Timeline for Decomposition of finite abelian groups of even order if there is an involution
Current License: CC BY-SA 4.0
9 events
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| May 20, 2023 at 11:28 | history | edited | Andrea Antinucci | CC BY-SA 4.0 |
added clarification
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| May 20, 2023 at 11:27 | comment | added | Andrea Antinucci | I am afraid that I am not able to understand your answer. The question is in very elementary terms, and the answer for $|G|$ odd is also extremely elementary: is it possible to describe the results you are mentioning in that level of simplicity? | |
| May 19, 2023 at 9:31 | comment | added | Emil Jeřábek | ... $\sigma$-subgroups of $(C_{2^k}\times C_{2^k},s)$. The latter can be classified in more detail, but this is not very illuminating. But in particular, the cyclic ones are $(C_{2^k},\alpha x)$ where $\alpha^2\equiv1\pmod{2^k}$, i.e., $\alpha=\pm1$ or $\alpha=2^{k-1}\pm1$. Any finite $\sigma$-group is thus a finite subdirect product of such finite sdi $\sigma$-groups. I do not know how much a decomposition into a direct product of directly indecomposable $\sigma$-groups will differ, but I suspect there do exist directly indecomposable finite $\sigma$-groups that are not sdi. | |
| May 19, 2023 at 9:26 | comment | added | Emil Jeřábek | ... Moreover, it is easy to see that $(Z(p^\infty)\times Z(p^\infty),s)\simeq(Z(p^\infty),x)\times(Z(p^\infty),-x)$ for odd $p$ (I’m using additive notation). Since quotients of $\sigma$-groups uniquely correspond to $\sigma$-subgroups, sdi $\sigma$-groups are closed under $\sigma$-subgroups, and it is straightforward to check that $(Z(p^\infty),\pm x)$ and $(Z(2^\infty)\times Z(2^\infty),s)$ are sdi. Thus, sdi $\sigma$-groups are exactly the $\sigma$-subgroups of $(Z(p^\infty),\pm x)$ and $(Z(2^\infty)\times Z(2^\infty),s)$; the finite ones are $(C_{p^k},\pm x)$ and ... | |
| May 19, 2023 at 9:20 | comment | added | Emil Jeřábek | Abelian groups with involution (I will write $\sigma$-groups for short) form a variety, hence any $\sigma$-group is a subdirect product of subdirectly irreducible (sdi) $\sigma$-groups. Now, what are these? Recall that sdi abelian groups are subgroups of the Prüfer groups $Z(p^\infty)$ (the finite ones being cyclic groups of prime power order). Every $\sigma$-group $(G,\sigma)$ embeds into $(G\times G,s)$ where $s(x,y)=(y,x)$ via $x\mapsto(x,\sigma(x))$, thus using a subdirect decomposition of $G$, $(G,\sigma)$ embeds into a product of $(Z(p^\infty)\times Z(p^\infty),s)$. ... | |
| May 17, 2023 at 11:45 | history | edited | Andrea Antinucci | CC BY-SA 4.0 |
corrected typo
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| May 16, 2023 at 17:47 | comment | added | Andrea Antinucci | If $g_1,g_2 \in A$, then $\sigma(g_1)=g_1^{-1}$, $\sigma(g_2)=g_2^{-1}$, and hence $\sigma(g_1g_2)=\sigma(g_1)\sigma(g_2)=g_1^{-1}g_2^{-1}=(g_1g_2)^{-1}$. Also if $g\in A$ $\sigma(g^{-1})=g=(g^{-1})^{-1}$, so $A\subset G$ is a subgroup. | |
| May 16, 2023 at 10:15 | comment | added | Derek Holt | In the special case when $G$ is elementary abelian,the only indecomposable modules for the action are the trivial module and the regular module of dimension $2$, so you can describe the action precisely. | |
| May 16, 2023 at 9:54 | history | asked | Andrea Antinucci | CC BY-SA 4.0 |