# Characterizations of the $p$-Prüfer group

I’m an undergrad student fairly keen on algebra. Over the different algebra courses I’ve taken, I’ve often encountered the so-called $p$-Prüfer group on exercises but somehow never got around to them. Now I’m trying to take care of that, but there are some statements I’ve seen about this group which I don’t know how to prove (maybe because I lack some more background in group theory, especially in the study of infinite abelian groups?)

Definition A $p$-group is a $p$-Prüfer group if it is isomorphic to $$C_{p^\infty}=\{e^{\frac{2k\pi i}{p^n}}:k\in \mathbb{Z}, n\in\mathbb{Z}^+\} \subset (\mathbb{C}^\times, \cdot)$$

What I’m having trouble to prove is:

The following are $p$-Prüfer groups:

1) An infinite $p$-group whose subgroups are totally ordered by inclusion,

2) An infinite $p$-group such that every finite subset generates a cyclic group,

3) An infinite abelian $p$-group such that $G$ is isomorphic to every proper quotient,

4) An infinite abelian $p$-group such that every subgroup is finite

Just for the record, what I (think I) could prove was that the following are $p$-Prüfer groups:

5) An injective envelope of $C_{p^n}$, for any $n\geq 1$,

6) A Sylow $p$-subgroup of $\frac{\mathbb{Q}}{\mathbb{Z}}$,

7) The direct limit of $0\subset C_p \subset C_{p^2}\subset …$

Here $C_{p^n}$ denotes a cyclic group of order $p^n$.

Any other characterizations of the $p$-Prüfer group are welcome.

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How one procedes to prove these probably depends a bit on which description of the $p$-Prüfer group one is most comfortable with. So my particular arguments may not seem very natural to you, in which case this will probably not help much.

1. Because the subgroups are ordered by inclusion, given any $x,y\in G$, either $x\in\langle y\rangle$ or $y\in\langle x\rangle$. Take an element $x_1\in G$. The subgroup it generates cannot be all of $G$ (it is finite, since $G$ is a $p$-group), so there exists $x_2\in G$ with $x_2\notin\langle x_1\rangle$. Therefore, $x_1\in\langle x_2\rangle$. But $x_2$ cannot generate all of $G$. So there exists $x_3\in G$, $x_3\notin \langle x_2\rangle$.

Continuing this way, you obtain a (countably infinite) collection of elements of $G$, $x_1,x_2,\ldots,$ such that $x_n\in\langle x_{n+1}\rangle$, $x_{n+1}\notin\langle x_n\rangle$. By introducing suitable elements into the sequence, we may assume that $[\langle x_{n+1}\rangle\colon\langle x_n\rangle]=p$.

Now prove that the subgroup generated by all the $x_i$ is isomorphic to the $p$-Prüfer group, and that it is equal to $G$.

2. Let $x_1\in G$. It does not generate all of $G$, so let $y\in G$ that is not in $\langle x_1\rangle$. Then $\langle x_1,y\rangle$ is cyclic, strictly larger than $\langle x_1\rangle$; let $x_2$ be a generator of this cyclic group. Since $x_2$ does not generate all of $G$, there exists $y\in G$ not in $\langle x_2\rangle$; but $\langle x_2,y\rangle$ is cyclic, so let $x_3$ be a generator. Lather, rinse, repeat, to get an infinite sequence as above, adding elements if necessary to get each step to increase the size of the cyclic group by $p$; that is, $x_n$ is of order $p^n$.

Again, prove the group generated by the $x_i$ is the $p$-Prüfer group and equal to all of $G$.

I’d written suitable arguments for 3 and 4, but Jack has posted them as well and they are essentially the same, so I’ll defer to him on them.

Arturo already explained the hint 1⇒7, 2⇒7 (and 1⇒2 is easy enough, I trust you can do it). I’ll explain how to use pG = { pg : g in G } to understand G for 3 and 4.

If every nonzero quotient of G is isomorphic to G, then what does G/pG look like? Well, it is an elementary abelian p-group, so a vector space over Z/pZ. If it is nonzero, then it has a one-dimensional quotient: Z/pZ. By the hypothesis on G, that would mean G itself is Z/pZ. Unfortunately such a group is not infinite, and so is not a group as in 3. Hence a group as in 3 must have G=pG. As it is a p-group, it is divisible, so a direct sum of Prüfer p-groups. However, such a direct sum always has a single Prüfer p-group as a quotient, and so G must itself be the Prüfer p-group.

If every proper subgroup of G is finite, then what does pG look like? If it is finite, then G/pG is infinite, and so infinite dimensional. Take a proper subspace. The preimage of that subspace in G is a proper subgroup that is infinite. Oops. So pG cannot be finite! So again G=pG, and G is a direct sum of Prüfer p-groups. How many? Well each summand is a subgroup, and so if there is more than one, then one has an infinite proper subgroup. So G must itself be a single Prüfer p-group.