# Z, subgroups and isomorphisms re-attempt

1. Nov 4, 2012

### Zondrina

1. The problem statement, all variables and given/known data

Show that Z has infinitely many subgroups isomorphic to Z.

2. Relevant equations

Two groups are isomorphic iff there exists a bijective map between them.

3. The attempt at a solution

I couldn't solve this before, I have a feeling that I have a stronger argument now so I'm hoping someone can help me.

So the first thing we have to prove is that a group of infinite order has infinitely many subgroups. Abusing some logic, this is equivalent to proving a group of finite order has finitely many subgroups.

So suppose that G is a finite group, we want to show G has finitely many subgroups. So there are two cases to consider.

Case : |g| = ∞ for some g in G.

If the order of g is ∞, then the cyclic subgroup generated by g, <g>, is isomorphic to Z. That is, <g> ≈ Z which has infinitely many subgroups. Thus G must have infinitely many subgroups contradicting the assumption.

Case : |g| = n for all g in G.

In this case, every element of g has finite order n. Since every group is a union of cyclic subgroups, G is also a union of the cyclic subgroups generated by its elements; each of which is finite.

In simple terms, since G has only a finite amount of subgroups, G must be a union of these finitely many subgroups and thus G must be finite.

Thus G is a finite group which has finitely many subgroups. Re-abusing our logic, we get that an infinite group has infinitely many subgroups. Hence Z must have infinitely many subgroups since |Z| = ∞.

I thought about that one for awhile while I was studying, I hope it makes sense.

2. Nov 4, 2012

### Dick

There's a number of things wrong with that. I'd say the worst is that proving a finite group has a finite number of subgroups DOES NOT prove an infinite group has an infinite number of subgroups. The second worst is that your proof ASSUMES Z has an infinite number of subgroups. Which is exactly what you wanted to prove to begin with. I think you best bet is to go back to the previous attempt and pick up where you left off. Why is <2> not equal to <3>?

Last edited: Nov 4, 2012
3. Nov 4, 2012

### Zondrina

So this is not good?

4. Nov 4, 2012

### Dick

Not good at all. I added to my previous comment.

5. Nov 4, 2012

### Zondrina

Because the two groups are not isomorphic to each other.

6. Nov 4, 2012

### Dick

They ARE isomorphic to each other! They are both isomorphic to Z. You were going to write down an isomorphism to prove that, remember? You want to show they aren't equal, not that they aren't isomorphic.

7. Nov 4, 2012

### Zondrina

Alright then, am I allowed to use the fact that any infinite cyclic group is isomorphic to Z?

Then I would say to consider the cases for Z :

Case : |a| = n for some a in Z

If |a| is finite, say n, then the cyclic subgroups which generate Z must also be finite. Now, Z is the union of these finite cyclic subgroups, implying that Z must be a finite set also with order n, but hence the contradiction since |Z| = ∞.

The |Z| being infinity and Z being a cyclic group tells us that at least one subgroup of Z has order of infinity.

8. Nov 4, 2012

### Dick

Stop trying to change the subject. I want to know why <2>≠<3> and you are going back to mumbo-jumbo. It's a simple question. <2>={...-4,-2,0,2,4,...}. <3>={...-6,-3,0,3,6,...}. I can see they aren't equal just by looking at them! Try and give me a mathematical sounding reason.

9. Nov 4, 2012

### Zondrina

They're different cosets of H?

I can consider 2<n> and 3<n> as the elements in a<n>?

Last edited: Nov 4, 2012
10. Nov 4, 2012

### Dick

What is H supposed to be? And no they aren't 'cosets'. I don't know why you aren't giving me a simple answer instead of pulling out some random group theory malarky. Forget group theory. Why aren't the two sets equal?? And forget about "mathematical sounding reason". Just go with "simple reason".

Last edited: Nov 4, 2012
11. Nov 4, 2012

### Zondrina

Okay so, since you're pushing this, I'll actually try to explain this.

<2>={...-6,-4,-2,0,2,4,6...}, <3>={...-9,-6,-3,0,3,6,9...}

So, <2> ≠ <3> because <2> is a subset of <3>, but the reverse inclusion does not hold. So <2> is a proper subset of <3>.

12. Nov 4, 2012

### Dick

<2> is NOT a subset of <3>, 2 is in <2>, 2 isn't in <3>. One more chance to rethink that and then I'll just tell you.

13. Nov 4, 2012

### Zondrina

This is honestly flying over my head for some reason.

Two sets aren't equal if they don't contain exactly the same elements.

14. Nov 4, 2012

### Dick

There you go! You are always trying to think of something complicated when a simple answer will do fine. If "Two sets aren't equal if they don't contain exactly the same elements" and "2 is in <2>, 2 isn't in <3>" doesn't that show <2>≠<3>? Now can you say why 2 isn't in <3>? <3> all integer multiples of 3, right?

15. Nov 4, 2012

### Zondrina

Yes, are you hinting at induction now or? I understand it for one case.

16. Nov 4, 2012

### Dick

If you want the big picture I want you to look at the set of subgroups {<1>,<2>,<3>,<4>,...} and show no two of them are equal. That would then be an infinite number of subgroups of Z, yes? No group theory gibberish, ok?

17. Nov 4, 2012

### Zondrina

So the set of all subgroups of Z, say S, is equal to the set of all cyclic subgroups of Z. In symbols : S = {<a> | a in Z }

Suppose we want to show that some $<a_i> = <a_j>$ for $a_i, a_j \in Z, i≠j$.

Notice that for $i≠j, \exists b \in <a_i> \space \wedge \space \exists c \in <a_j>$ such that b is not an element in <aj> and c is not an element in <ai>.

So it must be the case that <ai> = <aj> ⇔ i=j.

18. Nov 4, 2012

### Dick

I asked for no gibberish. And I'm disappointed. You haven't said what a_i is. How can i≠j mean much? Then you just started saying a bunch of stuff without proving it that's not even true. For example, <4> is a subset of <2>. There's not any element of <4> that's not in <2>. And <-2>=<2> but 2≠(-2).

You really haven't even given me a good reason why 2 isn't in <3> yet.