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Find some nonisomorphic groups that are direct limits of cyclic groups of orders $p,p^2,p^3, \cdots$.

I know that the Prüfer group of type $p^{\infty}$ is a direct limit of cyclic groups of oders $p,p^2,p^3, \cdots$. In order to find other direct limits, I think the only thing I can do is to redefine the homomorphisms between the groups.

Suppose the groups $G_i$ is generated by $x_i$, of order $p^i$, $i=1,2, \cdots$. For any positive integers $i$ and $j$, if $i \leq j$, define the map $\rho_{ij}: G_i \rightarrow G_j$, $x_i \mapsto x_j^{p^{j-i}}$. Then $D=\{ G_i, \rho_{ij} : i \leq j, i,j \in \mathbb{Z}_+ \}$ is a direct limit of $G_i$, which is the Prüfer group. As $(p+1)$ is prime to $p$, if the new homomorphism $\rho_{ij}': G_i \rightarrow G_j$ maps $x_i$ to $x_j^{p^{j-i}(p+1)^{j-i}}$, it is also injective. Then another direct limit can be obtained.

Are the two direct limits $D=\{ G_i, \rho_{ij} : i \leq j, i,j \in \mathbb{Z}_+ \}$ and $D'=\{ G_i, \rho_{ij}' : i \leq j, i,j \in \mathbb{Z}_+ \}$ isomorphic? If they are not, how can I prove? [I think the key problem lies in determining the possible homomorphisms between $D$ and $D'$.]

Many thanks.

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up vote 6 down vote accepted

You're right in that your only wiggle room is to choose a different set of connecting homomorphisms.

One thing that you might try is defining $\rho_{ij}:\mathbb{Z}/p^i \rightarrow \mathbb{Z}/p^j$ for $j\geq i$ to be the zero map if $i\neq j$ and the identity otherwise. This defines a directed system of cyclic $p$-groups for which the direct limit is the trivial group. But of course this is a pretty boring example.

Unfortunately, in your example, $D$ and $D'$ turn out to both be the Prüfer $p$-group $\mathbb{Z}[1/p]/\mathbb{Z}.$

To see this note that the Prüfer $p$-group is the direct limit of the system obtained from choosing a monomorphism $\rho_{i,i+1} : \mathbb{Z}/p^i \rightarrow \mathbb{Z}/p^{i+1}$ and defining $\rho_{i,j} = \rho_{i,i+1} \circ \ldots \circ \rho_{j-1,j} $ if $i<j$ and $\rho_{i,j} = id_{\mathbb{Z}/p^i}$ otherwise. Now there are many such monomorphisms but all yield isomorphic directed systems so the Prüfer $p$-group is well defined. Now note that both $D$ and $D'$ are direct limits of such directed systems. The former is generated by the inclusion $\rho_{i,i+1} : \mathbb{Z}/p^i \rightarrow \mathbb{Z}/p^{i+1}$ sending $x_i$ to $x_{i+1}^p$ and the latter by the inclusion $\rho_{i,i+1}' : \mathbb{Z}/p^i \rightarrow \mathbb{Z}/p^{i+1}$ sending $x_i$ to $x_{i+1}^{p(p+1)}$ for some choice of generators $x_i$ of $\mathbb{Z}/p^i$ and $x_{i+1}$ of $\mathbb{Z}/p^{i+1}.$ No sweat though, this is probably a pretty common error. To clear up your understanding it might be a helpful exercise to show that the Prüfer $p$-group is well defined i.e. that all such directed systems described above are isomorphic.

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In summary: a direct limit of cyclic p-groups is either a cyclic p-group (including the trivial group) or a Prüfer p-group. –  Jack Schmidt Aug 2 '11 at 17:44
    
Thank you very much. I was wrong because I thought $rho_{ij}$ must be injective, and that was why I get the Prüfer $p$-group again. Mr. @Jack Schmidt: Thank you very much for the summary. Now I am seeking the definition of homomorphisms to make a given cyclic $p$-group to be the direct limit of the system. –  ShinyaSakai Aug 4 '11 at 3:47
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