Let $G$ be a closed, torsion-free and divisible subgroup of locally compact abelian group $X$ such that $nX=G$ for some $n$. For $x\in X$, there exist $g\in G$ such that $nx=ng$. So we can define a homomorphism $f:X\to G$, $f(x)=g$. Now, my question: Is $f$ continuous?
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In your second sentence, do you mean for all $g\in G$, there exists $x\in X$ such that $nx = g$? It seems that's what $nX = G$ implies, unless I've mixed something up. – Stahl Nov 01 '13 at 16:34
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It might also be helpful to note that $X = G\oplus H$, as $G$ is divisible and $X$ is abelian. – Stahl Nov 01 '13 at 16:47
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$G$ need not to split in $X$! – Aliakbar Nov 01 '13 at 16:56
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The map $\mu_n\colon G\to G$ defined by $\mu_n(g)=ng$ is a bijection, since $g$ is torsion-free divisible. So you're basically asking if $\mu_n^{-1}$ is continuous. – egreg Nov 01 '13 at 17:04
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@aliakbar See this – Stahl Nov 01 '13 at 19:28
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@Stahl $G$ is an algebraically direct summand of $X$, no topologically direct summand. – Aliakbar Nov 02 '13 at 03:37