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Let $G$ be an infinite group, let $S_0\subseteq G$ be a subgroup and suppose that ${\frak C}$ is a collection of subgroups of $G$ such that

  1. $C \cong S_0$ for all $C\in {\frak C}$, and
  2. for all $C, C'\in {\frak C}$ we have $C\subseteq C'$ or $C'\subseteq C$.

It is easy to see that $\bigcup{\frak C}\subseteq G$ is a subgroup of $G$, but do we necessarily have $\bigcup{\frak C} \cong S_0$?

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    $\begingroup$ Obviously not. For instance if the chain has no maximum and each $C$ is infinite cyclic, then the union is not cyclic (the union cannot be finitely generated). For example $\mathbf{Z}[1/2]=\bigcup\frac1{2^n}\mathbf{Z}$ and $\mathbf{Q}=\bigcup\frac1{n!}\mathbf{Z}$. $\endgroup$
    – YCor
    Commented Oct 4, 2019 at 6:16

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This can fail on cardinality grounds. For instance if $G$ is a vector space of dimension $\aleph_1$ and $\mathfrak{C}$ is a chain of $\aleph_1$ subspaces each of dimension $\aleph_0$, then all of the groups in $\mathfrak{C}$ are isomorphic but their union is not isomorphic to any of them.

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    $\begingroup$ Thanks @jameshanson for this beautifully simple argument - it didn't occur to me to look into cardinalities $\endgroup$
    – Dominic van der Zypen
    Commented Oct 4, 2019 at 6:16
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    $\begingroup$ in this precise case, if the field is uncountable, the union is isomorphic as a group to the subspaces of countable dimension. But it works if the field is countable. $\endgroup$
    – YCor
    Commented Oct 4, 2019 at 6:22

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