Does Z Have Infinitely Many Isomorphic Subgroups?

[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??