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How many subgroups of index 10 are in Z(20) x Z(9) x Z(35)

  1. Nov 21, 2011 #1
    1. The problem statement, all variables and given/known data
    How many subgroups of index 10 are in [itex]\mathbb{Z}_{20} \times \mathbb{Z}_{9} \times \mathbb{Z}_{35}[/itex]?

    3. The attempt at a solution
    So I figure the order of this group would be 20 x 9 x 35 = 6300, and so any subgroup of index 10 would have order 630. By the fundamental theorem of finite commutative groups, you get that such a subgroup is isomorphic to one of these two:

    - [itex]\mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/itex]
    - [itex]\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/itex]

    On the other hand, the elementary divisor decomposition of the main group is [itex]\mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}.[/itex]

    I can see some of the elements in the decompositions match, but I don't know where to go from here. First I thought of just counting the number of elements of order 630, but I'm not sure this is the right approach, and I also don't really know how to account for all of them with so many different groups in the decomposition.

    Any thoughts on this?
     
  2. jcsd
  3. Nov 21, 2011 #2
    Hi Ryker,
    i am pretty sure you are in my group theory class. Anyways, I just started the problem and so far have taken the same approach as you, I will update when progress is made.
     
  4. Nov 21, 2011 #3

    Deveno

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    here is a hint:

    [tex]\mathbb{Z}_{20} \times \mathbb{Z}_9 \times \mathbb{Z}_{35} \cong \mathbb{Z}_5 \times \mathbb{Z}_{1260}[/tex] by the chinese remainder theorem.
     
  5. Nov 21, 2011 #4
    Seems so :smile:
    Hey Deveno, thanks for the hint. Unfortunately, even after putting some additional thought into this, I still don't know how to continue :frown: We're also given a hint to consider how many subgroups of order 5 [itex]\mathbb{Z}_{5} \times \mathbb{Z}_{5}[/itex] has (it seems the answer is 6), but after writing out some other similar groups, I can't see a pattern. Any further thoughts?
     
  6. Nov 21, 2011 #5

    Deveno

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    the hint you were given, and the hint i gave are "convergent".

    let's look at [itex]\mathbb{Z}_{5} \times \mathbb{Z}_{5}[/itex].

    for a subgroup of this to be of order 5, it has to be generated by an element of order 5. there are 24 elements of order 5, and we take out "4 at a time" since distinct subgroups of order 5 only intersect in the identity (0,0). can you explicitly list 6 generators?

    now in [itex]\mathbb{Z}_{5} \times \mathbb{Z}_{1260}[/itex], we have a similar situation.

    we can only get a subgroup of order 630, by consider one of two situations:

    a) 1x630-how many of these are there?
    b) 5x126-how many of these are there (hint: it's more than 1)?
     
  7. Nov 21, 2011 #6
    Yeah, so it's (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), and (0, 1). So in this specific case I see why this is so, but why would a group of order 630 have to be cyclic, as well, in that larger group? For example, [itex]\{(0,0), (0,1), (2,0), (2,1)\}[/itex] is a subgroup of [itex]\mathbb{Z}_{4} \times \mathbb{Z}_{2}[/itex], but it's not cyclic.

    a) Is it 144 elements of order 630, because that's how many invertible elements there are in the group of all invertible elements mod 630? (edit 2: Actually, I guess it should be 144 x 4 to take into account that the coordinate from [itex]\mathbb{Z}_{5}[/itex] can be either 0, 1, 2 or 3 (but not 4), right?)
    b) By the same reasoning, would it be 36, where in Z(5) we can only take 1, because 2 and 3 are divisible by 126, whereas lcm(4, 126) = 252?

    Also, why are these the only two options? Why not also 2x315 (edit 2: Oh, I guess it's because there is no element of order 2 in [itex]\mathbb{Z}_{5}[/itex]?)

    edit: Oh, and how would we now know how many "at a time" to take out for 630? I mean, in the case above for 5 it was clear, because the elements generated by the generators covered the whole set, but why would these now do the same? I don't know, it seems that when I get confused, I get confused hard, so I'm sorry if I seem a bit thick in regards to this :smile:
     
    Last edited: Nov 21, 2011
  8. Nov 21, 2011 #7
    I am also still having trouble with the question. Like Ryker I am having difficulty finding the parallel between [tex]\mathbb{Z}_5\times \mathbb{Z}_5[/tex] and [tex]\mathbb{Z}_5\times \mathbb{Z}_{1260}[/tex] (specifically,
    why must a supgroup of order 630 be cyclic?)

    Yet, The biggest difficulty I am having, is that I am not sure how I will relate back to the possible elementary divisor decompositions of H. We know that any subgroup of index 10 must be isomorphic to,
    [tex]\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/tex]
    or
    [tex]\mathbb{Z}_{2} \times \mathbb{Z}_{9}\times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/tex]

    but i am not sure how this will actually help us identify the number of subgroups of index 10? clearly, any such subgroup must be isomorphic to one of these groups, but I am having difficulty seeing where it ties in.
    If anyone could help me understand this (perhaps I have missed something obvious?), I would appreciate any thoughts.
     
  9. Nov 21, 2011 #8

    Dick

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    Z_20x{0}x{0}, {0}xZ_9x{0} and {0}x{0}xZ_35 are three subgroups of the whole group that generate the group. If H is a subgroup of order 630 and you can figure out the number of ways H can intersect those three subgroups, then you'll have it. Won't you?
     
  10. Nov 21, 2011 #9
    Hmm, so would this then be 70 (i.e. 2 x 5 x 7) + 18 (i.e. 2 x 9) from the first H decomposition plus 18 (i.e. 2 x 3 x 3) from the second decomposition? Or how exactly would you go about doing this?
     
  11. Nov 21, 2011 #10

    Dick

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    The product of the order of those three intersections must be 630, right? What are the possible orders of each intersection? For example what must be the order of the intersection with the Z_9 subgroup?
     
  12. Nov 21, 2011 #11
    It should be 9, I think. But I still don't see where that leads us.
     
  13. Nov 21, 2011 #12

    Dick

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    Ok, so the only subgroup of order 9 in Z_9 is Z_9 itself. So H contains {0}xZ_9x{0}. Now what about the Z_20 and Z_35 parts?
     
  14. Nov 21, 2011 #13
    It could contain [itex]\mathbb{Z}_{5} \times \mathbb{Z}_{7}[/itex] and a group of order 2 in [itex]\mathbb{Z}_{20}[/itex], I think.
     
  15. Nov 21, 2011 #14

    Dick

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    Or it could contain a subgroup of order 10 in Z_20 and a subgroup of order 7 in Z_35. Right?
     
  16. Nov 21, 2011 #15
    Yeah, sorry, forgot to put that in there, as well. So does that then mean that in both cases there are 10 possible combinations, i.e. 1 x 10 + 2 x 5, yielding 20 in total?
     
  17. Nov 21, 2011 #16

    Dick

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    Can you explain how you counted those? I think there is only one subgroup of order 2 in Z_20, for example. And one subgroup of order 35 in Z_35. But yes, now you just have to count the subgroup possibilities.
     
  18. Nov 21, 2011 #17
    Hmm, but there's also only one subgroup of order 10 in [itex]\mathbb{Z}_{20}[/itex], and only one of order 7 in [itex]\mathbb{Z}_{35}[/itex], isn't there? So that would only yield 2 subgroups of order 630? I don't know, you can probably tell I'm still not getting it, as I'm really unsure in all of my answers...
     
  19. Nov 21, 2011 #18

    Dick

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    I THINK that's right. Now I'm getting unsure of my answers, it's getting late here. What do you think?
     
  20. Nov 22, 2011 #19
    I can kind of see where you are going with this, but I think the reason we are uncertain of this method (or not necessarily of the method, but of the approach), is that the following hint was given:
    "Start by finding the elementary divisor decomposition of G and of such a subgroup of index 10 (done this). What is the order of a subgroup of index 10 (we used lagrange theorem to show it was 630)? How many subgroups of order 5 are there in [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex]? One way to answer the last question is to start enumerating the (cyclic) subgroups of order 5 of [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] until they have been exhausted, but this is not the most efficient way to proceed. To count such subgroups in a more efficient way, use the observation that if [itex]H_1,H_2[/itex] are two subgroups of order 5, then either they are equal or their intersection is the identity of [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex]."

    Accordingly, i have taken the approach that more or less follows the instructors suggestion. I have argued that there are 6 such subgroups of order 5 in [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] (all of which are cyclic) and now want to use this alongside the elementary divisor decompositions to show how many subgroups of G of order 630 there are. For this, I have noted that
    [itex]G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}[/itex].

    I am hoping to use this decomposition, alongside the fact that there are exactly 6 subgroups of order 5 in [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] and that a subgroup of order 630 is isomorphic to one of the two groups we previously listed to obtain the result. This is where I am stuck. My first concern, is does H (subgroup of order 630) have to be cyclic ( i don't see why it would)? If so, then my problem is much simpler, but it seems that I would be able to find more than 2 such cyclic subgroups alone (in fact, i see 6 already from the fact that [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] has 6 subgroups of order 5-but i could very well be mistaken!)
     
    Last edited: Nov 22, 2011
  21. Nov 22, 2011 #20
    It's late here, as well, and I guess while I can't find what exactly would be wrong with what you're suggesting, I feel as if it's understating the number of groups. Namely, what if you took your approach to the group [itex]\mathbb{Z}_{5} \times \mathbb{Z}_{5}[/itex], and try and determine the number of subgroups of order 5? Wouldn't it also yield only 2, instead of 6, groups?

    At this point, I'm also really curious what Deveno's approach, taken further, is. I probably answered some of his questions wrong, but that approach, too, would suggest more groups, I'm just not sure how many.
     
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