Timeline for Recognize this countably generated abelian group?
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Oct 30, 2015 at 20:37 | vote | accept | Aeryk | ||
Oct 30, 2015 at 20:37 | history | edited | Aeryk | CC BY-SA 3.0 |
Fixed isomorphism according to discussion in comments
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Oct 30, 2015 at 20:31 | comment | added | Aeryk | @JulianRosen: Egads, you're right! I was computing my factor group incorrectly. Instead of modding out by $<g_k>$, I was effectively modding out by $<g_k, g_{k+1}, \dots>$. I'll correct that above. Thanks! | |
Oct 30, 2015 at 19:57 | comment | added | Julian Rosen | If you believe my answer, then $G_{(p,q)}/\langle g_k\rangle \cong \mathbb{Z}[1/q]/\langle p^k/q^k \rangle$ is an infinite group, so cannot be $\mathbb{Z}/p^k\mathbb{Z}$. | |
Oct 30, 2015 at 19:29 | history | edited | Aeryk | CC BY-SA 3.0 |
Added more info about the group and the reasons behing the restrictions on p and q.
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Oct 30, 2015 at 19:29 | answer | added | Julian Rosen | timeline score: 10 | |
Oct 30, 2015 at 19:18 | comment | added | Aeryk | @AnthonyQuas: I thought something similar, but wouldn't $p$-adic integers with finite expansions (and a $+$ or $-$ sign) just be regular integers? | |
Oct 30, 2015 at 19:07 | comment | added | Anthony Quas | I think this is the subgroup of $p$-adic integers consisting of those $p$-adic integers with finite expansions. | |
Oct 30, 2015 at 19:01 | history | edited | Aeryk | CC BY-SA 3.0 |
edited body
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Oct 30, 2015 at 18:50 | history | asked | Aeryk | CC BY-SA 3.0 |