Divisible abelian groups. From Hungerford's Algebra (with slight changes)

1) Let $p$ be a prime and let $Z(p^{\infty})$ be the following subset of the group $\Bbb{Q}/\Bbb{Z}$ $$Z(p^{\infty}) = \{ \overline{a/b} \in \Bbb{Q}/\Bbb{Z} | a,b \in \Bbb{Z}, b=p^i \operatorname{for some} i \geq 0\}.$$ Show that for each prime $p$, $Z(p^{\infty})$ is a divisible group. 
2) No nonzero free abelian group is divisible. 

1) I find this a bit confusing. According to the textbook "An abelian group $D$ is said to be divisible if given any $y \in D$ and $0 \not= n \in \Bbb{Z}$, there exists $x \in D$ such that $nx=y$." 
So, for example, let's look at $p=3$. Let us choose $2/9 \in Z(3^{\infty})$ and $4 \in \Bbb{Z}$. Now we need to find an $x \in Z(3^{\infty})$ so that $4x = 2/9$. But then $x = 2/36$, and $36$ is not a power of $3$. How is this group divisible then? 
2) Let's assume, for contradiction, that there is a free abelian group is divisible. 
We have a lemma and a proposition in the textbook that states

Lemma. An abelian group $D$ is divisible if and only if $D$ is an injective (unitary) $\Bbb{Z}$-module. 
Proposition. A direct product of $R$-modules $\Pi_{i \in I} J_i$ is injective if and only if $J_i$ is injective for every $i \in I$. 

Since a free abelian group is a free $\Bbb{Z}$-module, the group must be isomorphic to a direct sum of copies of $\Bbb{Z}$. By the lemma, it must be injective; and by the proposition, the group is injective if and only if $\Bbb{Z}$ is injective. But, since $\Bbb{Z}$ is not injective, we have a contradiction. 
Is my answer correct? 
Thanks in advance.  
 A: When you want to divide $2/9$ by $4$, you need to find some $x \in Z(3^\infty)$ such that $4x = 2/9 \pmod 1$.
Since $2/9 = 2/9 + 2 = 20/9$, we can pick $x = 5/9$.
Let $Z(p^n) = \{a/b \in Z(p^\infty), b \mid p^n\} = p^{-n} (\Bbb Z / p^n \Bbb Z)$.
If $q$ is prime number distinct from $p$, then multiplication by $q$ is a bijection from $\Bbb Z/p^n\Bbb Z$ to itself, so it is also a bijection from $Z(p^n)$ to itself.
Since $Z(p^\infty)$ is the reunion of the $Z(p^n)$, multiplication by $q$ is a bijection from $Z(p^n)$ to itself.
Since $Z(p^\infty)$ is obviously $p$-divisible, it is $q$-divisible for all primes $q$, so ti its divisible.

Your answer for (2) is not completely correct, since a countably infinite direct sum is not a direct product. However you can still pick a basis for $G$, pick one basis element, and mention that this element it is not divisible by $2$.
A: 

*

*Let $p$ be a prime and let $Z(p^{\infty})$ be the following subset of the group $\Bbb{Q}/\Bbb{Z}$ $$Z(p^{\infty}) = \{ \overline{a/b} \in \Bbb{Q}/\Bbb{Z} \; |\; a,b \in \Bbb{Z}, b=p^i \operatorname{for some} i \geq 0\}.$$ Show that for each prime $p$, $Z(p^{\infty})$ is a divisible group.





*No nonzero free abelian group is divisible.


Those are actually items (a) and (c) of exercise 4 in Hungerford's Algebra (page 198).
Recall the definition: an abelian group $D$ is said to be divisible if given any $y \in D$ and $n \in \Bbb Z$, $n \neq 0$, there is $x \in D$ such that $nx=y$ (Hungerford's Algebra page 195).
Let us stat proving an easy (but useful) lemma.
Lemma: An abelian group $D$ is divisible if and only if given any $y \in D$ and $q \in \Bbb N$, such that $q$ is a prime number, there is $x \in D$ such that $qx=y$
Proof: The direction $\Rightarrow$ is trivial from the definition.  The direction $\Leftarrow$ is an easy consequence of $-1(-y) = y$, $1y=y$ and that factorization of non-null integers.
$\square$.
Prood of item 1: Using the lemma, let $q$ a prime. We have two cases:
Case 1: $q=p$. Given any $y \in Z(p^{\infty})$, then $y= \overline{a/p^i}$, for some $a \in \Bbb Z$, $0 \leq a <p^i$, and $i\geq 0$. Take $x= \overline{a/p^{i+1}}$. Then, we have $qx= p \overline{a/p^{i+1}} = \overline{a/p^i} =y$.
Case 2: $q \neq p$.  Given any $y \in Z(p^{\infty})$, then $y= \overline{a/p^i}$, for some $a \in \Bbb Z$, $0 \leq a <p^i$, and $i\geq 0$.  Since $q$ is prime and $q \neq p$, we have that $\psi_q: \Bbb{Z}/p^i\Bbb{Z} \to \Bbb{Z}/p^i\Bbb{Z}$ defined by $\psi_q([h]) = [qh]$ is a bijection. So there is $c$, $0 \leq c <p^i$, such that
$[qc] = [a]$. In means, $qc = a + kp^i$ for some $k \in \Bbb Z$. Take $x=\overline{c/p^i}$. It folows that
$$qx=q\overline{c/p^i}= \overline{qc/p^i}= \overline{(a+kp^i)/p^i}=\overline{(a/p^i) +k}=\overline{a/p^i}=y$$
So $Z(p^{\infty})$ is a divisible group.
$\square$
Proof of item 2: Let $F$ be any nonzero free abelian group. Then, by Theorem 1.1 (Hungerford's Algebra page 71), $F$ is isomorphic to a direct sum of copies of the additive group $\Bbb Z$ of integers. That is $F \simeq \bigoplus_{i \in I} \Bbb{Z}_i$, where, for each $i \in I$, $\Bbb{Z}_i$ is a copy of $\Bbb{Z}$.
So, to prove that $F$ is not divisible, it is enough to prove that $\bigoplus_{i \in I} \Bbb{Z}_i$ is not divisible.
Since $F$ is a nonzero free abelian group, $I \neq \emptyset$. Let $i_0$ be a fix element in $I$ and consider $y=(a_i)_{i \in I} \in \bigoplus_{i \in I} \Bbb{Z}_i$ such that $a_{i_0}=1$ and  $a_{i}=0$ for any $i \neq i_0$. There is no $x= (b_i)_{i \in I} \in \bigoplus_{i \in I} \Bbb{Z}_i$ such that $2x=y$. In fact, if such $x$ exist, we would have an integer $b_{i_0}$ such that $2b_{i_0} =1$.
So $\bigoplus_{i \in I} \Bbb{Z}_i$ is not divisible. So, $F$ is not divisible.
