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Supose that G is a finite abelian group that does not contain a subgro

  1. Apr 20, 2014 #1
    let us assume G is not cyclic. Let a be an element of G of maximal order. Since G is not cyclic we have <a>≠G. Let b be an element in G, but not in the cyclic subgroup generated by a.

    O(a) = m and O(b) = n where O() refers tothe orders. . then how can we use this to construct a subgroup of G isomorphic to Zp×Zp?

    Help will be appreciated. Thank you
     

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  3. Apr 20, 2014 #2

    Dick

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    Can you prove that if the prime factorization of |G| is ##p_1 p_2 p_3...p_n## and all of the primes ##p_i## are different, then G is cyclic? If two are the same, then what?
     
    Last edited: Apr 20, 2014
  4. Apr 20, 2014 #3
    If |G| =p1p2....pn and all of the primes are different, then |H| will be one of those primes. Since a group of prime order is cyclic, then H will be cyclic in that case . Infact all proper subgroups of G will be cyclic.
    but i am not sure how to prove G as cyclic with this data..
     
  5. Apr 21, 2014 #4

    micromass

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    What does the fundamental theorem of Abelian groups tell you?
     
  6. Apr 21, 2014 #5
    the book which i am reading ( GAllian ) has not introduced this topic as of yet. In fact, not even normal and factor groups. i am on the chapter on external direct products.
     
  7. Apr 21, 2014 #6

    micromass

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    Where exactly in Gallian is this?
     
  8. Apr 21, 2014 #7
    Supplementary exercise for chapters 5 - 8 . Question no. 50 . Gallian 7/e contemporary guide to abstract algebra
     
  9. Apr 21, 2014 #8
    and pg. no 50
     
  10. Apr 21, 2014 #9

    micromass

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    OK. So let ##G## be finite abelian group such that ##G## does not contain a subgroup isomorphic to ##\mathbb{Z}_p\times \mathbb{Z}_p## for any prime. Consider the prime factorization ##|G|= p_1^{a_1} ... p_k^{a_k}:= p_1^{a_1}m##.

    Define ##H = \{x\in G~\vert~\text{order of}~H~\text{divides}~p_1^{a_1}\}## and ##K = \{x\in G~\vert~\text{order of}~K~\text{divides}~m\}##.

    Start by proving the following:
    1) ##H## and ##K## are subgroups of ##G##.
    2) For any ##x\in G## and integers ##s## and ##t##, we have ##x^{sp^k}\in H## and ##x^{tm}\in K##. Deduce that ##G=HK##.
    3) Show that ##H\cap K = \{e\}##.
    4) Find ##|H|## and ##|K|##.
    5) Prove that ##H## is cyclic.
    6) Prove that ##G## is the product of cyclic groups
    7) Prove the result.
     
    Last edited: Apr 21, 2014
  11. Apr 21, 2014 #10

    Dick

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    Your definition of K doesn't make much sense. I assume you mean ##K = \{x\in G~\vert~\text{order of}~x~\text{divides}~m\}##.
     
  12. Apr 21, 2014 #11

    micromass

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    Yes, sorry!
     
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