# Characterising subgroups of Prüfer $p$-groups.

In my recent study of Artinian modules I have been looking at Prüfer $p$-groups (which I will denote by $\mathbb Z_{p^\infty}$ from now on).

I am attempting to prove that $\mathbb Z_{p^\infty}$ is an Artinian $\mathbb{Z}$-module which is not finitely generated.

To do this I am trying to show that every proper non-trivial submodule of $\mathbb Z_{p^\infty}$ is equal to $\left(\frac{1}{p^n}\mathbb{Z_{}}\right)/\mathbb{Z}$ for some $n$.

I am using the following definition of $\mathbb Z_{p^\infty}$: $$\mathbb Z_{p^\infty}=\left\{x\in \mathbb{Q}/\mathbb{Z}\mid \exists n:p^nx=0 \right\}$$ I have read in several places, Atiyah & MacDonald Example 6.3 and Wikipedia to name a few, that the subgroups of $\mathbb Z_{p^\infty}$ are of this form. This would demonstrate that $\mathbb Z_{p^\infty}$ is Artinian. However, I can't seem to find anywhere a proof or example showing why all the subgroups are of this form. I would like to understand this further, as it is not immediate to me that this would be the case, but my attempts to figure out why have so far been futile. I may be thinking too much and missing a simple explanation. I have looked at starting with an arbitrary subgroup $H$ such that $H\neq 0$ and $H\neq\mathbb Z_{p^\infty}$ and seeing if I can deduce anything from that. I got this far, but it doesn't seem to be going anywhere. I have also looked at Sylow's theorems, but I'm not sure if they will be any use to me as $|\mathbb Z_{p^\infty}|=\infty$ (unless I'm wrong about this!).

I would be extremely grateful if anybody could offer a few hints here as I am slowly losing my sanity trying to figure it out!

Thanks,

Andy.

• The proof is given on page $5$ here in the notes of Eric Jespers. We use B'ezout's lemma to obtain that every proper submodule is cyclic, so that the module is Artinian. – Dietrich Burde Nov 10 '16 at 15:43

Every element of ${\bf Z}_{p^\infty}$ is represented by some $a/p^n$, where $0\leq a<p^n$ (this is mostly immediate by your definition). I will simply identify this number with its class modulo ${\bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

Now, take any $G\leq {\bf Z}_{p^\infty}$. If $G$ is of finite exponent, say $p^m$, it follows that we can fix $n=m$ for elements of $G$, and have some $a/p^m\in G$, where $a$ is not divisible by $p$. Then by Euclidean algorithm (or the Chinese remainder theorem, whichever you prefer) we have also $1/p^m\in G$, and it clearly generates $G$.

If $G$ has infinite exponent, we have for any $n$ an element $a/p^n\in G$ with $a$ not divisible by $p$. By the same argument as in the preceding paragraph, we have for every $n$ that $1/p^n\in G$, and thus $G={\bf Z}_{p^\infty}$.

As a side remark on Sylow's theorems, they can be used for infinite profinite groups, but they are useless in this case, as ${\bf Z}_{p^\infty}$ is itself a $p$-group.

• Thanks @tomasz, extremely informative and clear. – andrew1601 Nov 10 '16 at 18:45

Consider the set $$P$$ consisting of the fractions $$a/p^n\in\mathbb{Q}$$, with $$a$$ any integer and $$n\ge0$$. This set is clearly a subgroup of $$\mathbb{Q}$$ and contains $$\mathbb{Z}$$.

By definition, $$\mathbb{Z}(p^\infty)=P/\mathbb{Z}$$.

Let $$H$$ be a subgroup of $$P$$ containing $$\mathbb{Z}$$. Let $$a/p^n\in H$$, for some $$a\in\mathbb{Z}$$ with $$\gcd(a,p)=1$$.

Then $$ax+p^ny=1$$, by Bézout; hence $$\frac{1}{p^n}=x\frac{a}{p^n}+\frac{y}{1}\in H$$ as $$H\supseteq\mathbb{Z}$$.

Thus we can consider the maximum $$k$$ such that $$1/p^k\in H$$, if it exists.

If $$b/p^m\in H$$, with $$\gcd(b,p)=1$$, then $$m\le k$$ and $$\frac{b}{p^m}=bp^{k-m}\frac{1}{p^k}$$ Therefore $$H$$ is cyclic with generator $$1/p^k$$. Note that $$k$$ is unique.

If the maximum exponent doesn't exist, then $$1/p^n\in H$$, for every $$n$$ and therefore $$H=P$$.

So if $$G=H/\mathbb{Z}$$ is a proper subgroup of $$\mathbb{Z}(p^\infty)$$, then $$G$$ is cyclic generated by $$1/p^k+\mathbb{Z}$$, for a unique $$k$$.