Injective hull of $\mathbb{Z} _p$ The $p$-primary component of the group $\mathbb{Q}/\mathbb{Z}$ is denoted by $\mathbb{Z}(p^{\infty})$, where $p$ is a prime.
Now I want to show that 

$\mathbb{Z}(p^{\infty})$ is an injective group, and the injective hull of $\mathbb{Z} _p$ is $\mathbb{Z}(p^{\infty})$.

Any hint? Thanks.
 A: Hint: Recall that Injective abelian groups are exactly the divisible abelian groups. 
Show that $\Bbb Z(p^{\infty})$ is a divisible group; let $n$ and $\frac{a}{p^i}$ with $(a,p)=1$, write $n=mp^j$ with $(m,p)=1$, there is $r<p^i$ with $mr\equiv a \pmod {p^i}$, so if $y=\overline{\frac{r}{p^{i+j}}}$, then $ny=\overline{\frac{a}{p^i}}$.
If $D$ is an abelian divisible group such that $\Bbb Z_p\subseteq D$, construct $x_1,x_2,\ldots\in D$ such that $px_1=\overline 1$, and $px_{n+1}=x_n$, it is not hard to see that the subgroup generated by $\bar 1$ and the $x_i's$ is isomorphic to $\Bbb Z(p^{\infty})$
A: Here are some general facts of algebra:
A module over a ring is called semisimple if it isomorphic to a direct sum of simple modules.  It is an exercise to check that any sum of semisimple submodules of a given module is again semisimple.  Thus any module contains a maximal semisimple submodule, called its socle.
If a module is Artinian, meaning that it satisfies the d.c.c. on submodules,
then if it is non-zero, it contains a non-zero simple submodule.  Hence an Artinian module is is non-zero if and only if its socle is non-zero.
Any non-zero Artinian module $M$ is an essential extension of its socle, i.e. any non-zero submodule has non-zero intersection with the socle of $M$. (Proof: Let $N$ be a non-zero submodule of $M$.  Since $M$ is Artinian, so is $N$, and thus if $N$ is non-zero, its socle $\mathrm{soc}(N)$ is non-zero.  But $\mathrm{soc}(N) = \mathrm{soc}(M) \cap N,$ and hence $N$ has non-zero intersection with $\mathrm{soc}(M)$.)
If $I$ is a non-zero injective module that is also Artinian, then $I$ is an injective envelope of its socle.  (Proof: An injective envelope of a module $M$ is characterized up to isomorphism as being an injective module that is also an essential extension of $M$.  Now $I$ is injective by assumption, and is an essential extension of $\mathrm{soc}(I)$ by the previous paragraph, since it 
is also assumed to be Artinian.)

Now consider the $\mathbb Z$-module $\mathbb Z(p^{\infty})$.  It is easy to prove that it is Artinian, and that its socle is cyclic of order $p$.  Thus, applying the preceding result, we find that it 
is an injective envelope of $\mathbb Z/p\mathbb Z$.
