Timeline for Center of p-groups
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Feb 6 at 7:12 | history | suggested | A_S | CC BY-SA 4.0 |
Correcting the index
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| Feb 6 at 6:34 | review | Suggested edits | |||
| S Feb 6 at 7:12 | |||||
| Jul 4, 2016 at 22:58 | answer | added | yakov | timeline score: -1 | |
| Sep 15, 2012 at 0:18 | vote | accept | i. m. soloveichik | ||
| Sep 15, 2012 at 0:18 | vote | accept | i. m. soloveichik | ||
| Sep 15, 2012 at 0:18 | |||||
| Sep 15, 2012 at 0:15 | vote | accept | i. m. soloveichik | ||
| Sep 15, 2012 at 0:18 | |||||
| Sep 14, 2012 at 19:57 | answer | added | Arturo Magidin | timeline score: 14 | |
| Sep 14, 2012 at 16:25 | comment | added | Arturo Magidin | It is easy to get examples with index $p^{2n}$ for any positive integer $n$ by using an extraspecial $p$-group of order $p^{2n+1}$ and using the construction given by Konstantin. | |
| Sep 14, 2012 at 15:23 | comment | added | user91132 | Yes, I was just about to add that. Both have the properties that they are generated by elements $x,y$ with $xy = yxz$ and $z$ central and nontrivial with $H / \langle z \rangle$ of order $p^2$. | |
| Sep 14, 2012 at 15:19 | comment | added | Will Sawin | As far as I can tell, you just have to replace the Heisenberg group with Hamilton's quaternions or $D_4$. | |
| Sep 14, 2012 at 14:58 | comment | added | user91132 | This is for $p > 2$. If $p = 2$, then I guess you have to be a little more clever. | |
| Sep 14, 2012 at 14:55 | comment | added | user91132 | I think the answer is "yes" if you ask for index $p^2$. Let $H$ be the Heisenberg group of uni-upper-triangular matrices with entries in the finite field $\mathbb{F}_p$ with $p$ elements. Then $H$ has order $p^3$ and it is generated by $3$ elements $x,y,z$ of order $p$ subject to the relation $xy = zyx$ say. Now let $g \in A$ be any element of order $p$, and let $G := (A \times H ) / \langle (g, z^{-1}) \rangle$. Then $G$ contains $A$ as a central subgroup of index $p^2$, and $Z(G)$ cannot strictly contain $A$ since then $Z(G)$ would have index $p$. So $A = Z(G)$ has index precisely $p^2$. | |
| Sep 14, 2012 at 14:34 | comment | added | i. m. soloveichik | I should have written index $p^2$. | |
| Sep 14, 2012 at 14:26 | vote | accept | i. m. soloveichik | ||
| Sep 15, 2012 at 0:15 | |||||
| Sep 14, 2012 at 14:09 | history | edited | Alexander Chervov |
tag p-groups
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| Sep 14, 2012 at 14:06 | comment | added | Alexander Chervov | If omit "of index p". Is it true that any abelian group can be center of of some p-group ? | |
| Sep 14, 2012 at 14:01 | comment | added | Gjergji Zaimi | The center of a p-group cannot have index p... | |
| Sep 14, 2012 at 14:01 | answer | added | user91132 | timeline score: 5 | |
| Sep 14, 2012 at 14:01 | answer | added | Charles Matthews | timeline score: 2 | |
| Sep 14, 2012 at 13:53 | history | asked | i. m. soloveichik | CC BY-SA 3.0 |