Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 18, 2016 at 20:53 | comment | added | Pax | There are two extensions given by $0\to \mathbb{Z}/(2)\to \mathbb{Z}/(4)\to \mathbb{Z}/(2)\to 0$ and the trivial one. Thus the extensions is isomorphic to one of these two, and $G=\mathbb{Z}/(2)^2, \mathbb{Z}/(4)$, which are abelian by computation. | |
| May 18, 2016 at 20:53 | comment | added | Pax | There is a more convoluted version of this fact. Assume that $G$ is nonabelian. Since it is a $p$-group, it has a nontrivial center, which must be $\mathbb{Z}/(2)$. This is normal, so we have an extension $1\to \mathbb{Z}/(2)\to G\to \mathbb{Z}/(2)\to 1$, and so an element of $H^2(\mathbb{Z}/(2), \mathbb{Z}/(2))=H^1(B\mathbb{Z}/(2); \mathbb{Z}/(2))=H^1(\mathbb{R}P^{\infty}; \mathbb{Z}/(2))=\mathbb{Z}/(2)$. | |
| Jan 6, 2014 at 20:52 | history | edited | Ant | CC BY-SA 3.0 |
added 6 characters in body
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| Jun 25, 2013 at 3:02 | review | Late answers | |||
| Jun 25, 2013 at 12:07 | |||||
| Dec 21, 2012 at 18:33 | comment | added | David Corwin | When I first learned character theory, I asked if one could prove the four-square theorem by exhibiting a group of every order with exactly four conjugacy classes. While it became obvious a second later that this was only a fantasy, I'm excited that someone else thought of the relation between character theory and the four-square theorem! | |
| Aug 20, 2012 at 9:32 | comment | added | James Cranch | No, Ostap. In general, there is not a group for every way of writing $n$ as a sum of squares. Indeed, every group of order 17 is abelian too, but $17 = 1^2 + 4^2 = 1^2 + \cdots + 1^2$. | |
| Aug 20, 2012 at 1:10 | comment | added | Todd Trimble | Well, 4 is already the sum of one square. To complete the proof, one should say "the only way to write 4 as a sum of squares, one of which is 1, coming from the trivial representation, is..." | |
| May 17, 2012 at 8:53 | comment | added | Ostap Chervak | I guess, you can make it more striking: "Using character theory, since any group of order 4 is abelian hence the only way to write 3 as a sum of squares is 3 =1^2 + 1^2+ 1^2" Right? | |
| Jan 29, 2011 at 4:26 | comment | added | Harry Altman | Well, any way that includes the required one copy of 1^2. Otherwise 2^2 would be a possibility... | |
| Jan 29, 2011 at 3:00 | history | answered | William Hale | CC BY-SA 2.5 |