7
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ How about $G_i = G \cong \mathbb Z$ for all $i$, and $H = \mathbb Q$ and $H_i = \frac{1}{i!} \mathbb Z$? $\endgroup$ – j.p. Jan 26 '11 at 13:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Guntram Jan 26 '11 at 13:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Someone Jan 26 '11 at 14:20
14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for the very informative answer! Much appreciated! $\endgroup$ – M.G. Jan 26 '11 at 14:29
  • $\begingroup$ 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? $\endgroup$ – M.G. Jan 26 '11 at 16:37
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Jan 26 '11 at 17:05
  • 1
    $\begingroup$ 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. $\endgroup$ – David Feldman Apr 17 '16 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.