A cyclic group proof

radou

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?

StatusX

Use the fact that the groups infinite.

radou

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.

matt grime

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.

Office_Shredder

Suppose a^k=e, k=/= 0. Then a^k+1 = a, and you just discovered your group is finite

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.

mathwonk

you have proved every cyclic groupis isomorphic to a quotient of Z. the only quotient that is infinite is the quotienht by {0}.

matt grime

Suppose not, then..... (this is something you must have proved on the first exercise sheet in the course)

radou

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 = <m>, where m is the least positive integer such that a^m = e.

Again, I'm not sure how to proceed.

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.

radou

OK, I understand now. I shouldn't have wasted so much time on this.

radou

Btw, I should have realized earlier how trivial it can be proved.

Assume Ker f =/= {0}. Ker f = <m>, 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.

radou

***

I don't want to start a new thread, so I'll simply post another thing I'm not sure about in this thread, hoping it will be noticed.

I have to prove that, an infinite group is cyclic iff it is isomorphic to each of its proper subgroups.

(=>) OK, let G = <a> = {a^n : n e Z} be an infinite cyclic group. Since G is cyclic, so is every of its subgroups. Let H < G, where H is a proper subgroup. Then H =<a^m> = {(a^m)^n : n e Z}, where m is the least positive integer such that a^m e H. Consider the mapping f: G --> H given with f : a^n --> (a^m)^n. It is a homomorphism, since f(a^p*a^q)=f(a^(p+q))= (a^m)^(p+q) = (a^m)^p*(a^m)^q = f(a^p)*f(a^q). Further on, it is a monomorphism, since (a^m)^p = (a^m)^q => p = q, since a^m is in G and the elements of G are all distinct. Clearly it is an epimorphism, since Im f = {(a^m)^n : n e Z} (since G is isomorphic to Z), which makes it an isomoprhism, so G is isomorphic to H.

(<=) This is the direction I'm having problems with, so I'd be grateful if someone checked my work and helped me out with a further hint/correction.

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. 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 <x> is isomorphic to G.

radou

It is a proved theorem in my book, so I used it.

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.)

StatusX

Let H be a non-trivial proper subgroup of G, and pick an element from H.

radou

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 = <a> 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.)