Additive group of rational numbers 
Let $\mathbb{Z}[\frac{1}{p}]$ be the additive group of rational numbers of the form $mp^n$ where $m$, $n$ are elements of $\mathbb{Z}$ and $p$ is a fixed prime. Describe $\text{End}(\mathbb{Z}[\frac{1}{p})]$ and $\text{Aut}(\mathbb{Z}[\frac{1}{p}])$.

Unfortunately I have never done this type of exercise and do not know where to start: I have great difficulties in general because I do not know how to describe these types of groups. Can you help?
Sorry if my English is not exactly correct.
 A: This answer is basically combining what Jack Schmidt has remarked and condensing Lubos Motl's answer. 
First we notice that $\mathbb{Z}[\frac{1}{p}]$ has a ring structure. This allows us to view $\mathbb{Z}[\frac{1}{p}]$ as a free $\mathbb{Z}[\frac{1}{p}]$-module of rank 1. We show that any endormorphism of $\mathbb{Z}[\frac{1}{p}]$ as an abelian group is in fact an endomorphism of  $\mathbb{Z}[\frac{1}{p}]$-modules as well.  For a commutative ring with unit, $\mathrm{End}_R(R)=R$ since it only depends on the image  of $1$.  
Last, $\mathrm{Aut}(\mathbb{Z}[\frac{1}{p}])$ is all invertible elements in $\mathrm{End}(\mathbb{Z}[\frac{1}{p}])$. In other words, we look for the units in the ring $\mathbb{Z}[\frac{1}{p}])$.  One can easily show that the units are $p^{\mathbb{Z}}\times \langle \pm 1\rangle$. 
A: ${\mathbb Q}_p = {\mathbb Z} [1/p]$, right? You changed the notation...
Hi, the endomorphisms $E$ have to map the numbers $mp^n$ to some other numbers $mp^n$. Because it's an additive group and we consider homomorphisms, it's enough to figure out what the generators $p^n$ are mapped to for various numbers of $n$. The multiples by $m$ are obtained by the homomorphism conditions.
Moreover, also because it is a homomorphism, 
$$E(p\cdot p^n)  = E(p^n+\cdots p^n)$$
must be equal to
$$E(p^n)+E(p^n)+\dots + E(p^n) = p\cdot E(p^n)$$
This proves that the information about $E(p^{n})$ and $E(p^{n+1})$ is not independent at all; the latter must be $p$ times the former. We may go in both directions; the former is $1/p$ times the latter.
It follows that the endomorphism is totally determined by $E(p^0)=E(1)$ which has to be an element of ${\mathbb Q}_p$. Note that ${\mathbb Q}_p$ was an additive group so $E(1)$ doesn't have to be $1$. It follows that the endomorphisms as a set are ${\mathbb Q}_p$ and the the endomorphisms, like always, are a semigroup, which is the multiplicative semigroup of these elements, ${\mathbb Q}_p^\times$. The superscript $\times$ means the multiplication. So ${\rm End} {\mathbb Q}_p \equiv {\mathbb Q}_p$.
The automorphisms are those endomorphisms that are one-to-one and therefore invertible. What can $E(1)$ be so that $E(mp^n)$ covers the whole ${\mathbb Q}_p$? If
$$E(1) = mp^n $$
then 
$$ p = (E(1)/m)^{1/n} $$
For which $m,n$ is the expression above in ${\mathbb Q}_p$ for any $E(1)$ in ${\mathbb Q}_p$?
The $n$th root is clearly too bad and sends us away from ${\mathbb Q}_p$. It can only work for $n=\pm 1$ if $m$ is allowed to be something else than a power of $p$. At any rate, I think that $E(1)$ has to be of the $\pm p^n$ form. The exponent $n$ is an integer so I think that
$${\rm Aut} {\mathbb Q}_p \equiv {\mathbb Z} \times {\mathbb Z}_2 $$
It's the direct product of the integers (with addition, the group shifting things by a power of $p$) and the two-element group (responsible for the overall sign). (Thanks for the fix, no semidirect here.)
