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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
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.
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
Oct 30, 2015 at 18:50 history asked Aeryk CC BY-SA 3.0