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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.

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This may sound weird at first, but this wikipedia page should be helpful:… – Myself May 19 '11 at 16:13
Don't worry about your English -- it's not only better than in most other questions on this site; it's flawless (after someone correct a typo). – joriki May 19 '11 at 16:22
Myself's comment is actually incredibly helpful. It might also help to notice that Qp has a natural ring structure, that is, it is closed under multiplication. – Jack Schmidt May 19 '11 at 16:33

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 $p^{\mathbb{Z}}\times \langle \pm 1\rangle$.

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${\mathbb Q}_p = {\mathbb Z} [1/p]$, 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.)

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Note also that the OP has said s/he is trying to learn basic group theory, so it is not optimally helpful to provide complete solutions (even if they are correct). – Pete L. Clark May 19 '11 at 17:55
(I have retyped my above comment to remove two typos): I'm not sure why you speak of semidirect products or what the $'$ in $\mathbb{Z} \times ' \mathbb{Z}_2$ is supposed to mean. The endomorphism ring of the commutative group $\mathbb{Z}[\frac{1}{p}]$ is itself, viewed as a commutative ring, so the automorphism group is the unit group of this ring, hence commutative. Specifically, for $\frac{a}{p^n}$ to be invertible in the ring $n$ can be an arbitrary integer and $a$ must be $\pm 1$, so the answer is the direct product $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. – Pete L. Clark May 19 '11 at 18:04
You're right, it's a direct product. – Luboš Motl May 19 '11 at 20:49

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