# Showing a Group cannot be finitely generated

1. Oct 18, 2014

### Bashyboy

Hello everyone,

I have to demonstrate that the two groups $(\mathbb{Q'}, \cdot )$ and $(\mathbb{R'}, \cdot )$, where $\mathbb{Q'} = \mathbb{Q} \setminus \{0\}$ and $\mathbb{R'} = \mathbb{R} \setminus \{0\}$.

While trying to solve this problem, a thought suddenly occurred to me. Here is the conjectured I formulated:

Let $G$ be some group with subgroup $H$. If $H$ cannot be finitely generated, then neither can $G$.

Proof: Let us suppose that this is false, that $H \subset G$ is not finitely generated, but $G$ is. Let that finite set which generates all of $G$ be $S$, so that $\langle S \rangle = G$. Because $H$ is a subgroup of $G$, then the set $H$ contains some of $G$'s elements. Furthermore, because $G$'s elements are generated by $S$, and because $H$'s contains some of the elements of $G$, then some, if not all, of the elements of $S$ generate $H$. Seeing as $S$ is a finite set, and any subset of it would also be finite, then $H$ must be finitely generated---which is a contradiction. Therefore, if $H \subset G$ is not finitely generated, then $G$ cannot be.

Supposing that this conjecture is correct, and that I have correctly proven it, I sought out to contrive a simple subgroup of $\mathbb{Q'}$. For instance, $\{\frac{1}{2},1, 2\}$ has an identity, and each element has an inverse. Then I figured that all I needed to do was enumerate all of the subsets of this set, and show that none of them could generate this subgroup.

After I listed all of them, I realized that I never verified if my subgroup had closure---which it does not...
I decided that, before I tread any further, I ought have some of my work verified, and make sure the methods I choose would avail me any. If not, what methods would you suggest?

2. Oct 18, 2014

### HallsofIvy

Staff Emeritus
This is awkwardly stated. It is NOT just that "H must contain some of G's elements", H must contain nothing but elements of G.

3. Oct 18, 2014

### Bashyboy

True, but does that fundamentally change my proof?

4. Oct 18, 2014

### dslowik

I think your conjecture is correct; and those elements of $S$ which generate $H$ are $S\cap H$. And once you've proven the result for But then your left with finding a subgroup which can't be finitely generated, so it would have to be infinite (or $H$ itself could finitely generate it). And even some countable infinite subgroups like

5. Oct 18, 2014

### Bashyboy

I am not entirely sure of what you are saying. Are you saying that I won't be able to find a finite subgroup of $\mathbb{Q'}$?

6. Oct 18, 2014

### dslowik

sorry, i was futzing with some Latex preview bug in trying to edit my reply..

I think your conjecture is correct; and those elements of $S$ which generate $H$ are $S\cap H$. And once you've proven the result for $\mathbb{Q'}$, you could use that conjecture to show it was true for $\mathbb{R'}$.

But then your left with finding a subgroup which can't be finitely generated, so it would have to be infinite (or $H$ itself could finitely generate it). And even some countable infinite subgroups like $2^{\mathbb{Z}}$ can be finitely generated. beyond that, i'm still thinking..

7. Oct 18, 2014

### gopher_p

The conjecture is wrong. There are finitely generated groups with subgroups which are not finitely generated; take $G=\langle a,b\rangle$ and $H=\langle\{a^nb^n\}_{n\in\mathbb{N}}\rangle$.

Furthermore, the generators of a group need not be generators for the subgroups, and vice versa; take $G=(\mathbb{Z},+)=\langle 1\rangle$ and $H=2\mathbb{Z}=\langle2\rangle$.

8. Oct 18, 2014

### dslowik

Here are two ideas that will lead to proofs, but I can't work through the details now.
1) Any rational, expressed in lowest terms (by dividing common factors top and bottom), num and den can be uniquely factored into primes. So any finite set thereof, has only a finite number of distinct primes in each of num and den. Any rational reachable as some product of integral(+/-) powers has num prime factorization containing only those distinct primes of original set. But there are an infinite number of primes so there is a rational that can't be reached.
So you got the rationals, and your conjecture proves it for the reals. Or,
2) Any product of integral(+/-) powers of a finite collection of N reals, which commute, can be viewed as a finite(N) product of countable(integers) spaces, which is countable, but the reals aren't.

if any questions on working these through let me know.

9. Oct 18, 2014

### dslowik

I agree with these two points. Interesting that to specify the structure/constraints of the subgroup in first point takes infinitely many generators, while the containing group has only two generators.

10. Oct 18, 2014

### Bashyboy

This appears to be different than what I said in my conjecture. I said that, if $H$ is a subgroup of $G$ and it is not finitely generated, then $G$ cannot be finitely generated either. Furthermore, if the conjecture is wrong, then why was I able to arrive the fact with ostensibly sound logic? Where is the error in my proof?

11. Oct 18, 2014

### gopher_p

Your claim is the contrapositive of (and therefore logically equivalent to) the statement "If $G$ is finitely generated, then every subgroup $H$ of $G$ is also finitely generated". I have provided a counterexample to both versions of that claim.

Your error is the part where you say
and interpret that (incorrect) claim as meaning that $H$ is generated by a subset of the generating set for $G$. I have provided a counterexample to that claim as well.

12. Oct 18, 2014

### Bashyboy

Ooh, very lovely. Thank you.