2010-06-29T20:42:05Z 2010-06-29T20:42:05Z In Article &quot;Complemented topologies on abelian groups&quot; All groups are abelian and hausdorff. In this article 3.6.Theorem followed as below: &quot;Let Let T1 be a topology on an infinite group G which is determined by some T-sequence. There exists a complement T2 of T1 on G which is determined by a T-sequence. Proof. Note that (G, T1) has a countable open subgroup. Now, apply Theorems 1.6 and 3.5 and Lemma 2.3.&quot; My question is: Why (G, T1) has a countable open subgroup?