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Direct Product of Cyclic Groups

  1. Nov 6, 2014 #1
    Hello everyone,

    I was wondering if the following claim is true:

    Let ##G_1## and ##G_2## be finite cyclic groups with generators ##g_1## and ##g_2##, respectively. The group formed by the direct product ##G_1 \times G_2## is cyclic and its generator is ##(g_1,g_2)##.

    I am not certain that it is true. If I make the following stipulation

    Let ##G_1## and ##G_2## be finite cyclic groups with generators ##g_1## and ##g_2##, respectively, and the group formed by the direct product ##G_1 \times G_2## is cyclic, then it has the generator ##(g_1,g_2)##.

    this might be true. However, I would like to hear from you before I try to go prove something that is false.
     
  2. jcsd
  3. Nov 6, 2014 #2

    WWGD

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    The product is cyclic iff the orders are relatively-prime. In the direct product, you have:

    ## |G_1 \times G_2|=|G_1| \times |G_2| ## . Then, if you can find an element in the product with that
    order, you are done.
     
  4. Nov 6, 2014 #3
    More importantly, I am interested in knowing if the generator of ##G_1 \times G_2## is based off the generators of the individual groups ##G_1## and ##G_2##.

    Or is this not true in general?
     
  5. Nov 6, 2014 #4

    WWGD

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    Notice that the generator of a cyclic group is not necessarily unique.

    RE your question on relation between individual generators and generators of the product: yes, it is. Let ## g_1, g_2 ## be generators for ##G_1, G_2 ## respectively. Then there are positive integers ##m,n## with

    ##g_1^n=e_{G_1} , g_2^m = e_{G_2}##. What is then the order of ## (g_1, g_2)## ?
     
  6. Nov 7, 2014 #5
    Let's see if I understand this correctly: Let ##|G_1| = x## and ##|G_2| = y##, and let both be both cyclic with the generators alluded in the above posts. If this is so, then ##\langle g_1 \rangle = G_1 \implies |\langle g \rangle | = x##, with a similar thing being true of ##G_2##. Furthermore, ##|G_1 \times G_2 | = xy##. There is a theorem that I proved which shows that the direct product of two subgroups produces a subgroup of the direct product of the larger group(s). Therefore, ##\langle g_1 \rangle \times \langle g_2 \rangle## is a subgroup, and its order is ##| \langle g_1 \rangle \times \langle g_2 \rangle | = xy##. Thus, the element ##(g_1,g_2)## has an order of ##xy##, and it must generate ##G_1 \times G_2##.

    Does this sound right?
     
  7. Nov 9, 2014 #6
    The order need not be ##xy##. Take ##Z_2## which is cyclic with generator ##1##. Then in the group ##Z_2 \times Z_2##, the element ##(1,1)## has order two since ##(1, 1) \neq (0,0)##, and ##(1, 1) + (1, 1) = (0, 0)##, yet ##Z_2 \times Z_2## has order ##4##, so ##(1,1)## does not generate ##Z_2 \times Z_2##. Note ##Z_2 \times Z_2## is isomorphic to the Klein 4-group which is not cyclic. I think perhaps a look at the previous post is a good idea as it really says it all. What is the order of ##(g_1, g_2)##? It is worth looking at. LaTeX fixed, thanks Greg:)
     
    Last edited: Nov 10, 2014
  8. Nov 10, 2014 #7
  9. Nov 10, 2014 #8

    WWGD

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    No, the order of the product equals the product of the orders only if the orders are relatively-prime to each other.
    Otehrwise, you use the LCM.
     
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