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## A cyclic group proof

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?
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 Recognitions: Homework Help Use the fact that the groups infinite.

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 Quote by StatusX Use the fact that the groups infinite.
OK, but I'm not really sure how. The book says that, since Ker f is a subgroup of Z, and if it's non trivial, then we have Ker f = <m>, where m is the least positive integer such that a^m = e. I understand that, but I don't see how it proves anything.

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## A cyclic group proof

The group is cyclic, so it has a generator, x, so every element is equal to x^n for some integer n. x has infinite order, since the group is infinite, hence x^n=x^m if and only if n=m. Hence the group has an obvious isomorphism with Z.
 Blog Entries: 1 Recognitions: Homework Help Suppose a^k=e, k=/= 0. Then a^k+1 = a, and you just discovered your group is finite

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 Quote by matt grime hence x^n=x^m if and only if n=m
I'm probably missing something, but where does this fact about injectivity come from?

I know the isomorphism is obvious, but I don't understand some formal details.
 Recognitions: Homework Help Science Advisor you have proved every cyclic groupis isomorphic to a quotient of Z. the only quotient that is infinite is the quotienht by {0}.

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 Quote by radou I'm probably missing something, but where does this fact about injectivity come from? I know the isomorphism is obvious, but I don't understand some formal details.

Suppose not, then..... (this is something you must have proved on the first exercise sheet in the course)
 Recognitions: Homework Help Okay, so suppose there exist integers p and q, p=/=q, such that a^p = a^q. Then we have a^(p-q) = e, so p - q belongs to Ker f = , where m is the least positive integer such that a^m = e. Again, I'm not sure how to proceed.
 Recognitions: Homework Help Science Advisor No, you're doing this the wrong way. I said x (which you relabelled a) was a generator of this notional infinite cyclic group. Obivously x^m is never e for a generator of an infinite cyclic group unless m=0. Look, the infinite cyclic group is {x^n : n in Z}, so the isomorphism sends x^n to n. It is trivially an isomorphism, the question really has no content: it is just taking logs base x.

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 Quote by matt grime No, you're doing this the wrong way. I said x (which you relabelled a) was a generator of this notional infinite cyclic group. Obivously x^m is never e for a generator of an infinite cyclic group unless m=0. Look, the infinite cyclic group is {x^n : n in Z}, so the isomorphism sends x^n to n. It is trivially an isomorphism, the question really has no content: it is just taking logs base x.
OK, I understand now. I shouldn't have wasted so much time on this.
 Recognitions: Homework Help Btw, I should have realized earlier how trivial it can be proved. Assume Ker f =/= {0}. Ker f = , where m is the least positive integer such that a^m = e. Hence, a^(m+1) = a^m*a = e*a = a , a^(m+2) = a^(m+1)*a = a^2 , ... , a^(2m) = a^m*a^m = e, etc. Therefore we have a contradiction with the fact that G is an infinite cyclic group, so Ker f = {0}, which makes f a monomorphism, and hence an isomorphism.
 Recognitions: Homework Help For the first direction, have you proved every subgroup of a cyclic group is cyclic? This is the main part of the proof, but you seemed to have skipped this and focused on the more trivial details. Once you get this, you can just use the fact you showed above that all infinite cyclic groups are isomorphic to Z, and so to each other. For the other direction, find an x in G that isn't a generator, and use the hypothesis that is isomorphic to G.

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 Quote by StatusX For the first direction, have you proved every subgroup of a cyclic group is cyclic? This is the main part of the proof, but you seemed to have skipped this and focused on the more trivial details.
It is a proved theorem in my book, so I used it.

 Quote by StatusX For the other direction, find an x in G that isn't a generator, and use the hypothesis that is isomorphic to G.
OK, so let the infinite group G be isomorphic to each of its proper subgroups. Let H = <a> be a nontrivial cyclic subgroup of G. But, how do I know a isn't a generator of G? (I can only assume, but that's not enough, I suppose.)
 Recognitions: Homework Help Let H be a non-trivial proper subgroup of G, and pick an element from H.
 Recognitions: Homework Help Let a e H, where H < G. Since G is isomorphic to H, for every a^n where n is an integer, there exists a unique element from G mapped to that element, and since G is infinite H = is an infinite cyclic subgroup, and hence isomorphic to Z, which makes G isomorphic to Z, so G is an infinite cyclic group. (I feel there is something wrong with my logic somewhere.)