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I am wondering what are the cyclic subgroups of $(5)$ in ($\mathbb{R}$,$\times$,$1$, $^{-1}$).

Any help will be highly appreciated.

Maths enthu
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  • By $(5)$ do you mean the subgroup generated by $5\in\Bbb R$? This itself is a cyclic group. – Berci May 09 '22 at 20:04
  • Yes. I am trying to find its cyclic subgroups now. – Maths enthu May 09 '22 at 20:15
  • Well any cyclic subgroup has a generator - what are the possible generators? What makes two groups with different generators the same? – Mark Bennet May 09 '22 at 20:27
  • @MarkBennet ..you mean I need to use the concept $(a, b)$ = $gcd(a, b)$ – Maths enthu May 09 '22 at 20:29
  • Well you are working with subgroups of the multiplicative reals, so you can't assume you are working with integers and in fact you can't be, because the inverse of $5$ is $\frac 15$. You'd love to be working with integers and addition, though. And that is a big hint. – Mark Bennet May 09 '22 at 20:37
  • @MarkBennet, so, ${5^0} = 0$, ${5^1}=5$, ${5^2}= 5 + 5 =10$...is that correct? – Maths enthu May 10 '22 at 04:28
  • To change from multiplication to addition you really need the idea of a logarithm - you add and subtract the powers so $5^2 \time 5^{-6}=5^{-4}$ just as $2-6=-4$ – Mark Bennet May 10 '22 at 08:17

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Every number $5^k$, $k\in\mathbb{N}$, generates such a subgroup.

Przemysław Scherwentke
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