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A cyclic group proof

  1. Apr 25, 2007 #1

    radou

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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?
     
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  3. Apr 25, 2007 #2

    StatusX

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    Use the fact that the groups infinite.
     
  4. Apr 25, 2007 #3

    radou

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    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.
     
  5. Apr 25, 2007 #4

    matt grime

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    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.
     
  6. Apr 25, 2007 #5

    Office_Shredder

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    Suppose a^k=e, k=/= 0. Then a^k+1 = a, and you just discovered your group is finite
     
  7. Apr 25, 2007 #6

    radou

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    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.
     
  8. Apr 25, 2007 #7

    mathwonk

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    you have proved every cyclic groupis isomorphic to a quotient of Z. the only quotient that is infinite is the quotienht by {0}.
     
  9. Apr 25, 2007 #8

    matt grime

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    Suppose not, then..... (this is something you must have proved on the first exercise sheet in the course)
     
  10. Apr 26, 2007 #9

    radou

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    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.
     
  11. Apr 26, 2007 #10

    matt grime

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    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.
     
  12. Apr 26, 2007 #11

    radou

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    OK, I understand now. I shouldn't have wasted so much time on this.
     
  13. Apr 28, 2007 #12

    radou

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    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.
     
  14. Apr 29, 2007 #13

    radou

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    ***

    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.
     
  15. Apr 29, 2007 #14

    StatusX

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    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.
     
  16. Apr 29, 2007 #15

    radou

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    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.)
     
  17. Apr 29, 2007 #16

    StatusX

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    Let H be a non-trivial proper subgroup of G, and pick an element from H.
     
  18. Apr 29, 2007 #17

    radou

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    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.)
     
  19. Apr 29, 2007 #18

    StatusX

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    This is true, but irrelevant.

    What does this have to do with anything?

    How do you know this?

    H is not necessarily <a>. Also, there's an unnecesary excursion here to Z and then back again.

    I think you're overthinking this, it's a one or two-liner.
     
    Last edited: Apr 29, 2007
  20. Apr 29, 2007 #19

    radou

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    Well, it is known that G is an infinite group (see my first post on this problem).

    If I pick a e H, where H < G, I can generate another subgroup <a> of H, right? Now, I don't know anything about it, only that it is isomorphic to the infinite group G. If what I wrote above doesn't make any sense, then how should I reason? Does, in general, an isomorphism between an infinite group (G in our case) and another group, necessarily make that group infinite, too?
     
  21. Apr 29, 2007 #20

    StatusX

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    I missed that line about G being infinite. I thought you wanted to show something like this:

    "Show that a group G is infinite cyclic iff it is isomorphic to all of its non-trivial proper subgroups (and such that there's at least one such subgroup, to exclude groups of prime order)."

    Then the proof is pretty much the same, except you have the extra step that if G is isomorphic to a proper subgroup, then as a set it has the same cardinality as a proper subset, and so both must be infinite.

    And yes, an isomorphism between groups means they differ essentially only in notation (although the exact nature of the isomorphism is often important, or else automorphism groups wouldn't contain any information). So one is cyclic iff the other is, one is infinite iff the other is, etc.
     
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