[Math] any subgroup of $(\mathbb{Q},+)$

abstract-algebragroup-theory

Any subgroup of $(\mathbb{Q},+)$ is _______

  1. cyclic and finitely generated but not abelian and normal,
  2. cyclic and abelian but not finitely generated and normal,
  3. abelian and normal but not cyclic and finitely generated, or
  4. finitely generated and normal but not cyclic and abelian.

I know

  1. that any finitely generated subgroup of ($\mathbb{Q},+)$ is cyclic and hence abelian and hence normal,

  2. $(\mathbb{Z},+)$ is cyclic, abelian, generated by $\{1,-1\}$ and of course normal.

But I am not able to figure out or tackle the above statements separately.

Best Answer

That's a very strangely worded question, but I think it's probably meant to be a fill-in the blank multiple choice question which would more intelligibly be stated as:

Every subgroup of $(\mathbb{Q}, +)$ is ____________________.

A) cyclic and finitely generated, but not necessarily abelian or normal.

B) cyclic and abelian, but not necessarily finitely-generated or normal.

C) abelian and normal, but not necessarily cyclic or finitely-generated.

D) finitely generated and normal, but not necessarily cyclic or abelian.

Since $(\mathbb{Q}, +)$ is an abelian group, every subgroup is obviously abelian and normal, which rules out A, B, and D, so the answer is C. Of course you want to actually be able to put your hands on a non-finitely generated subgroup of $(\mathbb{Q}, +)$, so as anon says in the comments, consider the additive group of $\mathbb{Z}[\frac{1}{2}]$, i.e. the subgroup of $(\mathbb{Q}, +)$ consisting of elements where the denominator is a power of 2.

(Note that a finitely-generated subgroup of $(\mathbb{Q}, +)$ has an upper bound on its denominators, and is in fact cyclic.)

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