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What does Z6 x Z3 look like?

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Find all cyclic subgroups of Z6 x Z3.

    2. Relevant equations



    3. The attempt at a solution
    I understand how to find a cyclic subgroup of a simpler group such as Z4, but having trouble understanding what subgroups look like in a direct product of integer spaces, let alone cyclic subgroups. Also, having trouble understanding what makes a direct product subgroup cyclic. Is it cyclic when (a1, a2)^n = (e1, e2)? Please help!!
     
  2. jcsd
  3. Jun 18, 2012 #2
    A group is cyclic if it is generated by a single element. So find subgroup generated by each element (some elements might generate the same subgroup).
     
  4. Jun 18, 2012 #3
    Direct product of two groups [itex]G[/itex] and [itex]H[/itex], is the group [itex]G\times H = \{ (g,h) | g \in G, h \in H \}[/itex].

    If [itex]*[/itex] is the operation of G and H, [itex](g,h)*(g_1,h_1) = (g*g_1,h*h_1)[/itex]. Similarly the inverse [itex](g,h)^{-1} = (g^{-1},h^{-1})[/itex].

    Now can you find any element [itex](g,h) \in \mathbb{Z}_6\times \mathbb{Z}_3[/itex] such that each element in [itex]\mathbb{Z}_6\times \mathbb{Z}_3[/itex] can be represented in the form [itex](g^n,h^n)[/itex]?
     
    Last edited: Jun 18, 2012
  5. Jun 18, 2012 #4
    Ok so I think I found all the subgroups generated by each element of Z6 x Z3. It looks like (0,0), (0,1), (1,0), (1,1), (3,0), (3,1), (4,0), (4,1), (5,0), (5,1) each generate only themselves, while each of the other elements of Z6 x Z3 only generate 2 elements. So, it appears that there are no cyclic subgroups, unless i computed the subgroups incorrectly.
     
  6. Jun 19, 2012 #5
    If your notation includes zero, you must be using the additive notation. When you say (0,1) only generates itself, do you mean it generates a subgroup of one element? There's two things wrong with that.

    One is, every group, and subgroup, must contain the identity by definition. Under your implicit choice of addition as the operation, the identity is (0,0).

    The second problem is, 5 is relatively prime to 6, so for instance 5+...+5=5*5=25=1+24=1 mod 6. So (5,0) generates the same group (1,0) does.

    Perhaps you do not know what it means for an element to generate a subgroup. Under addition, all multiples, under multiplication, all powers (a sometimes tricky distinction).

    Please be more careful, and propose different groups.
     
  7. Jun 19, 2012 #6
    OK thanks. I got a little confused because the other post said I should generate by using (a^n, b^n). I think I have my subgroups correct now.
     
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