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I wanted to prove the following proposition: Let H be an open divisible subgroup of an abelian topological group G. Then G is topologically isomorphic to H x G/H. As for the proof, using extension of the identity homomorphism of H to itself, I used a theorem that can extend the homomorphism to G. And used for the second coordinate, the canonical map of G onto G/H. I managed to show that it is an isomorphism, but could not show it is neither open/closed nor continuous. Any help in this topic would be much appreciated.

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Recapping what you have apparently already done. $H$ is divisible and abelian, hence an injective object in the category of abelian groups. Thus the identity mapping $id_H:H\to H$ can be extended to a homomorphism $\phi:G\to H$ such that $\phi(h)=h$ for all $h\in H$. General facts about groups then imply that as groups $G\cong H\times \ker\phi$. We can thus use $D=\ker\phi$ as a set of representatives of cosets of $H$ in $G$.

Let $U\subseteq H$ be any open set. Because $H$ itself is open in the topology of $G$, so is $U$. Consider the preimage $$\phi^{-1}(U)=\bigcup_{d\in D}dU.$$ Here all the sets $dU,d\in D,$ are open as translates of the open subset $U$. Therefore $\phi^{-1}(U)$ is open in the topology of $G$. Because $U$ was arbitrary, this implies that $\phi$ is continuous.

As a subspace of $G$ the group $D\cong G/H$ inherits the discrete topology. This is because any singleton $\{d\},d\in D,$ is the intersection of $D$ and the open subset $dH$.

It follows then easily that the basic open subsets of the product topology of $H\times D$ are open in $G$. The remaining task is to show that any open subset $V$ of $G$ is also open in the product topology. This follows from the fact that $V$ is the disjoint union of open sets $$ V=\bigcup_{d\in D}(dH\cap V). $$ For a fixed $d\in D$ here $dH\cap V=d(H\cap d^{-1}V)$ is a basic open subset of the product topology because $H\cap d^{-1}V$ is open in $H$, and $\{d\}$ is open in $D$.

Summary: $G$ is isomorphic and homemorphic to $H\times D$, where $D\cong G/H$ is given the discrete topology, and $H$ keeps the topology it got as a subspace of $G$.

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  • $\begingroup$ Took me some time, but very good :). Thank you @Jyrki $\endgroup$ – User666x May 7 '15 at 8:10

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