Does Z Have Infinitely Many Isomorphic Subgroups?

[STEMucator]

Homework Statement

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

Homework Equations

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

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.

 

[hedipaldi]

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

 

[STEMucator]

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>

 

[hedipaldi]

Yes but tou asked about subgroups of Z.

 

[micromass]

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

 

[STEMucator]

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

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

&lt;1&gt; = \left\{{ n | n \in \mathbb{Z}} \right\}
&lt;-1&gt; = \left\{{ -n | n \in \mathbb{Z}} \right\}
&lt;2&gt; = \left\{{ 2n | n \in \mathbb{Z}} \right\}
&lt;-2&gt; = \left\{{ -2n | n \in \mathbb{Z}} \right\}
&lt;3&gt; = \left\{{ 3n | n \in \mathbb{Z}} \right\}
&lt;-3&gt; = \left\{{ -3n | n \in \mathbb{Z}} \right\}

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

 

[micromass]

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

&lt;1&gt; = \left\{{ n | n \in \mathbb{Z}} \right\}
&lt;-1&gt; = \left\{{ -n | n \in \mathbb{Z}} \right\}
&lt;2&gt; = \left\{{ 2n | n \in \mathbb{Z}} \right\}
&lt;-2&gt; = \left\{{ -2n | n \in \mathbb{Z}} \right\}
&lt;3&gt; = \left\{{ 3n | n \in \mathbb{Z}} \right\}
&lt;-3&gt; = \left\{{ -3n | n \in \mathbb{Z}} \right\}

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

Yes. But you're not done yet. Some of these groups are equal to each other. 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}.

 

[STEMucator]

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

&lt;a&gt; = \left\{{ na | n \in \mathbb{Z}} \right\}
or
&lt;-a&gt; = \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?

 

[micromass]

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

&lt;a&gt; = \left\{{ na | n \in \mathbb{Z}} \right\}
or
&lt;-a&gt; = \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?

Yes.

 

[Dick]

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

&lt;a&gt; = \left\{{ na | n \in \mathbb{Z}} \right\}
or
&lt;-a&gt; = \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?

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.

 

[STEMucator]

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.

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

 

[micromass]

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

That's not right.

 

[STEMucator]

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.

 

[Dick]

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.

Is -a in <a>?

 

[STEMucator]

Is -a in <a>?

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

 

[Dick]

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

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

 

[STEMucator]

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

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\inZ } 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> = <a^j>

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

 

[Dick]

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\inZ } 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> = <a^j>

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

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).

 

[STEMucator]

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).

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.

 

[Dick]

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.

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??

 

[STEMucator]

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>

 

[Dick]

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>

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.

 

[STEMucator]

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.

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.

 

[Dick]

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.

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>?

 

[STEMucator]

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 ^

 

[Dick]

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 ^

I don't mean to be offensive or demeaning but you are being completely tone deaf about the difference between a proof and a nonproof. That doesn't make any sense at all. Just for practice, give me a proof that <2>≠<3>. Just tell me why in words that are not at all similar to the 'proof' you just gave. Please?

 

[micromass]

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.

Take a=1 and b=-1. You have just "proved" that <1>=<-1> implies 1=-1...

 

[STEMucator]

I think I'm going to save this until tomorrow when my brain is capable of processing simple information.

<2> is not <3> because every element in <2> is not contained in <3> and vice versa.

 

[micromass]

I think I'm going to save this until tomorrow when my brain is capable of processing simple information.

<2> is not <3> because every element in <2> is not contained in <3> and vice versa.

6 is an element of <2> that is contained in <3>.

 

[Dick]

I think I'm going to save this until tomorrow when my brain is capable of processing simple information.

<2> is not <3> because every element in <2> is not contained in <3> and vice versa.

I'd agree with that nap time. You definitely want to give it more thought.

 

[STEMucator]

Okay, work is done, mind is fresh. I thought about this while I was working and I think contradiction was definitely the way to go initially. I'm not really seeing the method that you guys tried showing me happen.

So suppose the contrary that Z has finitely many subgroups, say n subgroups so that |Z| = n. So Z must also have a finite number of cyclic subgroups, say m cyclic subgroups where m≤n.

We know that Z is generated by the cyclic subgroups <1> and <-1> so that Z is cyclic, but the |<1>| = |<-1>| = ∞.

Since Z is finite, but it can be generated by an infinite cyclic group, our hypothesis about Z having finitely many subgroups must be false because |Z| = ∞ and thus Z has infinitely many subgroups.

Trying this new approach, it seems to make much more sense to me.

 

[micromass]

Okay, work is done, mind is fresh. I thought about this while I was working and I think contradiction was definitely the way to go initially. I'm not really seeing the method that you guys tried showing me happen.

So suppose the contrary that Z has finitely many subgroups, say n subgroups so that |Z| = n.

Are you saying that Z has n elements?? How does that follow from the assumption that Z has n subgroups?

So Z must also have a finite number of cyclic subgroups, say m cyclic subgroups where m≤n.

We know that Z is generated by the cyclic subgroups <1> and <-1> so that Z is cyclic, but the |<1>| = |<-1>| = ∞.

Since Z is finite,

Z is finite?? Why??

but it can be generated by an infinite cyclic group, our hypothesis about Z having finitely many subgroups must be false because |Z| = ∞ and thus Z has infinitely many subgroups.

Trying this new approach, it seems to make much more sense to me.

 

[STEMucator]

Z is finite because I'm assuming something false from contradiction and then showing it's false. I think you may be reading what I said wrong? Otherwise I think I'm really just going to ask my prof about this one rather than waste anymore time on it.

 

[micromass]

Z is finite because I'm assuming something false from contradiction and then showing it's false. I think you may be reading what I said wrong? Otherwise I think I'm really just going to ask my prof about this one rather than waste anymore time on it.

OK, so you used the following:

"If a group G has finitely many subgroups, then G is finite"

I think you still need to prove this statement.

 

[STEMucator]

OK, so you used the following:

"If a group G has finitely many subgroups, then G is finite"

I think you still need to prove this statement.

My only question is can I let G = Z or do I simply prove it on the basis that G is a finite group. If I start with G being a finite group, I think I may know how to do this by letting a₁,...,a_n be the elements in G and using a little union trick.

 

[micromass]

My only question is can I let G = Z or do I simply prove it on the basis that G is a finite group. If I start with G being a finite group, I think I may know how to do this by letting a₁,...,a_n be the elements in G and using a little union trick.

You need to prove that G is finite. Why do you think it is ok to start of from the assumption that G is finite then??