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Infinite cyclic group only has two generators
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[QUOTE="fresh_42, post: 6029912, member: 572553"] No. That's a good example why I favor to write all dependencies explicitly. E.g. people write "##\forall \, \varepsilon \,\exists \,N ## such that ..." Wrong! It has to be "##\forall \, \varepsilon \,\exists \,N(\varepsilon) ## such that ..." because the choice of ##N## depends on the choice of ##\varepsilon \,.## It is the same in what you wrote: ##\forall \,b\in \mathbb{Z} \,\exists \, m(b)\in \mathbb{Z} \ldots ## We thus have ##b=m(b)\cdot a##. So what? This is no surprise as it is exactly what we started with. There is an easy way to check your proof, because ##(H,\cdot)=(\mathbb{Z},+)##. The exponents of the multiplicative ##x## are precisely the integers, so that is the isomorphism. Just assume ##a \neq \pm 1\,.## How could we ever reach points between ##1## and ##a\,?## Only if we had a relation, but ##x## doesn't have a relation, it is a free group. Edit: Correction. Now I got your argument. ##a \,\mid \,b## for all ##b## and thus ##a## is a unit, hence ##a= \pm 1\,.## Sorry. It's a bit late here. [/QUOTE]
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Infinite cyclic group only has two generators
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