- #1
Bashyboy
- 1,421
- 5
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?
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?