# Pontrjagin duality for profinite and torsion abelian groups

I'm having trouble proving exercise 6.11.3 of "Introduction to homological algebra" by Weibel. I need to show that the category of torsion abelian groups is dual to the category of profinite abelian groups. It also gives a hint to show that $A$ is a torsion abelian group iff $\hom(A,\mathbb{Q}/\mathbb{Z})$ is a profinite group.

I'm stuck with the hint. I've proved that the torsion abelian group part implies that $$\hom(A,\mathbb{Q}/\mathbb{Z}) = \lim_{\leftarrow} \hom(H,\mathbb{Q}/\mathbb{Z}),$$ with $H$ going through all finite subgroups of $A$ with restriction maps as homomorphisms in the obvious way. I have absolutely no idea how to proof the other implication. I also don't see how this is going to help to associate a torsion abelian group to a profinite abelian group to make the duality.

Any thoughts ?

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The hint suggests that to a torsion $A$, associate the group ${\rm Hom}(A,{\mathbf Q}/{\mathbf Z})$. To a profinite abelian group $G$, look at ${\rm Hom}(G,{\mathbf Q}/{\mathbf Z})$. (Maybe you should also review what Pontryagin duality is generally saying beyond the confines of the torsion and profinite cases.) –  KCd Mar 31 '12 at 15:18
Thanks, but I'm still stuck. I must admit, I'm not very good at this subject (I'm trying to solve as many exercises as I can about Galois Cohomology in this book, not always with success). Why is $\hom(G,\mathbb{Q}/\mathbb{Z})$ a torsion abelian group ? –  KevinDL Apr 1 '12 at 8:48

I've done some reading about Pontryagin duality and this is what I've come up with: Pontryagin duality says that $\hom(\hom(G,\mathbb{R}/\mathbb{Z}),\mathbb{R}/\mathbb{Z}) = G$ in case of $G$ being a locally compact abelian group and $\hom$ standing for all continuous group homomorphisms. Now, in case of $G$ being a torsion abelian group or a profinite abelian group, we can change $\mathbb{R}$ by $\mathbb{Q}$ (If $G$ is torsion, this is trivial, since every image of $g \in G$ has to have finite order. If $G$ is profinite, this follows from $$\hom(G,\mathbb{R}/\mathbb{Z}) = \hom(\varprojlim G_i,\mathbb{R}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{R}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})\\ = \hom(G,\mathbb{Q}/\mathbb{Z})$$, since all $G_i$ are finite). Now, since $\varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})$ is a quotient $\oplus_i\hom(G_i,\mathbb{Q}/\mathbb{Z})$, which is clearly a torsion abelian group, we are done.
The third equality above is non-trivial, i.e., it is not immediately obvious that you can exchange the two limits. There is a natural injective (assuming $G\rightarrow G_i$ is surjective, which you can) map from each $\mathrm{Hom}(G_i,\mathbb{R}/\mathbb{Z})$ to $\mathrm{Hom}_{cts}(G,\mathbb{R}/\mathbb{Z})$, but to prove the induced map from the direct limit is surjective, you need to use that $G$ is profinite. Why must a continuous homomorphism from a profinite group $G$ to $\mathbb{R}/\mathbb{Z}$ factor through a finite quotient of $G$? This is because –  Keenan Kidwell Apr 4 '12 at 3:30
$\mathbb{R}/\mathbb{Z}$ has "no small subgroups," meaning that there is a neighborhood $U$ of the identity in $\mathbb{R}/\mathbb{Z}$ which contains no non-trivial subgroups. The inverse image in of $U$ in $G$ is an open neighborhood of the identity in $G$, so, because $G$ is profinite, there is an open normal subgroup $N$ of $G$ contained in the inverse image. It follows that $N$ is contained in the kernel of $G$ (its image is a subgroup of $U$, and so must be trivial), so your homomorphism factors through the finite quotient $G/N$. You might also want to check that the algebraic isomorphism –  Keenan Kidwell Apr 4 '12 at 3:32
In the second comment above this one, it should say "...kernel of the homomorphism" instead of "...kernel of $G$." –  Keenan Kidwell Apr 4 '12 at 3:53