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Question:

Let $p$ be a prime number. Let $G_n=C_{p^n}$ be the cyclic group of order $p^n$ with generator $x_n$. We define $\varphi:G_n \rightarrow G_{n+1}$ by $\varphi(x_n^a)=x_{n+1}^{pa}$. Using the above construction we obtain a group $G$, called a quasicyclic group and usually denoted $C_{p^\infty}$. Prove that $C_{p^\infty} \cong H$ where $H=\{ z \in \mathbb{C}^\times: \exists n \: z^{p^n}=1 \}$.

The lecturer didn't include the "above construction" and I can't find any more detail on the net. Anyone know how to construct this group from the homomorphisms given?

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    @Qiaochu: I'd found that web page and didn't really understand what it was saying. @Lubos: I don't think I made my question clear enough. I'd already found that isomorphism (between $C_{p^n}$ and $H_n$) and made the observation given in your last paragraph. What I need help with is the limit bit, which I still don't understand. I don't know anything about this "limit" idea, not even enough for me to see how it's valid that two sequences of isomorphic groups have isomorphic limits. This is what I'd like explaining, please!2011-05-26
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    Dear Kate, if two sequences are isomorphic one element after another, then all properties of these two sequences including the limit have to be isomorphic as well. "Isomorphic" really means "the same thing when it comes to any behavior of them".2011-05-26

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