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

  • Thread starter Ryker
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  • #26
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... looking at the decomposition [itex]G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}[/itex] yields that there is exactly one subgroup of G of order 630 corresponding to each subgroup of [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex] of order 5 ...
Is that because there is only one subgroup of order 126 in [itex]\mathbb{Z}_{252}[/itex]? How would we know that, though? I mean, I guess both the subgroup of order 126 (which would be cyclic) and [itex]\mathbb{Z}_{252}[/itex] have the same number of elements of order 126, so you could not have two such subgroups of that order, but how do you show a group has a certain number of elements of a particular order? With all the other homework to do, I'm just tuning in and out of this, so I'm always really puzzled when I get back to it :smile:
 
  • #27
Deveno
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here is a basic fact about cyclic groups you need to know:

if k|n, then [itex]\mathbb{Z}_n[/itex] has a UNIQUE subgroup of order k.

clearly if k|n, then if d = n/k, <d> has order k. so we know at least one such subgroup exists.

suppose <m> is another. since the order of m is k, we must have km = 0 (mod n).

so km is a multiple of n: km = tn = tkd, so m = td, hence <m> is a subgroup of <d>. but <m> has the same order as <d>, so <m> = <d>.
 
  • #28
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Hmm, alright, so I guess I'll take the following approach. I'll use Theorem.'s advice on showing a subgroup of order 630 must be cyclic, and then I'll find the number of elements of order using Deveno's approach (or at least a modified version). However, I think you missed one possibility, and this is 630 x 5, that is, combining an element of order 630 from [itex]\mathbb{Z}_{1260}[/itex] with an element of order 5 from [itex]\mathbb{Z}_{5}[/itex]. I think I messed up a bit replying to your post on the first page, but I did this all again, and to me it seems this would result in total in 6 x 4 x 36 elements of order 630 in total. Since the subgroups are cyclic, they all contain the same number of elements of a particular order, and since [itex]\mathbb{Z}_{630}[/itex] itself contains 4 x 36 such elements, then every subgroup of order 630 must also contain that amount of those elements, as it is isomorphic to [itex]\mathbb{Z}_{630}[/itex]. Thus, since the total amount of those elements is 6 times that, there are 6 subgroups of order 630, i.e. of index 10.

Do you see anything wrong with my approach? If yes, please say so, but otherwise I think I'm just going to go with it. Also, Deveno, Dick and others, what do you think of this problem? Is it not supposed to be as hard as I seemed to think it is? I guess it always depends on what you cover in class before getting it as a homework problem, but coming up with all of that (i.e. showing it must be cyclic, and then finding how many such subgroups there are) doesn't really seem trivial and easy to me :smile:
 
  • #29
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You seem to have the right idea about the whole question now, this is a very similar approach to what I have taken. Once you have gotten past the part where we conclude that such a subgroup must be cyclic, it is not so hard to show there are only 6 such possibilities. Furthermore, there seems to be many ways of doing this, you can compare the decomposition of such a subgroup with the decomposition of G, or you can write the decomposition of G in a different way, which is what I have done. This allowed me to pinpoint what conditions needed to be satisfied in order for a subgroup to have order 630. The problem basically reduces to the fact that we have shown there are 6 distinct subgroups of order 5 in [itex]\mathbb{Z}_5\times \mathbb{Z}_5[/itex]
 

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