Looking for a Group Theorist (Math)

[Yiab]

I was talking with my supervisor a while ago and I came up with the following example of a group and subgroup meeting some particular criteria (which I've now forgotten).

[latex]Let $A = \mathbb{Z}^{\omega}$ be the countably infinite direct product of the $1$-generated free abelian group.\\
Let $B = \{ (a_n)_{n<\omega} : \{ n : a_n\neq 0 \} \mbox{ is finite} \}$ be the subgroup of $A$ isomorphic to the countably infinite direct sum of the $1$-generated free abelian group.\\
Question: Is $A/B$ isomorphic to $A$?[/latex]

It seems clear that these two should not be isomorphic, certainly B is not isomorphic to A, but I can't figure out how to prove that A/B and A are not isomorphic. Not being a group theorist and being more familiar with finite groups than infinite ones, I may not have all the tools necessary to do so.

Is there anyone out there who can help?

 

[andyandy]

I was talking with my supervisor a while ago and I came up with the following example of a group and subgroup meeting some particular criteria (which I've now forgotten).

[latex]Let $A = \mathbb{Z}^{\omega}$ be the countably infinite direct product of the $1$-generated free abelian group.\\
Let $B = \{ (a_n)_{n<\omega} : \{ n : a_n\neq 0 \} \mbox{ is finite} \}$ be the subgroup of $A$ isomorphic to the countably infinite direct sum of the $1$-generated free abelian group.\\
Question: Is $A/B$ isomorphic to $A$?[/latex]

It seems clear that these two should not be isomorphic, certainly B is not isomorphic to A, but I can't figure out how to prove that A/B and A are not isomorphic. Not being a group theorist and being more familiar with finite groups than infinite ones, I may not have all the tools necessary to do so.

Is there anyone out there who can help?

I can't see how finite groups could be isomorphic to infinite ones....but given that B is apparently isomorphic to an infinite group my hunch would be that A/B is isomorphic to A - but hunches don't come with proofs

maybe this is useful?
http://www.math.uiowa.edu/~jsimon/COURSES/M201Fall04/Groups2a.pdf

sorry don't know....i'm sure one of the JREF maths bods'll be able to help

 

[andyandy]

....if not then the sosmaths forum is worth a try http://www.sosmath.com/CBB/

 

[Yiab]

I can't see how finite groups could be isomorphic to infinite ones....but given that B is apparently isomorphic to an infinite group my hunch would be that A/B is isomorphic to A - but hunches don't come with proofs

Not only are none of the groups I mentioned finite, none of them are finitely generated.

Here are some simpler descriptions of the groups I'm looking at:
A consists of all finite and countably infinite sequences of integers, addition is defined coordinate-wise.
B consists of all finite sequences of integers, addition is defined coordinate-wise.
B is clearly a subgroup of A and since A is abelian, B is normal in A so we can talk about A/B.
A/B consists of all countably infinite sequences of integers, but if two sequences are the same on all but finitely many coordinates they are considered the same sequence.

In generator/relation notation, we get the following:
[latex]$A = < \{x_c\}_{c\subseteq \mathbb{N}} | \{x_{c}^{-1}x_{d}^{-1}x_cx_d\}_{c,d\subseteq \mathbb{N}} \cup \{x_{c\cup d}^{-1}x_{c\cap d}^{-1}x_cx_d\}_{c,d\subseteq \mathbb{N}}>\\
B = <\{x_n\}_{n\in \mathbb{N}} | \{x_{n}^{-1}x_{m}^{-1}x_nx_m\}_{n,m\in\mathbb{N}}>$[/latex]

Unfortunately, I don't know of a good way to represent A/B in this form or I might be able to figure this question out.

Also, thank you for the sosmath link.

 

[Jarom]

Sorry to bump an older thread, but did you ever find an answer to this? When you first posted it, it looked easy, and I worked on it for about an hour, but got nowhere (so not, in fact, easy).

It stayed at the back of my brain and I'm pretty sure I have an idea for an approach that will work (to prove non-isomorphism), but it will take me a couple hours to work out and I don't want to duplicate effort if you've already found a published result along these lines.

 

[Yiab]

No, I still haven't found an answer for this. Even if you don't feel like taking the time to work it through I'd be interested in hearing your idea for an approach to it.