17

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.

Alexander Gruber
  • 28,037
Bruno Stonek
  • 12,973
  • Can you clarify what your specific question is? Are you asking for hints on showing each of 1,2,3,4 implies the definition? Can you see 1⇒2⇒7? For 3 and 4, you might look at pG (which is both a quotient and a subgroup). – Jack Schmidt Feb 20 '11 at 22:31
  • @Jack: The question is, how can I show that 1,2,3,4 are isomorphic to $C_{p^\infty}$. Of course one is allowed to use the fact that isomorphism is a transitive relation. – Bruno Stonek Feb 20 '11 at 23:02
  • More than four years later, I should edit that seventh item to change those inclusions to "multiply by $p$" maps, but I don't want to bump the question to the front page. – Bruno Stonek Jun 03 '15 at 16:14

2 Answers2

10

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 Magidin
  • 417,286
  • What homomorphism gives us the quotient $\frac{G}{pG}$? –  Jul 09 '19 at 18:58
  • 1
    @MathematicalMushroom: I have absolutely no idea what you mean. Are you aware that the Prufer $p$-group is $p$-divisible, so that $pG=G$ and so $G/pG$ is trivial? What exactly are you trying to ask/say/inquire? – Arturo Magidin Jul 09 '19 at 23:00
  • Sorry, a bit of context would help suppose, It seems your powers fall just short of psychic. In Jack Schmidt's answer below your answer, he investigates the quotient $\frac{G}{pG}$, which means that $pG$ is a normal subgroup of $G$ in this case (or maybe even in general?). Anyway, it is my understanding that every normal subgroup can be exhibited as the kernel of some homomorphism, and I was wondering what that homomorphism was in this case. –  Jul 10 '19 at 00:36
  • 1
    @MathematicalMushroom: If your query is prompted by someone else’s answer, why would you not ask it of that someone instead of someone else? In any case: this is an abelian group. In an abelian group, any subgroup is normal. In an abelian group, the set of all $n$th powers (if written multiplicatively) or $n$-times-each-element (if written additiviely) is always a subgroup (and hence normal). $pG$ is reserved for abelian groups written additively. It is the kernel of the quotient map, for one. – Arturo Magidin Jul 10 '19 at 01:39
  • 1
    @MathematicalMushroom: As for general groups, written multiplicatively, the set of $n$th powers (or even $p$th powers) is not necessarily a subgroup, but $G^n$ represents the subgroups generated by the $n$th powers, and that one is always normal (in fact, characteristic; in fact, fully invariant). This should be easy for you to verify, if you know and have mastered the basics of group theory. Otherwise, this question is well above your paygrade and you need to do a lot more studying/reviewing of basic stuff. – Arturo Magidin Jul 10 '19 at 01:41
  • How do you know here that $\langle x_1,y \rangle$ is cyclic? –  Aug 02 '19 at 18:25
  • 1
    @MathematicalMushroom: You shouldn't have said "here", given that this is about mutliple definitions of the Prufer $p$-group. I am guessing you mean in paragraph 2. Well, I know by hypothesis! Do me the favor of reading what condition 2 in the statement says. In general, it is good practice to bother reading through before jumping and asking questions that are explicitly answered in the text. – Arturo Magidin Aug 02 '19 at 18:29
6

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.

Jack Schmidt
  • 56,967
  • @Jack: Oops. I was just adding those two. Let me yank them out and refer to yours – Arturo Magidin Feb 20 '11 at 23:24
  • Hehe, I assumed by "had to go" you meant you'd be back tomorrow! I'm fine if you just give the whole answer. I can delete this one just as well :) – Jack Schmidt Feb 20 '11 at 23:28
  • @Jack: There were a few things that needed to get done, so I went to do them. That's why I said "for a bit". (-: The only difference between what I wrote and what you did is that you are invoking the description of divisible abelian groups, and I was just saying "show that $G$ is equal to the subgroup you already found". – Arturo Magidin Feb 20 '11 at 23:49
  • Hmm, yes, I've recently become aware of the description of divisible abelian groups, but I wouldn't be comfortable using it yet. – Bruno Stonek Feb 21 '11 at 00:25
  • Dear Jack,

    It is true that we can identified Z(p^infinity) as infinite union of all T_n where T_n={z in C: z^(p^n)=1 for n>0 and a prime number p}. Does this view simplify proving 1 and 3 or 4 above? Thanks for your help.

     – Mikasa Mar 21 '11 at 10:50
  • hello. I have a quick question abut 4). Does infinite direct product or sum of cyclic group of order p satisfy it? thx! – scsnm Dec 31 '21 at 02:24