Let $(G,+)$ a finely generated abelian group. My problem is to prove that $G$ is direct sum of cyclic subgroups, each of which is infinite, or has order the potency of a prime number.
Suppose that $G$ is not a periodic group, and let $G=<x_1,...,x_t>$. We proceed by induction $t$, for $t=1$ the assertion is obvious. Let $t>1$, and consider
$$H:=\lbrace x \in G : mx \in <x_1>, m \in \mathbb{N} \rbrace$$
$H \leq G$, $x_1 \in H$, and $H$ is not periodic (because $G$ is not periodic? is it correct?). Now $G/H$ is $(t-1)$-generated (why is it $(t-1)$-generated?) and $G$ not periodic implies also that $G/H$ is not periodic in the sense that $G/H$ defines a partition of $G$? In particular, by inductive hypothesis, $G/H$ is free abelian group, i.e. there is $K \leq G$ such that $G=H \bigoplus K$.
$H/<x_1>$ is finely generated and periodc group (why is it finely generated and periodic?), and by another result is finite, let $card(H/<x_1>)=n$, and consider the following homomorphism
$$\varphi: h \in H \longrightarrow nh \in <x_1>$$
$H$ not periodic implies that Ker$(\varphi)=\lbrace 0 \rbrace$ and $H$ is isomorphic to a subgroup of $<x_1>$, and then $H$ is infinite cyclic (why $H$ is infinite?).
My doubts are during the demonstration, thanks for every reply.
$H$ is not periodic, because it is a subgroup of an aperiodic group, you are right there. $G/H$ is $(t-1)$-generated, because clearly $x_1 \in H$, and so the classes of $x_2,...,x_t$ mod $\langle x_1 \rangle$ generate $G/H$.
$G/H$ is not periodic, because if $n\cdot (y+ H) = 0$, then $n\cdot y \in H$, which implies that there exists $m$ such that $mn\cdot y \in \langle x_1 \rangle$, which in turn, by definition, implies $y\in H$, which implies $y+ H = H$, which is the zero of $G/H$.
The argument as to why there exists $K$ such that $K\oplus H = G$ is not very clear. I think one way to see it is to do it is using Zorn's lemma, but I'm not sure it's what your teacher meant.
$H/\langle x_1\rangle$ is periodic because if $y\in H$ then for some $m$, $m\cdot y \in \langle x_1\rangle$, that is $m\cdot (y+\langle x_1 \rangle) = 0$ mod $\langle x_1\rangle$. It is finitely generated because $H$ is.
$H$ is infinite because it is not perdiodic (a finite group is periodic)