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Two Finite Abelian Group Problems

  1. Jun 17, 2008 #1
    The problem statement, all variables and given/known data
    1. Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.

    2. Suppose that G is a finite Abelian group. Prove that G has order pn, where p is prime, if and only if the order of every element of G is a power of p.

    The attempt at a solution
    1. By the Fundamental Theorem of Finite Abelian Groups, G must equal [itex]Z_{n_1} \oplus \cdots \oplus Z_{n_m}[/itex] where the n's are power's of primes. I also know that [itex]Z_{n_1} \oplus \cdots \oplus Z_{n_m}[/itex] is cyclic if the n's are relatively prime. I don't see how subgroups come into play here.

    2. Since the order of every element in G must divide |G|, the order of every element must be some power of p where the power divides n. That takes care of one direction. In the other direction, if every element of G is a power of p, since these powers must divide |G|, then |G| = pn where n is a multiple of all the powers of the orders of the elements of G right? Is this enough to prove the statement?
     
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  3. Jun 17, 2008 #2

    matt grime

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    Hints:

    1) Does take Z_2 + Z_2, this has 4 elements. It doesn't have an element of order 4. Generalize.

    2) No, it doesn't suffice. All you've shown, in the second half, is that some power of p divides the order of G. There is a fact about groups: if p is a prime divisor of |G| then there is an element of order p.
     
  4. Jun 17, 2008 #3
    Is the generalization this: If p and q are not relatively prime, then Z_p + Z_q does not have an element of order pq. I'm not sure how this helps.

    My book calls that fact Cauchy's theorem for Abelian groups. How does this help in demonstrating that the order of G is a power of p?
     
  5. Jun 17, 2008 #4

    matt grime

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    Sorry misread 1. Will try to think of a decent hint.

    If |G| is not a power of p then there is another ????? dividing |G| other than p. What word does ????? represent?

    Better hint for 1: OK if NOT cyclic then Z_r and Z_s exist as given (you used p and q, but they are usually reserved for primes and prime powers of p) where some prime p divides r and s. Now what do you know about the subgroups of Z_r and Z_s? Can you find two that are isomorphic and distinct?
     
    Last edited: Jun 17, 2008
  6. Jun 17, 2008 #5
    The word is prime. Hmm...so if another prime q divides |G|, does that mean there must be an element of order q in G contradicting the fact that all the elements of G have orders which are powers of p? This seems unlikely but it's the only thing I can think of at the moment.

    In general, any subgroup of Z_n is cyclic and the order of the subgroup must divide n (by Langrange's Theorem). Since p divides r and s, by Cauchy's theorem, there are elements of order p in both Z_r and Z_s, say a and b (resp.) and since <a> and <b> are isomorphic to Z_p, they must be isomorphic to each other. If a = b, then...I can't think of anything contradictory here.
     
  7. Jun 17, 2008 #6

    matt grime

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    At this point it becomes moot as to what is help, and what is giving you the answer. Let me write out what you yourself have admitted to knowing,

    In (2) you now know that if |G| is not p^n for prime p, then there is another prime call it b such that b divides |G|. But the hypothesis on G means that there is an element of order b but all elemnts have order a power of p....


    In (1), if G is not cyclic, then it has a subgroup of the form Z_r + Z_s where p a prime divides both r and s. Now, you admit this means that you know Z_r has subgroup of order p, and so does Z_s. So G has two (unequal) subgroups of order p.

    Is the real problem that you don't understand that

    A =>B

    is the same as

    not B => not A?

    I ask only because what you've written is sufficient to prove all the things you're asking about, and you don't seem to realize.
     
  8. Jun 17, 2008 #7
    Sorry for being dense. I know what the contrapositive is. I just don't understand what the flow of logic is.

    For (2), all that was left to prove was: if the order of all the elements of G are powers of p, then the order of G is a power of p. If I understand correctly, the argument is one by contradiction: Given the antecedent and the negation of the consequent, then there is a another prime b that divides |G|. By Cauchy's theorem, then G has an element of order b but this is a contradiction. Right?

    For (1), by the Fundamental Theorem of Finite Abelian Groups, G must equal [itex]Z_{n_1} \oplus \cdots \oplus Z_{n_m}[/itex] where the n's are power's of primes. G is cyclic if the n's are relatively prime so suppose G isn't cyclic, i.e. for some i ≠ j, r = ni and s = nj are not relatively prime. Without loss of generality, suppose i = 1 and j = 2. Now consider the subgroup {(a, b, 0, ..., 0) : a in Zr and b in Zs}. Since r and s are not relatively prime, there's a prime p that divides r and s and by Cauchy's theorem, Zr and Zs have an element of order p and thus a subgroup of order p. However, it remains to show that these two subgroups are unequal and that p is a divisor of |G|. Otherwise, there's no contradiction.
     
  9. Jun 17, 2008 #8

    matt grime

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    Sorry, but now I give up: you've shown that if the conclusion is false, then there is an element of order not a power of p, for 1. (Order of G not a power of p implies not all elements have order a power of p, is the contrapositive of what statement?)

    And in 2 you've found two distinct subgroups of order p (different p from the previous paragraph) if the group is not cyclic. (To labour the point: G not cyclic implies what?)

    There is nothing else I can do - it is up to you to understand what you've done.
     
    Last edited: Jun 17, 2008
  10. Jun 17, 2008 #9
    Yes, and this is a contradiction so the statement is in fact true. No need for the contrapositive. (By the way, this is for 2.)

    No I haven't found two distinct subgroups of order p. That they're distinct remains to be shown. That p divides |G| also remains to be shown. (By the way, this is for 1.)
     
  11. Jun 17, 2008 #10

    matt grime

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    You *have* found two distinct subgroups of order p, for 1 - the intersection of the two subgroups is the identity element. You can't get more distinct than that, for subgroups. That p divides |G| is automatic since by assumption Z_r +Z_s is a subgroup of G.

    The definition, in the above usage, of Z_r + Z_s (for groups) is that they only intersect in the identity.
     
    Last edited: Jun 17, 2008
  12. Jun 17, 2008 #11
    I don't understand how the intersection of them is just the identity. What I'm looking for is a proof by contradiction: Suppose the two subgroups are the same. Yada yada yada, contradiction. Ergo, the subgroups are distinct.

    I understand know why p divides |G|: If p divides the order of a subgroup H of G, then for some integer n > 0, np = |H| and since |H| divides |G| (by Lagrange's Theorem), there is some integer m > 0 such that |G| = m|H| so |G| = mnp. Thus p divides |G|.
     
  13. Jun 17, 2008 #12

    matt grime

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    It's the *definition* of the direct sum: additively written the intersection of Z_r and Z_s is 0. Both of these subgroups Z_r and Z_s contain a subgroup isomorphic to Z_p (p=/=1), and by *definition* these two subgroups of G of order p can only intersect in 0 (additively written), so they must be distinct.
     
  14. Jun 17, 2008 #13
    I didn't know that. My book doesn't define direct sum, only direct products using [itex]\oplus[/itex] as the product symbol. I haven't read anything about the intersection of the terms in the direct sum being trivial.

    How is it even possible for the intersection to be trivial? The intersection of Zr and Zs share at least 1 besides 0 right?
     
  15. Jun 17, 2008 #14

    Dick

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    You are driving matt crazy. Direct sum and direct product mean the same thing in this context. The elements of Zs+Zr have the form (s,r) where s is in Zs and r is in Zr. The subgroup Zs of that has elements of the form (s,0). The subgroup Zr has elements of the form (0,r). What do you mean they have 1 in common? They only have (0,0) is common.
     
  16. Jun 18, 2008 #15
    You're thinking of them as tuples. I was thinking Zr = {0, 1, ..., r - 1} and Zs = {0, 1, ..., s - 1}. I think the notation is hindering my understanding.
     
  17. Jun 18, 2008 #16

    matt grime

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    We were thinking of them as subgroups of G, and I used 0 for the additive identity in G. I wasn't think of them as 2-tuples at all.

    I think you should not think of "Z_r = {0,1,..,r-1}". Z_r is just a cyclic group of r elements. The residues mod r that you've written above are such an example. As is the rotation group of the regular r-gon in the plane. Clearly I can identify these groups (i.e. write down an isomorphism).

    So G has a subgroup *isomorphic* to Z_r x Z_s (x for direct product) which is the same as Z_r + Z_s (direct sum).
     
    Last edited: Jun 18, 2008
  18. Jun 18, 2008 #17
    I understand now. Thanks a lot, esp. for putting up with my laggardness.
     
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