# Homework Help: Z, subgroups and isomorphisms

1. Nov 2, 2012

### Zondrina

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

Show that Z has infinitely many subgroups isomorphic to Z. ( Z is the integers of course ).

2. Relevant equations

A subgroup H is isomorphic to Z if $\exists \phi : H → Z$ which is bijective.

3. The attempt at a solution

So I didn't really know how to approach this one, I'm guessing I might want to try a proof by contradiction? So I would suppose that Z does not have infinitely many subgroups isomorphic to it.

Not quite sure how to start this one.

2. Nov 2, 2012

### hedipaldi

Any infinite cyclic group is isomorphic to Z.nZ (the integers divisible by n) are such.

3. Nov 2, 2012

### Zondrina

Yes I know that |Z| = ∞. I also know that 1 and -1 are cyclic generators of Z under addition and that |<1>| = |1| = ∞ as well as |<-1>| = |-1| = ∞.

So since <1> and <-1> are cyclic generators, Z = <1> = <-1>

4. Nov 2, 2012

5. Nov 2, 2012

### micromass

For every $n\in \mathbb{Z}$, you can look at the subgroup generated by n.

6. Nov 2, 2012

### Zondrina

Wait, so $\forall a \in \mathbb{Z}$ there is a subgroup generated by a. So for example ( In additive notation ) :

$<1> = \left\{{ n | n \in \mathbb{Z}} \right\}$
$<-1> = \left\{{ -n | n \in \mathbb{Z}} \right\}$
$<2> = \left\{{ 2n | n \in \mathbb{Z}} \right\}$
$<-2> = \left\{{ -2n | n \in \mathbb{Z}} \right\}$
$<3> = \left\{{ 3n | n \in \mathbb{Z}} \right\}$
$<-3> = \left\{{ -3n | n \in \mathbb{Z}} \right\}$

Each of these cyclic subgroups has order infinity, but only 1 and -1 are generators of Z.

7. Nov 2, 2012

### micromass

Yes. But you're not done yet. Some of these groups are equal to eachother. For example: <2>=<-2> and <3>=<-3> and so on.

To prove that you have infinitely many subgroups you got to prove that <n> and <m> are only isomorphic is n=m or n=-m. This needs to be proven, because for all we know, we might have <2>=<3>=<4>=...

Furthermore, you got to show that each <n> is isomorphic to $\mathbb{Z}$.

8. Nov 2, 2012

### Zondrina

Hang on a moment here... so any subgroup of Z has the form :

$<a> = \left\{{ na | n \in \mathbb{Z}} \right\}$
or
$<-a> = \left\{{ n(-a) | n \in \mathbb{Z}} \right\}$

Would that mean we could define a bijective mapping :
$\phi : n \mathbb{Z} → \mathbb{Z}$

and then showing that mapping is indeed isomorphic would be sufficient to show that Z has infinitely many subgroups?

9. Nov 2, 2012

### micromass

Yes.

10. Nov 2, 2012

### Dick

Yes, it should be pretty easy you to write down a mapping $\phi : n \mathbb{Z} → \mathbb{Z}$ and prove it's an isomorphism. But as micromass has been saying, that doesn't prove there are an infinite number of groups. For example you keep writing <a> and <-a> as though they were two different groups. They aren't. They are the same group. You need to figure out when <a> and <b> can represent the same group.

11. Nov 2, 2012

### Zondrina

The only time a and b can be the same group is when a=1 and b=-1 if I'm not mistaken?

12. Nov 2, 2012

### micromass

That's not right.

13. Nov 2, 2012

### Zondrina

Okay so, I have to show that <a> ≈ <b> if a=b or a=-b for integers a and b.

If a = b, then we are done because both cyclic groups are the same.

If a = -b, I'm not quite sure how to show this one.

14. Nov 2, 2012

### Dick

Is -a in <a>?

15. Nov 2, 2012

### Zondrina

Yes -a is in <a>. Since n can be any integer including -1.

16. Nov 2, 2012

### Dick

Or because <a> is an additive group. Fine. So you shouldn't have any trouble showing <a>=<-a>.

17. Nov 2, 2012

### Zondrina

Ohh I see. Okay, so let me sum up all these thoughts into a single post here.

"Show that Z has infinitely many subgroups isomorphic to Z"

First off, we must show that Z has infinitely many subgroups. Notice that |Z| = ∞ and for all a in Z, we have any subgroup of Z having the form :

<a> = { na | n$\in$Z } such that |<a>| = |a| = ∞.

Some of the groups generated by a are equal, namely <a> and <-a>. So let us show that <a> = <-a>.

My problem here is that a has infinite order. So i can't use the criterion for <ai> = <aj>

Would I prove this using double inclusion then? Hope this looks good so far.

18. Nov 2, 2012

### Dick

I would skip all the summary until you understand all of the parts and just cut to the problem at hand. If by double inclusion you mean you are planning to show <a> is a subset of <-a> and <-a> is a subset of <a> that will work great. Do it. Then tell me why <2>≠<3> and <3>≠<5> etc. Show <a>=<b> iff a=b or a=(-b).

19. Nov 2, 2012

### Zondrina

Okay first i'll try to show <a> = <-a>.

Case : <a> $\subseteq$ <-a>

Choose some b in <a> so that b = na for some integer n... The additive notation is throwing me off completely when I try to manipulate my equation now.

20. Nov 2, 2012

### Dick

I don't know why you are getting thrown so easily. b=na=(-n)(-a). Look, <a>=Za={...,-2a,-1a,0,a,2a,...}, <-a>=Z(-a)={...,2a,1a,0,-1a,-2a,...}. Same thing, right??

Last edited: Nov 2, 2012
21. Nov 2, 2012

### Zondrina

I've never actually had an example with additive notation before, usually our binary operation is multiplication in the book. Also I'm terrible at algebra.

So for case 1 :

b = na
b = (-n)(-a) Hence b is in <-a>

Case 2 : <a> $\supseteq$ <-a>

Suppose b is in <-a> so that b = n(-a). Then :

b = n(-a)
b = (-n)(a) hence b is in <a>

Okay so that shows that <a> = <-a>

22. Nov 2, 2012

### Dick

It sure does. Now tell me why some subgroups must NOT be equal like <2>≠<3> and <3>≠<5> etc. Try and do it in a way that will generalize to any <a> and <b> where a=b or a=(-b) doesn't hold.

23. Nov 2, 2012

### Zondrina

Some subgroups are not equal to each other because the elements contained in some might not be the same elements in others.

So for example, suppose for two integers a and b, we have two cyclic groups <a> and <b>. Then <a> = <b> ⇔ a = b or a = -b. Otherwise the two groups are not equal.

Case : a = b

If a = b, we are done. <a> = <b> is the same group.

Case : a = -b

If a = -b, then we want to show <a> = <-b>. Then I would appeal to the work just done in showing that <a> = <-a> ( Double inclusion ) by choosing c in <a> so it has the form c = na. Then :

c = na
c = n(-b) Hence c is in <-b>

Also showing the reverse inclusion we pick c in <-b> so it has the form c = n(-b). Then :

c = n(-b)
c = na Hence c is in <a>

Thus <a> = <b> ⇔ a = b or a = -b and <a> ≠ <b> otherwise.

24. Nov 2, 2012

### Dick

You are studying math, correct? So far you have shown that (a=b or a=(-b))->(<a>=<b>). You haven't shown (<a>=<b>)->(a=b or a=(-b)). Repeating the proof that (a=b or a=(-b))->(<a>=<b>) doesn't show it. A->B does not show B->A. You have GOT to know this. You might need a different idea! Don't let abstract group notation distract you. You are working over the INTEGERS. You know all about them. You are free to use any property of the integers you might know. Why is <2>≠<3>?

25. Nov 2, 2012

### Zondrina

Oh whoops, I totally forgot it was an iff. My bad. So now we start by assuming <a> = <b> and we want to show a=b or a=-b.

Case : <a> = <b> implies a = b

If <a> = <b>, then surely we have that a = b.

Case : <a> = <b> implies a = -b

If <a> = <b> we know that some c in <a> can be written as c = na, but we also know that c can be written as c = nb since <a> and <b> are equal.

So :

c = c
na = nb
na = (-n)b
na = n(-b)

and by the cancellation law in groups

a = -b.

I'm hoping that's correct now ^