Proving an isomorphism between finitely generated non-trivial subgroup

Question

Okay, I feel like I start every question this way, but I have an idea of the concepts and need some help actually putting it into practice. I'm working on this question from Groups and Symmetry by Mark A. Armstrong:

Prove that a finitely generated non-trivial subgroup of $\mathbb{R}\setminus \{0\}$ must be isomorphic to $\Bbb Z_2$ or to $\Bbb Z^s$ or $\Bbb Z_2 \times \Bbb Z^s$ for some positive integer $s$.

I know how to prove an isomorphism in general - show that a bijection exists and that multiplication carries from one group to another - but I'm not sure about the steps in this particular case. We have not yet covered automorphisms, either, which I see a lot in discussions of isomorphisms and kernels.

Answer 1

First note that $\Bbb Z/2\Bbb Z \times (0,\infty) \cong \Bbb R^\times$ via $(i,t)\mapsto (-1)^i t$. As $\Bbb Z/2\Bbb Z$ has no nontrivial subgroups it suffices to consider subgroups of $(0,\infty)$. In fact, the isomorphism $\Bbb R \cong (0,\infty)$ given by $k \mapsto e^k$ shows that it suffices to consider subgroups of $\Bbb R$. Since this is an abelian group, every (in our case finitely generated) subgroup $G$ is abelian. Now a minimal generating set $\{g_1,...,g_n\}$ yields a surjective homomorphism $$\Bbb Z^n \rightarrow G \text{ via } \sum\limits_{i=1}^n k_ie_i \mapsto \sum\limits_{i=1}^n k_ig_i.$$ If we can show that this map has trivial kernel, we are done.

Suppose we have $(k_1...k_n) \in \Bbb Z^{\oplus n}$ with $k_1g_1+...+ k_ng_n=0$. Assuming that all but one $k_i$ are zero yields $k_ig_i= 0$ hence $g_i=0$ contradicting minimality of the generating set. So we can assume wlog $$0\neq k_ng_n = -\sum \limits_{i=1}^{n-1} k_ig_i.$$ But now setting $\widetilde{g_i} = \frac{g_i}{k_n}$ for $i=1...n-1$ we find a smaller generating set $\{\widetilde{g_1},...,\widetilde{g_{n-1}}\}$ of $G$. A contradiction.

So we are done.

A more advanced way to show the exercise is by noting that by the classification of finitely generated abelian groups $G$ is of the form $\Bbb Z^{\oplus n} \oplus \Bbb Z/q_1\Bbb Z \oplus ... \oplus \Bbb Z/q_k\Bbb Z$. As every element in $(0,\infty)$ has infinite order, we find $q_i = 1$ for all $i=1...k$, so $G \cong \Bbb Z^{\oplus n}$.

Answer 2

First things first : before you can establish the existence of an isomorphism between a finitely generated subgroup of $\mathbb{R}^*$ and either $\mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^s$ or $\mathbb{Z}^s$, you must first know how the former look like. To this extent, let's first look at the subgroups of $\mathbb{R}^*$ generated by a single element :

Claim a non-trivial subgroup of $\mathbb{R}^*$ generated by a single element is isomorphic either to $\mathbb{Z}/ 2\mathbb{Z}$ or to $\mathbb{Z}$.

Let $a\in \mathbb{R}^*$, and $G := \langle a \rangle$ the subgroup generated by $a$.

• If $a = 1$, then $G$ is trivial.
• If $a = -1$, then $G = \{-1, 1\}$. The function $f : G \mapsto \mathbb{Z}/ 2\mathbb{Z}$ defined by $f(1) = 0$ and $f(-1) = 1$ is easily seen to be an isomorphism.
• Otherwise, $|a| \neq 1$, and $G = \{ a^n \ \big| \ n\in \mathbb{Z} \}$. The map $f : \mathbb{Z} \mapsto G$ defined by $f(n) = a^n$ is an isomorphism (left to the reader).

From here on, we use induction on the number $n$ of generators. Induction start is the claim above. So let's move to the induction step :

Let $n > 0 \in \mathbb{N}$, suppose that we have already established that for any $a_1, \dots, a_n \in \mathbb{R}^*$, the subgroup $\langle a_1, \dots, a_n \rangle$ generated by $a_1, \dots, a_n$ is ither trivial or isomorphic to $\mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^s$ for some $s$ (possibly $0$) or isomorphic to $\mathbb{Z}^s$ for some $s \neq 0$.
Let $a_1, \dots, a_{n+1} \in \mathbb{R}^*$. We write $G := \langle a_1, \dots, a_n \rangle$ and $H:=\langle a_1, \dots, a_{n+1} \rangle$.

WLOG $a_{n+1} \notin H$ otherwise $G=H$ and there's nothing to do.
This implies that $a_{n+1}^k \notin H$ for any $k \neq 0$. Hence, $\langle a_{n+1} \rangle \cap H = \{1\}$. $\langle a_{n+1} \rangle$ and $H$ are hence complement subgroups of $G$ and we get that $G$ is isomorphic to $H \times \langle a_{n+1} \rangle$ since $G$ is commutative. Now by induction hypothesis there are three cases :

• Case $H = \{1\}$ : we get $G = \langle a_{n+1}\rangle$, hence $G$ is either trivial or isomorphic either to $\mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^0$ or to $\mathbb{Z}^1$ by the claim above.
• Case $H \cong \mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^s$ : Since $a_{n+1} \notin \{-1, 1\}$, $\langle a_{n+1}\rangle \cong \mathbb{Z}$ by the claim above.
  Hence, $G \cong \mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^{s+1}$.
• Case $H \cong \mathbb{Z}^s$ : There are two possibilities depending on wether $a_{n+1} = -1$ or not :
  If $a_{n+1} = -1$, then $G \cong \mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}^s$.
  Otherwise, $G = \mathbb{Z}^{s+1}$.