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Does every locally compact group (second countable and Hausdorff) topological group $G$ that is not compact have a nontrivial continuous homomorphism into $\mathbb{R}$?

Obviously for compact groups it is not possible since continuous functions send compact sets to compact sets, and there is only one (trivial) compact subgroup of $\mathbb{R}$.

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  • $\begingroup$ Maybe you want to consider homomorphisms into $\mathbb{C}$. $\endgroup$ – egreg Jul 15 '16 at 20:13
  • $\begingroup$ @egreg Any nontrivial homomorphism into $\mathbb{R}^2$ has a nontrivial vector in its image, hence we may postcompose with projection onto the line containing that vector to get a nontrivial homomorphism into $\mathbb{R}$. $\endgroup$ – arctic tern Jul 15 '16 at 20:22
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    $\begingroup$ $\mathrm{SL}_2(\mathbb{R})$ is an example (it's only proper nontrivial normal subgroup is $\{\pm I\}$). $\endgroup$ – arctic tern Jul 15 '16 at 20:35
  • $\begingroup$ @arctictern Sorry, I was thinking to $\mathbb{C}^\times$, the multiplicative group $\endgroup$ – egreg Jul 15 '16 at 21:24
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    $\begingroup$ $\mathrm{SL}_2(\mathbb{R}$ (or any other connected semisimple Lie group) doesn't have nontrivial homomorphisms into any abelian group, in particular $\mathbb{C}^\times$. $\endgroup$ – Cronus Aug 19 '16 at 23:42
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Suppose $G$ is infinite, discrete, and every element has finite order. Then any homomorphism $h : G \to \mathbb R$ would map each element to something of finite order since $h(g)^n = h(g^n) = h(e)=e$ for some positive $n \in \mathbb N$. But only the zero element of $\mathbb R$ has finite order. This implies $h$ is the trivial homomorphism. An example of such a $G$ is the group of all permutations of $\mathbb N$ that fix all but finitely many elements.

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  • $\begingroup$ So maybe add the hypothesis $G$ is connected too. $\endgroup$ – arctic tern Jul 15 '16 at 20:17
  • $\begingroup$ Great example, should have noticed this sooner. $\endgroup$ – nullUser Jul 15 '16 at 20:48
  • $\begingroup$ Do you know the answer if the group is connected? $\endgroup$ – Daron Jul 16 '16 at 11:52
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For a connected example one can take $G=\mathrm{SL}_2(\mathbb{R})$ (or any connected simple Lie group). If $f:G\to \mathbb{R}$ is a continuous homomorphism then $f(G)$ is a connected simple subgroup of $\mathbb{R}$, hence trivial.

EDIT: I see now someone already mentioned this in the comments.

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