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Let $\hat{\mathbb{Z}}$ the projective limit of $\mathbb{Z}/n\mathbb{Z}$ and $H$ a subgroup of finite index. Let $K=\hat{\mathbb{Z}}/H$ (we can do it because $\hat{\mathbb{Z}}$ is commutative). Since $K$ is abelian and finite, i know that it is a product of cyclic groups.

Is it true that $K$ is in fact a cyclic group itself?

Angelo Rendina
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1 Answers1

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Use the fact that $\widehat{\mathbb{Z}}$ is the product of each of the $p$-adic integer rings, $\widehat{\mathbb{Z}}\cong\prod \mathbb{Z}_p$. What must a quotient by a finite index subgroup look like? Then use the Chinese remainder theorem.

Sal
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  • What if i see $\hat{\mathbb{Z}}$ as the Galois group of $\overline{\mathbb{F}_p}$ over $\mathbb{F}_p$, the finite group of $p$ elements? Does your hint apply too? – Angelo Rendina Nov 30 '14 at 21:39
  • Indeed. The fact that a quotient by a finite index subgroup is cyclic is just saying that the Galois group $\mathrm{Gal}(K/\mathbb{F}p)$ is cyclic for any finite subextension $\overline{\mathbb{F}}_p\supset K\supseteq \mathbb{F}_p$, which is obviously true because any such $K$ is just some $\mathbb{F}{p^n}$. – Sal Nov 30 '14 at 21:47
  • True. But how do I know that any quotient of finite index subgroup is the Galois group of such extension? – Angelo Rendina Nov 30 '14 at 21:51
  • The infinite-degree case of the fundamental theorem of Galois theory (Wikipedia link). – Sal Nov 30 '14 at 21:52
  • I get it, but my problem is: let's say $[\hat{\mathbb{Z}}:H]=n$, so $K=\hat{\mathbb{Z}}/H$ and $K$ is abelian of $n$ elements. If $K$ is cyclic, then $K=\mathbb{Z}/n\mathbb{Z}$ and $K$ is the Galois group of $\mathbb{F}_{p^n}$ over $\mathbb{F}_p$. But if $K$ is not cyclic, then $K$ can't be any such Galois group. How do I get that $K$ HAS to be cyclic? I mean, both $\mathbb{Z}_4$ and the Klein group have degree 4, but the second one is not cyclic. – Angelo Rendina Nov 30 '14 at 21:58
  • My original hint is for the purpose of proving why such a $K$ is cyclic. Also, the FTGT says that the quotient by any closed subgroup is the Galois group of some subextension and vice versa, so if you prove that a finite index subgroup must be closed and take as known that the only finite degree subfields of $\overline{F}p$ are $\mathbb{F}{p^n}$ then that's another proof of what you want. – Sal Nov 30 '14 at 22:04