Let $G$ be a non-Abelian infinite group. Can $G$ admit more than one (inequivalent) non-compact locally compact metrizable second countable topologies that make it a topological group?

Thank you.


Turning my comment into an aswer: yes, it is possible for such a $G$ to have more than one topology with your requirements.

Consider $(\Bbb R,+)$ with its usual topology. As a group it is isomorphic to $(\Bbb C,+)$, so there is also a topology in which $(\Bbb R,+)$ is a topological group homeomorphic to $\Bbb R^2$, both topologies obviously satisfy your requirements.

The only issue is that $\Bbb R$ is Abelian, but that can be fixed by taking a direct product with any locally compact, second countable, metrizable, nonabelian topological group, since all of this properties are preserved under finite products.

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    $\begingroup$ Although "direct product with any locally compact, second countable, metrizable, nonabelian topological group" will preserve the distinctness of the two topologies, preserving non-homeomorphism seems to require being more specific about the nonabelian factor. (Any finite, discrete, nonabelian factor will do.) $\endgroup$ – Andreas Blass Feb 8 at 16:21
  • $\begingroup$ MO recommends avoiding comments of gratitude, but I like thanking people for their help. Thank you, this was enlightening. In this example there exists no continuous group isomorphism (not necessarily open) between the two. Is it possible to answer the same question but with the additional requirement that there is a continuous abstract group isomorphism between the two? If you think that this makes things very different, I will set up a new question. Thanks again. $\endgroup$ – Bedovlat Feb 8 at 16:26
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    $\begingroup$ @Bedovlat I don't see an easy way to construct a pair with such an isomorphism, I'm not sure whether it's a much harder question or whether there is another neat construction $\endgroup$ – Alessandro Codenotti Feb 8 at 21:06
  • $\begingroup$ A continuous bijection from compact to Hausdorff is a homeomorphism. From locally compact to Hausdorff is a similar situation, so I agree that it is probably impossible (or at least much harder). $\endgroup$ – Bedovlat Feb 8 at 21:11

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