Show that $G\{m\}$ is finite for all non-zero $m\in \mathbb{Z}$ Let $G$ be an abelian group, and let $m$ be an integer, then we define $G\{m\} := \{a\in G:ma=0_G\}$.
Now, suppose that $G$ is an abelian group that satisfies the following properties:
(i) For all $m\in \mathbb{Z}$, $G\{m\}$ is either equal to $G$ or is of finite order;
(ii) for some $m\in \mathbb{Z}$, $\{0_G\}\subsetneq G\{m\}\subsetneq G$.
Show that $G\{m\}$ is finite for all non-zero $m\in \mathbb{Z}$.
Edit: I've solved the problem, but my solution (posted below) is somewhat messy. Any simpler solutions will be appreciated.
 A: After working long and hard on this problem I think I solved it (with some help from @nik).
Howerver, I'm not very satisfied with this solutions, as it seems a bit clumsy to me, and I would appereciateit if someone could provide a more elegant solution.
Solution:
Let $G$ be an abelian group that satisfies properites $(i)$ and $(ii)$. We proceed with a proof by contradiction. Assume that there is some non-zero integer $m$ such that $G\{m\}$ is of infinite order. since $G\{m\}=G\{-m\}$ there must be a minimal positive integer $m^{*}$ for which $G\{m^*\}$ is of infinite order. Now, let $n\in\mathbb{Z}$ be such that $\{0_G\}\subsetneq G\{n\}\subsetneq G$. By property $(i)$ we see that $G=G\{m^*\}$ and $G\{n\}$ is finite. We define $d:=gcd(m^*,n)$, and let $\ 0_G\neq x\in G\{n\}$. If $d=1$ then there exist $u,v\in \mathbb{Z}$, such that $um^*+vn=1$ and hence $x=(um^*+vn)x=u(m^*x)+v(nx)=0_G$ - a contradiction. Thus we must have $d > 1$. We define $m':=m^*/d$. Note that $m'G\subseteq G\{d\}\subseteq G\{n\}$, since for every $g\in m'G\{m^*\}=m'G$, there exists $h\in G$ such that $g=m'h$ and $m^*h=0_G$, hence, $dg=m^*h=0_G$, so $g\in G\{d\}$. We conclude that $m'G$ is finite. Now, observe the $m'$-multiplication map on $G$:
$$
\begin{align}
\rho: & G\rightarrow G \\
& x\mapsto m'x 
\end{align}
$$
One can easily see that $Im\ \rho = m'G$, and $\ker\ \rho=G\{m'\}$. Thus by the first isomorphism theorem we have $G/G\{m'\}\cong m'G$, and hence $\left|G/G\{m'\}\right|= \left|m'G\right|$. Since $m'G$ is finite, but $G$ is not, we must conclude that $G\{m'\}$ is not finite, which contradicts the minimality of $m^*$.
Edit:
Here is an alternative solution that only uses the material that was covered in the book from the question is taken.
The solution uses the following two easily verifyable facts:
$$
G\{k\}+G\{m\}=G\{\text{lcm}(k,m)\} \\
G\{k\}\cap G\{m\}=G\{\gcd(k,m)\} 
$$
Indeed, if $d:=\gcd(k,m)$, then $sk+tm=d$ for some integers $s,t$, and so if $x\in G\{k\}\cap G\{t\}$ then $dx=skx+tmx=0$. Thus $G\{d\}\subseteq G\{k\}\cap G\{t\}$. The other direction is trivial, and the other equality is proven in a similar fashion.
If $G$ is finite there is nothing to prove, so assume $G$ is infinite. Let $n$ be the exponent of $G$. If $n=0$ then we again have nothing to prove, so assume $n>1$ ($n=1$ is impossible, since it would imply that $G$ is finite).
Let $p_1^{e_1}\cdots p_r^{e_r}$ be the prime decomposition of $n$. We have $G=G\{n\}=G\{p_i^{e_i}\}+\cdots+G\{p_i^{e_i}\}$, and therefore at least one of the subgroups $G\{p_i^{e_i}\}$ must be infinite. Hence, by property (i) we have $m=p^e$ for some prime $p$ and integer $e$. From property (ii) we know there is some integer $0<f<e$ such that $G\{p^f\}$ is finite, and hence $e\ge2$. In particular, Since $G\{p^{e-1}\}$ is a non-trivial subgroup of $G$, and so by property (i), it must be finite. We note that all of the elements in $G\{p^e\}\setminus G\{p^{e-1}\}$ have order $p^e$, so there are infinitely many such elements. Since every $x\in G\{p^e\}\setminus G\{p^{e-1}\}$ maps into some non-zero element in $G\{p^{e-1}\}$ when multiplied by $p$, we can find an infinite sequence $\{x_i\}_i$ of distinct elements such that $px_i=px_j$ for all $i,j$ (via the pigeonhole principle). We examine the quotient group $H:=G\{p^e\}/G\{p^{e-1}\}$. This group contains infinitely many elements of order $p$, and among them are the elements $[x_i]$. Therefore we can find two indices $i,j$ such that $\langle[x_i]\rangle\cap\langle[x_j]\rangle=0_H$. However, $[x_i] - [x_j] = 0$, which implies that $[x_i]\in \langle[x_i]\rangle\cap\langle[x_j]\rangle$ - a contradiction.
A: Short version: the set $S$ of $m$ for which $G\{m\}$ is not finite is always a union of prime ideals, and the set $P$ of $m$ such that $G\{m\}=G$ is an ideal, so if $P=\mathbb{Z}\setminus S$, then either $S=\mathbb{Z}$ or for some prime $p$, $G=G\{p\}$ is an infinite elementary abelian group.
The ideas of multiplicatively closed sets and prime ideals are extremely fundamental topics in algebra and number theory. They are generally covered in the undergraduate algebra sequence in the US.

Definition: For an abelian group $G$, set $S(G) = \{ m : G\{m\} \text{ is finite } \}$ and $P(G) = \{ m : G\{m\} = G \}.$
Lemma: For any abelian group, $S(G)$ is a saturated multiplicative closed set.
Proof: Let $G$ be an abelian group and $a,b$ be integers with $G\{a\}$ and $G\{b\}$ finite. Then consider the function $f:G\{ab\} \to G\{b\} : x \mapsto ax$. It has the given codomain, and its kernel is $G\{ab\}\{a\}=G\{a\}$. Since both its image and kernel are finite, so is its domain. In other words, $S=\{ n : G\{n\} \text{ is finite} \}$ is multiplicatively closed and non-empty. Since $G\{a\} \subseteq G\{ab\}$, we also get that $S$ is saturated, $ab \in S$ implies $a \in S$. $\square$
Lemma: For any abelian group, $P(G)$ is either empty or an ideal.
Proof: Let $a,b \in P(G)$, so that $G\{a\}=G\{b\}=G$. If $g \in G$, then $(a+b)g = ag+bg = 0 + 0 = 0$, since $g\in G\{a\}$ and $g\in G\{b\}$. Hence $g \in G\{a+b\}$ and $G=G\{a+b\}$, so that $a+b \in P(G)$ as well. If $n \in \mathbb{Z}$, $a \in P(G)$, and $g\in G$, then $(na)g =n(ag) = n(0) = 0$, so again $g \in G\{na\}$ and $G\{na\} = G$ and $na \in P(G)$. $\square$
Proposition: If $P(G)=\mathbb{Z}-S(G)$, then $P(G)$ is either empty or a prime ideal, say $(p)$, and in the latter case $G\{p\}=G$ is an infinite elementary abelian $p$-group.
Proof: Assume $P$ is not empty, so that it is an ideal. Let $ab \in P$. If $a \notin P$, then we must show $b \in P$. However, this follows from $\mathbb{Z}\setminus P(G) = S(G)$ being multiplicatively closed: If $a \notin P$ and $b \notin P$, then $a \in S$ and $b\in S$, so $ab\in S$, and $ab\notin P$, a contradiction. Hence $P$ is a prime ideal, generated by a prime number, say $p$. In particular, $p \in P$, so that $G\{p\}=G$ is an elementary-abelian $p$-group. Since $p \notin S$, $G$ is not finite. $\square$
Corollary: Let $G$ be an abelian group such that for each integer $m$, either $G\{m\}$ is finite or $G\{m\}=G$. If for some integer $m$, $0 \neq G\{m\} \neq G$, then $G\{m\}$ is finite for all $m$.
Proof: The first condition says that $S(G) \cup P(G) = \mathbb{Z}$. If $m \in S(G) \cup P(G)$, then $G=G\{m\}$ is finite, so we are done. Otherwise, if the intersection is empty, then $P(G) = \mathbb{Z} \setminus S(G)$. The proposition gives that either $P(G)$ is empty (and $S(G)=\mathbb{Z}$ as desired) or $G=G\{p\}$ for some prime $p$. However, in the latter case we have that $G\{m\} = 0$ if $m$ is relatively prime to $p$, and $G\{m\} = G$ is $m$ is divisible by $p$, violating the second hypothesis. $\square$

Generalization: Let $R$ be a commutative, unital, associative ring, and $X$ be a class of $R$-modules closed under submodules and extensions. For an $R$-module $M$, define the $X$-support of $M$ to be $\operatorname{supp}_X(M) = \{ r \in R : \{ m \in M : rm = 0 \} \notin X \}$. The first lemma shows that $\operatorname{supp}_X(M)$ is a union of prime ideals. The second lemma says that the annihilator ideal of $M$ is an ideal. The proposition says that that if $M=M\{r\}$ for all $r \in \operatorname{supp}_X(M)$, then either $\operatorname{supp}_X(M)$ is empty, or a single prime ideal, $\operatorname{Ann}(M)$, and $M \cong (R/\operatorname{Ann}(M))^{(I)}$ for some index set $I$.
