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If we are given $G = \mathbb{R}$ and the operation isn't explicitly given, is it assumed to be the $+$ operator?

Michael Hardy
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spitfiredd
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    Probably, but I guess it depends on context. – Seth Mar 27 '14 at 23:52
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    YES. And if they mention a topological property without specifying a topology, it should be assumed to be the usual topology. And if they mention measure without spedifying which measure, it should be assumed to be Lebesgue measure. And if they say $\mathbb R$ is a field, without mention of which operations, then assume the usual operations. Etc. – GEdgar Mar 27 '14 at 23:59
  • Assuming that other context doesn't hint at the possibility of a different group operation: Yes! Or if not, then your teacher is being abusive. – Jyrki Lahtonen Mar 28 '14 at 06:09

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For any odd integer $n$, you can define $a \# b = \sqrt[n]{a^n+b^n}$. This operation on $\mathbb{R}$ makes it a group however, the usual operation is addition.

ah11950
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  • +1 - nice answer! The OP is referring to the general case - yours is a special case, not used very often. –  Mar 27 '14 at 23:55
  • Indeed. Hence "the usual" - just illustrating that other operations exist. – ah11950 Mar 27 '14 at 23:56
  • The question we were given is G = $\mathbb{R}$, H = ${\log a| a \in \mathbb{Q}, a > 0}$; determine if H is a subgroup. Obviously $0 \not\in H$. – spitfiredd Mar 28 '14 at 00:07
  • As remarked by @GEdgar above, the operation in your question will be addition. I was just giving an example of some less usual operations one could define! Can you prove/disprove the question with this in mind? – ah11950 Mar 28 '14 at 00:10
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    @spitfiredd $0\in H$! It is of the form $\log a$ with $a\in\mathbb{Q}$, after all (take $a=1$)... – Steven Stadnicki Mar 28 '14 at 01:16
  • Whoops you are correct!! I even plotted this on an excel spreadsheet and I guess I overlooked it. Thanks for the catch! – spitfiredd Mar 28 '14 at 01:22
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This is not just about mathematics; it is a useful guiding principle, or a thumb rule in the absence of any other information. See below:

http://en.wikipedia.org/wiki/Occam%27s_razor

P Vanchinathan
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  • It could just as easily be multiplication, if you're not acquainted with the restriction that specifying a group by its underlying set requires it to be a group over the whole set. – Loki Clock Mar 28 '14 at 01:11
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Usually that is what it would be, although I think I'd say so explicitly.

Michael Hardy
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