Timeline for Invertible elements in a group algebra
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2016 at 21:20 | comment | added | YCor | Related: math.stackexchange.com/questions/544093, mathoverflow.net/questions/79702 | |
| Feb 2, 2016 at 21:16 | comment | added | YCor | It's indeed a classical conjecture of Kaplansky (en.wikipedia.org/wiki/Kaplansky%27s_conjecture) that if $K$ is a field , the units of $KG$ are the multiples of elements in the standard basis, and is well-known and easy when $G$ is abelian. In particular if $K$ has 2 elements, the obvious homomorphism $G\to KG^\times$ is an isomorphism (which is the right statement, just saying that they are isomorphic is weaker and much less interesting). | |
| Feb 2, 2016 at 20:44 | comment | added | Yonatan Harpaz | I just realized it doesn't matter. If $x \in \mathbb{K}[H]$ is invertible then there exists a finitely generated subgroup $H' \subseteq H$ such that $\mathbb{K}[H'] \subseteq \mathbb{K}[H]$ contains both $x$ and $x^{-1}$, and so we may as well assume that $H$ is finitely generated. The answer hence is yes. | |
| Feb 2, 2016 at 20:32 | comment | added | Ofra | @YonatanHarpaz No, I'm not assuming that H is finitely generated. | |
| Feb 2, 2016 at 20:31 | comment | added | Yonatan Harpaz | Are we assuming $H$ to be finitely generated or not necessarily? In the finitely generated case the answer is yes. | |
| Feb 2, 2016 at 19:55 | review | Close votes | |||
| Feb 2, 2016 at 23:34 | |||||
| Feb 2, 2016 at 19:42 | history | edited | Ofra | CC BY-SA 3.0 |
added 18 characters in body
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| Feb 2, 2016 at 19:42 | comment | added | Ofra | oops thanks, I will change my question. | |
| Feb 2, 2016 at 19:37 | comment | added | Johannes Hahn | No, obviously not, because $\mathbb{K}^\times \leq (\mathbb{K}[H])^\times$. | |
| Feb 2, 2016 at 19:27 | history | asked | Ofra | CC BY-SA 3.0 |