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Groups that cannot be the direct product of subgroups

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to show that neither ##Z_{p^n}## nor ##\mathbb{Z}## can be written as any family of its proper subgroups.

    2. Relevant equations


    3. The attempt at a solution

    First, I believe this solution (http://www.auburn.edu/~huanghu/math7310/7310-hw2-answer.pdf see problem 6) is falls short of what it purports to prove, since it only proves that ##Z_{p^n}## and ##\mathbb{Z}## cannot be the direct product of two of its subgroups. Am I right? If so, here are my proofs.

    First we deal with ##Z_{p^n}##. By way of contradiction, suppose that it is the direct product of subgroups ##H_1 \times ... \times H_k##, each of which must be cyclic since ##Z_{p^n}## is cyclic. Then the order of this product is ##p^n##. Moreover, ##|H_i|## divides the order of the product, and therefore ##|H_i| = p^{n_i}## for some ##n_i =1,...,n##. But this means that the orders of the subgroups are not pairwise relatively prime and therefore cannot be cyclic, which is a contradiction.

    Here's the only problem I can identify with the above proof. Although I have already proven that a product of cyclic groups is cyclic iff their orders are pairwise relatively prime, this fact hasn't yet appeared at the point of this problem in the book, so strictly speaking it isn't available. So, how would one proof it without appeal to this theorem.

    Here is a proof of the same fact concerning ##\mathbb{Z}##. By way of contradiction, suppose that it is the direct product ##\prod_{n \in I} n \mathbb{Z}##. Now, since ##\mathbb{Z}## is an infinite cyclic group, each proper subgroup is cyclic to ##\mathbb{Z}##. Hence, ##\prod_{n \in I} n \mathbb{Z} \simeq \prod_{n \in I} \mathbb{Z}##. But this product is not cyclic, for if ##(z_n)_{n \in I}## is a generator of it, and ##e_k \in \prod_{n \in I} \mathbb{Z}## has ##1## in its ##k##-th coordinate and zeros elsewhere, then ##(pz_n)_{n \in I} = e_k## implies ##z_i = 0## for every ##i \neq k##. But since this holds for every ##k \in I##, we must have ##z_i = 0## for every ##i \in I##. But zero cannot generate the product ##\prod_{n \in I} \mathbb{Z}##, which is a contradiction.

    Does this seem right?
     
  2. jcsd
  3. May 16, 2017 #2
    Drats! I just read somewhere that isomorphism is not necessarily preserved by products of groups, so my proof for ##\mathbb{Z}## may fail...Is this true? Is there a way to fix it?

    EDIT: Never mind. I products should preserve isomorphism.
     
    Last edited: May 16, 2017
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