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I was thinking about how to prove $\operatorname{Br}(K)\cong H^2(\operatorname{Gal}(\bar{K}/K),\bar{K}^*)$ without having to introduce inductive limits and all the profinite stuff. So, I started wondering if the conditions of a direct system could be weakened for the category of abelian groups in a way that isomorphisms would be still preserved. This brought me to the following general question:

Let $G$ and $H$ be two abelian groups, not necessarily finite, $I$ an index set and $(G_i) _{i\in I}$ and $(H_i)_{i\in I}$ families of subgroups respectevely of $G$ and $H$ such that

(1) $\forall i\in I: G_i \cong H_i$ and

(2) $\bigcup_{i\in I}G_i=G$ and $\bigcup_{i\in I}H_i=H$.

Question 1: Can we conclude that $G\cong H$?

Question 2: If yes, can we drop "abelian"?

EDIT: I forgot to mention that the $G_i$ (and $H_i$) are also assumed to be distinct subgroups.

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How about $G_i = G \cong \mathbb Z$ for all $i$, and $H = \mathbb Q$ and $H_i = \frac{1}{i!} \mathbb Z$? – j.p. Jan 26 '11 at 13:46
Consider $G_i=G=\mathbf{Z}/2\mathbf{Z}$ and $H=G \times G$ and $H_1=<(1,0)>, H_2=<(0,1)>$ and $H_3=<(1,1)>$, so you need some further assumptions. – Guntram Jan 26 '11 at 13:46
Even $I$ countable and $G_i$ finite abelian for all $i\in I$ doesn't help: take $G_0=H_0=0$, $G_{i+1} = G_i \times Z_{p^i}$ and $H_i$ to be the subgroup $p\cdot H_{i+1}$ ($H_{i+1} \cong G_{i+1}$). $G$ has then an element of order $p$ that is not a $p$-th power, whereas every element of $H$ has a $p$-th root. – Someone Jan 26 '11 at 14:20
Thank you for the insightful comments! – efq Jan 26 '11 at 14:30
up vote 13 down vote accepted

The answer is no.

For a counterexample, let $G_i=\mathbb{Z}$ be the integers and let $H_i=\frac1i\mathbb{Z}$, for positive natural numbers $i$. The union $\bigcup_i G_i=\mathbb{Z}$, but $\bigcup_i H_i=\mathbb{Q}$.

For the revised question, where you want $G_i$ and $H_i$ distinct, there are still counterexamples, such as $G_i=i\mathbb{Z}$ and $H_i=\frac1i\mathbb{Z}$.

On a positive note, if you have a bit more coherence in your isomorphisms, then you can make the affirmative conclusion. That is, if we can find particular isomorphisms $\pi_i:G_i\cong H_i$ which agree on their common domains, then they will build together into an isomorphism of $G$ and $H$. That is, what you want is not merely that $G_i\cong H_i$, but rather that the way that $G_i$ sits inside $G$ is the same as the way $H_i$ sits inside $H$. More generally, if $I$ is not just a naked index set, but is a directed set, such that when $i\lt j$ in this order then we have maps $G_i\to G_j$ and $H_i\to G_j$ and the isomorphisms $G_i\cong H_i$ make a commutative system, then the direct limit $G$ of the $G_i$'s will be isomorphic to the direct limit $H$ of the $H_i$'s by universal property arguments.

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Thank you for the very informative answer! Much appreciated! – efq Jan 26 '11 at 14:29
Just one question: In the less general case, where the isomorphisms $\pi_i$ agree on their common domains, how do they build into an isomorphism $G\cong H$ constructively? – efq Jan 26 '11 at 16:37
I meant that if we assume also that any two objects in the union appear together in some piece $G_i$, then the set-theoretic union of the piece-wise isomorphisms is literally an isomorphism of the unions. This is because being an isomorphism on the union sets is a local property, which will be preserved to unions, since $\pi(a+b)=\pi(a)+\pi(b)$ can be witnessed once you have $a$ and $b$ together, and similarly for injectivity and surjectivity. – Joel David Hamkins Jan 26 '11 at 17:05
I believe one can get by with less. We simply need the homomorphisms $H_i \rightarrow G_j$ to agree on any particular element of $H$ for $i$ sufficiently large in the sense if the directed index set $I$. But then the graph of the desired homomorphism will emerge from an intersection of unions. – David Feldman Apr 17 at 0:54

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