I'm currently going through Hungerford's book "Algebra", and the first proof I found a bit confusing is the proof of the theorem which states that every infinite cyclic group is isomorphic to the group of integers (the other part of the theorem states that every finite cyclic group of order m is isomorphic to the group Zm, but I understood that part).(adsbygoogle = window.adsbygoogle || []).push({});

So, if G = <a> is some infinite cyclic group, then the mapping f : Z --> G with the formula f(k) = a^k is clearly an epimorphism. To prove that it's an isomorphism (i.e. that Z is isomorphic to G), one needs to show that f is a monomorphism, too. That's where I'm stuck. I know that f is injective iff Ker(f) = {0}, so I somehow need to show that Ker(f) = {0}.

So, by definition, Ker(f) = {k in Z | a^k = e}. Suppose Ker(f) is non-trivial. Ker(f) is a subgroup of the integers Z, and hence it is cyclic, infinite and generated with m, where m is the least positive integer in it, so Ker(f) = <m>.

Thanks in advance.

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# Infinite cyclic groups isomorphic to Z

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