Timeline for Is there a "direct" proof of the Galois symmetry on centre of group algebra?
Current License: CC BY-SA 4.0
17 events
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| Feb 3, 2022 at 11:33 | comment | added | Chris H | By a generators and relations argument I would think some symbolic argument that holds in any torsion group of exponent coprime to $n$, building $a,b$ from $g,h$ using these assumptions and the assumed $n$th root function. If one could give a group of this type where the result fails, it would rule out such an argument, which would also be very interesting. | |
| Feb 3, 2022 at 10:26 | comment | added | YCor | What do you mean by "a generators-and-relations argument for... $(gh)^n=ag^na^{-1}bh^nb^{-1}$"? I doubt there exist such $a,b$ in the free group over $g,h$. For $n=2$ there are none: any $x,y,z$ in a free group satisfying $x^2y^2=z^2$ must commute (it follows there are none whenever $n$ is even). | |
| Feb 3, 2022 at 10:16 | comment | added | Tom WIlde | A slightly different (but definitely character theoretic) proof of this interesting fact uses the structure constants $a_{KLM}=\frac{|K||L|}{|G|}\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(x_K)\chi(x_L)\overline{\chi(x_M)}}{\chi(1)}.$ If $K^\prime$ etc. are the classes containing the respective $n^{th}$ powers and if $\sigma$ is Chris' automorphism then $a_{K^\prime L^\prime M^\prime}=\frac{|K||L|}{|G|}\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi^\sigma(x_K)\chi^\sigma(x_L)\overline{\chi^\sigma(x_M)}}{\chi(1)}=a_{KLM}^\sigma,$ so $a_{K^\prime L^\prime M^\prime}=a_{KLM}^\sigma=a_{KLM}.$ | |
| Feb 2, 2022 at 11:20 | comment | added | Geoff Robinson | There are a number of similar facts which seem non-obvious from a group-theoretic point of view, but are easy to prove with characters: for example, the number of times $ x \in G$ is expressible as a commutator is unchanged if we replace $x$ by a different generator of $\langle x \rangle.$ | |
| Feb 2, 2022 at 6:16 | comment | added | Fedor Petrov | @spin ah, thanks | |
| Feb 2, 2022 at 5:52 | comment | added | spin | @FedorPetrov: Note the assumption $\gcd(n, |G|) = 1$. So if $g^n = 1$, then $g = 1$. | |
| Feb 2, 2022 at 4:54 | comment | added | Chris H | The Frobenius group of order $20$ is an example, trivial outer automorphism group, with nonrational characters, take $n=3$. | |
| Feb 2, 2022 at 4:23 | comment | added | David E Speyer | Basic question: What is an example of a finite group $G$ and an integer $n$ prime to $|G|$ such that there is no automorphism of $G$ which induces the permutation $\text{Conj}(g) \mapsto \text{Conj}(g^n)$ on conjugacy classes? | |
| Feb 2, 2022 at 0:58 | history | edited | Sam Hopkins |
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| Feb 1, 2022 at 23:58 | comment | added | darij grinberg | Wow, that's a beautiful find. I am getting a slightly cohomological whiff from it (as, e.g., in the proof of the transfer's homomorphism property). | |
| Feb 1, 2022 at 22:44 | comment | added | LSpice | There is a theory of products and powers of conjugacy classes; see Beltrán, Felipe, and Melchor - Some problems about products of conjugacy classes in finite groups for what appears to be a recent survey. It seems to be in a different direction, but I wonder if that might be the place to look. | |
| Feb 1, 2022 at 22:23 | history | edited | Chris H | CC BY-SA 4.0 |
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| Feb 1, 2022 at 22:18 | comment | added | Chris H | Its true, and follows from the fact that $g\mapsto g^n$ is an algebra automorphism of $\mathbb{Z}[G]^G$, by expanding the product of the classes of $g$ and $h$. I don't know of a non character theoretic proof of why this map is actually an algebra automorphism however. | |
| Feb 1, 2022 at 22:08 | comment | added | LSpice | Is the quoted equality about $(g h)^n$ true, but you don't know how to prove it, or you're not sure if it's true? | |
| Feb 1, 2022 at 21:57 | history | edited | LSpice | CC BY-SA 4.0 |
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| Feb 1, 2022 at 21:47 | history | edited | Chris H | CC BY-SA 4.0 |
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| Feb 1, 2022 at 21:38 | history | asked | Chris H | CC BY-SA 4.0 |