Timeline for Factor group of direct product by restricted direct product
Current License: CC BY-SA 4.0
8 events
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| Aug 18, 2021 at 12:44 | comment | added | Joseph Van Name | The condition that $|F_{i}|<|F_{i+1}|$ is redundant. If $\prod_{i\in\omega}G_{i}$ is a product of finite groups and infinitely many of these groups are non-trivial, then there is an increasing sequence $(n_{k})_{k\in\omega}$ with $n_{0}=0$ such that if $F_{k}=G_{n_{k}}\times\dots\times G_{n_{k+1}-1}$, then $|F_{k}|<|F_{k+1}|$ for all $k$, and here $\prod_{k\in\omega}F_{k}/\text{Fin}\simeq\prod_{i\in\omega}G_{i}/\text{Fin}$. | |
| Aug 4, 2021 at 10:15 | comment | added | YCor | If the $F_i$ have unbounded exponent then the near product ($\prod F_i/\bigoplus F_i$) is also not residually finite. I even think it's not an iff condition: fix an odd prime $p$ and take the Heisenberg group of order $p^{2n+1}$ (central product of $n$ copies of the group of order $p^3$ of exponent $p$): the near product might fail to be residually finite. | |
| Aug 3, 2021 at 23:51 | answer | added | markvs | timeline score: 4 | |
| Aug 3, 2021 at 22:14 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 3, 2021 at 22:13 | comment | added | YCor | Take $F_i=\mathbf{Z}/p_i\mathbf{Z}$, where $p_i$ is prime and $p_i\to\infty$. Then your quotient contains $\mathbf{Q}$ as subgroup, hence is not residually finite. | |
| Aug 3, 2021 at 21:35 | history | edited | IGT | CC BY-SA 4.0 |
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| Aug 3, 2021 at 20:57 | review | First posts | |||
| Aug 3, 2021 at 21:07 | |||||
| Aug 3, 2021 at 20:55 | history | asked | IGT | CC BY-SA 4.0 |