I have a process which generates infinite sequences $S=S_0\rightarrowtail S_1 \rightarrowtail S_2\ldots$ of finite Abelian groups $S_i$ connected by monomorphisms $S_i\rightarrowtail S_{i+1}$ (injective group homomorphisms). Each sequence thus corresponds to a functor $S:\omega\rightarrow Ab$ from the usual linear order on the counting numbers $\omega=\{0,1,2,\ldots\}$ to the category of Abelian groups. Since it is guaranteed that the minimal number of generators of $S_i$ is strictly smaller than of $S_{i+1}$, the direct limit $\varinjlim S$ is not finitely generated.
I am interested in sufficient conditions for the elements $S_i^A$ and $S_i^B$ of two such sequences $S^A$ and $S^B$ such that the existence of an isomorphism between the direct limits can be excluded. I am rather new to infinitely generated Abelian groups, thus, any reference to standard approaches/sufficient theorems I could try out would be helpful.
Additional information
- Every sequence $S^A:\omega\rightarrow Ab$ factors as $S^A=G F^A$, $F^A:\omega\rightarrow P$ and $G:P\rightarrow Ab$, where $P$ is a poset. The functor G is thereby the same for every sequence.
- There is a forgetful functor $U:P\rightarrow Set$ from $P$ to convex bounded subsets of $\mathbb{R}^2$ which maps morphisms to standard set inclusions, i.e. $U(P_1\subseteq P_2)=U(P_1)\subseteq U(P_2)$.
- The sequences $S^A=G F^A$ are constructed such that $\varinjlim U F^A=\mathbb{R}^2$.
- I don't have a closed formula for the structure of the finite Abelian groups $S^A_i$. However, for small enough $U F_i^A$, I am able to computationally decompose $S^A_i$ into the direct product of cyclic groups of prime power order.
- I have some limited control over the construction of the sequences $S$. Specifically, I can construct $S$ such that $S_i$ and $S_{i+1}$ contain cyclic subgroups $C_i\subseteq S_i$ and $C_{i+1}\subseteq S_{i+1}$, where $|C_i|$ divides $|C_{i+1}|$, such that the monomorphism $S_i\rightarrowtail S_{i+1}$ maps $C_i$ into $C_{i+1}$.
- When only concentrating on these subgroups, and by setting $s_0=1$, $s_{i+1}=\frac{|C_{i+1}|}{|C_{i}|}$ , I get the direct system $\mathbb{Z}/\mathbb{Z} \rightarrowtail \mathbb{Z}/(s_1\mathbb{Z}) \rightarrowtail \mathbb{Z}/(s_1s_2\mathbb{Z}) \rightarrowtail\ldots$ equivalent to $\mathbb{Z}/\mathbb{Z} \rightarrowtail \frac{1}{s_1}\mathbb{Z}/\mathbb{Z} \rightarrowtail \frac{1}{s_1s_2}\mathbb{Z}/\mathbb{Z} \rightarrowtail\ldots$.
- I can now define two sequences $S^A$ and $S^B$ such $S^A_0=S^B_0$, and such that $s^A_i = i$ and $s^B_i=p=const$. Then, the direct limits of these subgroups are given by $\bigcup\left(\frac{1}{s_i!}\mathbb{Z}\right)/\mathbb{Z} = \mathbb{Q}/\mathbb{Z}$, respectively by $\bigcup\left(\frac{1}{p^i}\mathbb{Z}\right)/ \mathbb{Z} = \mathbb{Z}[1/p] / \mathbb{Z} = \mathbb{Z}(p^\infty)$, with $\mathbb{Z}(p^\infty)$ the Prüfer p-group.
- This makes me believe that the direct limits of $S^A$ and $S^B$ should be not isomorphic, too. However, currently I lack understanding of what would be sufficient to show additionally to be able to prove this.