Timeline for Quotient of subgroups by center.
Current License: CC BY-SA 3.0
13 events
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May 20, 2011 at 3:01 | comment | added | Arturo Magidin | @Stuart: If $N\triangleleft G$ and $H$ is any subgroup of $G$, then $(H\cap N)\triangleleft H$, because $hNh^{-1}=N$ and $hHh^{-1}=H$ for all $h\in H$. | |
May 19, 2011 at 17:57 | vote | accept | Stuart | ||
May 19, 2011 at 17:51 | comment | added | Stuart | Umm... yeah, clearly because if it is in the center it commutes with everything... thanks again! | |
May 19, 2011 at 17:47 | comment | added | Stuart | So if I follow correctly, because of that, it follows that ($H \cap Z_G$) is normal in $H$ because it is a subgroup of the center? | |
May 19, 2011 at 17:42 | vote | accept | Stuart | ||
May 19, 2011 at 17:45 | |||||
May 19, 2011 at 17:42 | comment | added | Stuart | I didnt realize $Z_H \geq (H \cap Z_G)$. Then that means that $HZ_G/Z_G$ is actually a subgroup and everything is fine. Thanks!! | |
May 19, 2011 at 17:38 | comment | added | Arturo Magidin | @Stuart: Oh, I see what the problem was; a typo. I meant "the third term of the lower central series of $F_k$; fixed. | |
May 19, 2011 at 17:38 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
typo
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May 19, 2011 at 17:37 | vote | accept | Stuart | ||
May 19, 2011 at 17:38 | |||||
May 19, 2011 at 17:31 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
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May 19, 2011 at 17:28 | comment | added | Stuart | If G is nilpotent of class 2, then the third term of the lower central series is trivial, is it not? | |
May 19, 2011 at 17:25 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
added 334 characters in body
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May 19, 2011 at 17:16 | history | answered | Arturo Magidin | CC BY-SA 3.0 |