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

[Ryker]

Homework Statement

How many subgroups of index 10 are in $\mathbb{Z}_{20} \times \mathbb{Z}_{9} \times \mathbb{Z}_{35}$?

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:

- $\mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$
- $\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$

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

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?

 

[Theorem.]

Homework Statement

How many subgroups of index 10 are in $\mathbb{Z}_{20} \times \mathbb{Z}_{9} \times \mathbb{Z}_{35}$?

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:

- $\mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$
- $\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$

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

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?

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.

 

[Deveno]

here is a hint:

$$\mathbb{Z}_{20} \times \mathbb{Z}_9 \times \mathbb{Z}_{35} \cong \mathbb{Z}_5 \times \mathbb{Z}_{1260}$$ by the chinese remainder theorem.

 

[Ryker]

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.

Seems so

here is a hint:

$$\mathbb{Z}_{20} \times \mathbb{Z}_9 \times \mathbb{Z}_{35} \cong \mathbb{Z}_5 \times \mathbb{Z}_{1260}$$ by the chinese remainder theorem.

Hey Deveno, thanks for the hint. Unfortunately, even after putting some additional thought into this, I still don't know how to continue We're also given a hint to consider how many subgroups of order 5 $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$ has (it seems the answer is 6), but after writing out some other similar groups, I can't see a pattern. Any further thoughts?

 

[Deveno]

Seems so Hey Deveno, thanks for the hint. Unfortunately, even after putting some additional thought into this, I still don't know how to continue We're also given a hint to consider how many subgroups of order 5 $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$ has (it seems the answer is 6), but after writing out some other similar groups, I can't see a pattern. Any further thoughts?

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

let's look at $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$.

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 $\mathbb{Z}_{5} \times \mathbb{Z}_{1260}$, 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)?

 

[Ryker]

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

let's look at $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$.

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?

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, $\{(0,0), (0,1), (2,0), (2,1)\}$ is a subgroup of $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$, but it's not cyclic.

now in $\mathbb{Z}_{5} \times \mathbb{Z}_{1260}$, 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)?

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 $\mathbb{Z}_{5}$ 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 $\mathbb{Z}_{5}$?)

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

 

[Theorem.]

I am also still having trouble with the question. Like Ryker I am having difficulty finding the parallel between $$\mathbb{Z}_5\times \mathbb{Z}_5$$ and $$\mathbb{Z}_5\times \mathbb{Z}_{1260}$$ (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,
$$\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$$
or
$$\mathbb{Z}_{2} \times \mathbb{Z}_{9}\times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$$

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.

 

[Dick]

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?

 

[Ryker]

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?

 

[Dick]

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?

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?

 

[Ryker]

It should be 9, I think. But I still don't see where that leads us.

 

[Dick]

It should be 9, I think. But I still don't see where that leads us.

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?

 

[Ryker]

It could contain $\mathbb{Z}_{5} \times \mathbb{Z}_{7}$ and a group of order 2 in $\mathbb{Z}_{20}$, I think.

 

[Dick]

It could contain $\mathbb{Z}_{5} \times \mathbb{Z}_{7}$ and a group of order 2 in $\mathbb{Z}_{20}$, I think.

Or it could contain a subgroup of order 10 in Z_20 and a subgroup of order 7 in Z_35. Right?

 

[Ryker]

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?

 

[Dick]

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?

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.

 

[Ryker]

Hmm, but there's also only one subgroup of order 10 in $\mathbb{Z}_{20}$, and only one of order 7 in $\mathbb{Z}_{35}$, 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...

 

[Dick]

I THINK that's right. Now I'm getting unsure of my answers, it's getting late here. What do you think?

 

[Theorem.]

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 $\mathbb{Z}_5\times \mathbb{Z}_5$? One way to answer the last question is to start enumerating the (cyclic) subgroups of order 5 of $\mathbb{Z}_5\times \mathbb{Z}_5$ 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 $H_1,H_2$ are two subgroups of order 5, then either they are equal or their intersection is the identity of $\mathbb{Z}_5\times \mathbb{Z}_5$."

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 $\mathbb{Z}_5\times \mathbb{Z}_5$ (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
$G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}$.

I am hoping to use this decomposition, alongside the fact that there are exactly 6 subgroups of order 5 in $\mathbb{Z}_5\times \mathbb{Z}_5$ 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 $\mathbb{Z}_5\times \mathbb{Z}_5$ has 6 subgroups of order 5-but i could very well be mistaken!)

 

[Ryker]

I THINK that's right. Now I'm getting unsure of my answers, it's getting late here. What do you think?

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 $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$, 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.

 

[Theorem.]

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 $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$, 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.

I think his approach would also yield (at least) 6 or something close ( I haven't punched it out, just looked through details), but this is the same approach that I have listed, he just decomposed G slightly different, I chose the decomposition above just to stress $\mathbb{Z}_5\times\mathbb{Z}_5$, the only problem i have still with both what Deveno and I myself have written down is that we really haven't considered the case where H is not cyclic? I feel like we should be using the decomposition possibilities for H somewhere in here too.

 

[Dick]

I think his approach would also yield (at least) 6 or something close ( I haven't punched it out, just looked through details), but this is the same approach that I have listed, he just decomposed G slightly different, I chose the decomposition above just to stress $\mathbb{Z}_5\times\mathbb{Z}_5$, the only problem i have still with both what Deveno and I myself have written down is that we really haven't considered the case where H is not cyclic? I feel like we should be using the decomposition possibilities for H somewhere in here too.

Yes, of course, you guys are correct. There are six order 5 subgroups of Z(5)xZ(5). My argument is wrong. So it could be that H intersects Z(5)x{0} and {0}xZ(5) only in {0}x{0}. Ooops.

 

[Ryker]

I think his approach would also yield (at least) 6 or something close ( I haven't punched it out, just looked through details), but this is the same approach that I have listed, he just decomposed G slightly different, I chose the decomposition above just to stress $\mathbb{Z}_5\times\mathbb{Z}_5$, the only problem i have still with both what Deveno and I myself have written down is that we really haven't considered the case where H is not cyclic? I feel like we should be using the decomposition possibilities for H somewhere in here too.

Yeah, I also have a hunch we should be using the decomposition somehow, even though what Deveno suggests seems like a good method for finding cyclic groups (if we could follow it through). I don't know what to do with non-cyclic groups, though. I feel as if there aren't any and that this could somehow be deduced from the fact that the group we are given is isomorphic to $\mathbb{Z}_{5} \times \mathbb{Z}_{1260}$, where all elements in $\mathbb{Z}_{5}$ have order either 1 or 5, but I just can't quite get to it yet.

 

[Theorem.]

Yeah, I also have a hunch we should be using the decomposition somehow, even though what Deveno suggests seems like a good method for finding cyclic groups (if we could follow it through). I don't know what to do with non-cyclic groups, though. I feel as if there aren't any and that this could somehow be deduced from the fact that the group we are given is isomorphic to $\mathbb{Z}_{5} \times \mathbb{Z}_{1260}$, where all elements in $\mathbb{Z}_{5}$ have order either 1 or 5, but I just can't quite get to it yet.

Yes, i have talked to the professor today briefly and the only part I was really missing is showing that any subgroup of index 10 has to be cyclic. this is indeed the case since any subgroup of index ten cannot be isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}$ You can see this is true by comparing the elements of order 3 in this decomposition to those of order 3 in G. You will see that this decomposition contains more elements of order 3 than G itself does, which is impossible. Therefore all the subgroups of index 10 are isomorphic to the other decomposition, and therefore cyclic.

 

[Theorem.]

Once we have gotten to this point, the problem is not Hard. In fact, since there are 6 distinct subgroups of order 5 in $\mathbb{Z}_5\times \mathbb{Z}_5$, looking at the decomposition $G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}$ yields that there is exactly one subgroup of G of order 630 corresponding to each subgroup of $\mathbb{Z}_5\times \mathbb{Z}_5$ of order 5 and thus there are exactly 6 such subgroups of G of index 10( this is not hard to explain at this point)

 

[Ryker]

... looking at the decomposition $G\cong (\mathbb{Z}_5\times \mathbb{Z}_5)\times \mathbb{Z}_{252}$ yields that there is exactly one subgroup of G of order 630 corresponding to each subgroup of $\mathbb{Z}_5\times \mathbb{Z}_5$ of order 5 ...

Is that because there is only one subgroup of order 126 in $\mathbb{Z}_{252}$? How would we know that, though? I mean, I guess both the subgroup of order 126 (which would be cyclic) and $\mathbb{Z}_{252}$ 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

 

[Deveno]

here is a basic fact about cyclic groups you need to know:

if k|n, then $\mathbb{Z}_n$ 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>.

 

[Ryker]

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 $\mathbb{Z}_{1260}$ with an element of order 5 from $\mathbb{Z}_{5}$. 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 $\mathbb{Z}_{630}$ 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 $\mathbb{Z}_{630}$. 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

 

[Theorem.]

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 $\mathbb{Z}_5\times \mathbb{Z}_5$