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I came across a theorem the proof of which I don't quite understand.
The theorem states that every infinite cyclic group is isomorphic to the additive group Z.
So, the mapping f : Z --> G given with k |--> a^k, where G = <a> is a cyclic group, is an epimorphism, which is quite obvious. Now, in order to show that it's an isomorphism, one should show that it's a monomorphism, i.e. that the kernel is trivial. So, we have Ker f = {k from Z such that a^k = e}. Now, Ker f contains 0, but how do I show that it only contains 0?
The theorem states that every infinite cyclic group is isomorphic to the additive group Z.
So, the mapping f : Z --> G given with k |--> a^k, where G = <a> is a cyclic group, is an epimorphism, which is quite obvious. Now, in order to show that it's an isomorphism, one should show that it's a monomorphism, i.e. that the kernel is trivial. So, we have Ker f = {k from Z such that a^k = e}. Now, Ker f contains 0, but how do I show that it only contains 0?