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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 3, 2012 at 10:09 vote accept Mark Grant
Mar 21, 2011 at 15:04 history edited Mark Grant CC BY-SA 2.5
added link to new question
Mar 18, 2011 at 10:49 history edited Mark Grant CC BY-SA 2.5
Asked follow-up question
Mar 14, 2011 at 7:56 answer added agt timeline score: 1
Mar 13, 2011 at 13:25 comment added Mark Grant Thanks Giuseppe. In my applications $G$ is connected, but your comment shows I was being too optimistic (as does Jack's connected answer).
Mar 13, 2011 at 13:21 answer added Jim Humphreys timeline score: 2
Mar 13, 2011 at 13:21 history edited Mark Grant CC BY-SA 2.5
typo
Mar 13, 2011 at 10:40 answer added Jack Schmidt timeline score: 4
Mar 13, 2011 at 10:19 comment added agt If you consider groups whose cardinality is less than or equal to $\aleph_0$ and endow them with the discrete topology, then you get 0-dimensional Lie groups. Now take $G=S_n$ the symmetric group of $\{1,\ldot,n\},$ and $H_i$ the subgroup of the permutations fixing $i,$ for $i=1,\ldots,n$ these are obviously conjugate each other. Trivially $H_1\cap H_2$ is not conjugate to $H_1$. Excuse me if this answer is not what you wanted.
Mar 13, 2011 at 10:01 history edited Mark Grant CC BY-SA 2.5
typos
Mar 13, 2011 at 9:56 comment added agt Excuse me, but in the last line, shouldn't there be ${}^aH\cap{}^bH$ instead of ${}^aH={}^bH$ ? or not?
Mar 13, 2011 at 9:24 history asked Mark Grant CC BY-SA 2.5