# infinite abelian group with finite proper subgroups

Discussion in 'Undergraduate Math' started by Jon Gerard, Jun 26, 2003.

1. ### Jon GerardGuest

Let G be an infinite abelian group such that every proper subgroup of
G is finite. Show that G is isomorphic to Z(p^infty), where

Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }

I've been looking at this one for a week and nothing. Can somebody
help! It's driving me crazy.

Jon Gerard, Jun 26, 2003

2. ### Arturo MagidinGuest

You say you have gotten nothing, so some comments, which may help
getting you started; I haven't thought it through in detail, so I
could be wrong in some things. But it seems right to me:

First, note that G is a torsion group; as such, it can be decomposed
into a direct sum of its p-parts, that is

G = G_2 \oplus G_3 \oplus G_5 \oplus ...

where G_p = {g in G; g^{p^m}=e for some m>0}.

Now, let's consider only the non-trivial p-parts; so we have

G = G_{p_1} \oplus G_{p_2} \oplus ...

Note that there cannot be an infinite number of non-trivial p-parts;
if there were, then the subgroup G_{p_2}\oplus ... would be a proper
infinite subgroup. So

G= G_{p_1} \oplus G_{p_2} \oplus ... \oplus G_{p_m};

since G is infinite, at least one of the G_{p_i} is infinite. But
then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
G=G_{p_i}. That is, G is an infinite p-group.

Now consider the quotients G/pG, pG/p^2G, p^2G/p^3G,...
these quotients are vector spaces over F_p, hence they have a
well-defined dimension. Prove that the dimensions are non-increasing,
and that all of them are finite and nonzero. From that, you should be
able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
cyclic; and then that (b) all of them are cyclic; using the fact that
G does not have infinite proper subgroups.

Once you have that all of them are cyclic, it should not be hard to
write down an isomorphism.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Jun 26, 2003

3. ### J. WoodwardGuest

Hmmm...I think I follow you, but I still don't see the actual isomorphism...

J. Woodward, Jun 27, 2003
4. ### Timothy MurphyGuest

But is every proper subgroup of this finite?
Perhaps the question was: every proper subgroup of finite index?

Timothy Murphy, Jun 27, 2003
5. ### Arturo MagidinGuest

Well, assuming everything I said was correct, and you have shown that
G is a p-group, and moreover, that each of p^iG/p^{i+1}G is cyclic,
what does that mean? Can you show that G/p^iG is cyclic for all i>0?

Now, take G/pG. This is cyclic of order p, say with generator
x+pG. Since G/p^2G is also cyclic, you should be able to see that it
is generated by an element which, when multiplied by p, is equal to
x. Call it x/p. Then you need to find one which when multiplied by p
is x/p; call it x/p^2. Etc.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Jun 27, 2003
6. ### Arturo MagidinGuest

Probably because the final two paragraphs are complete bollocks... All
the quotients given will be trivial.

I screwed up; I used the p-powers subgroups instead of the subgroups
of elements of exponent the p-powers...

Here's a better way to proceed; we know that G is an infinite
group. First, consider, pG. If pG were trivial, then G would be of
exponent p, hence an infinite dimensional vector space over F_p, which
clearly has infinite proper subgroups; so pG is nontrivial. If pG
is finite, then G/pG is an infinite dimensional vector space over F_p,
and you can lift an infinite proper subgroup to G. So pG is infinite,
hence pG=G. So G is divisible (being a p-group and p-divisible).

At this point you could invoke classification theorems for divisible abelian
groups, which will tell you that G must be a bunch of copies of
Z_p^{infty}, and thereby conclude that it must be equal to one of
them. But here's a way to proceed without invoking those
classification theorems:

For each n>0, let G[p^n] = {g in G : g^{p^n}=e}, the subgroup of
elements of epxonent p^n.

G[p] is nontrivial; let x in G[p] be different from e1. Since G is
p-divisible, there is an element x_2 such that px_2 = x; an element
x_3 such that px_3=x_2; ...; an element x_{n+1} such that px_{n+1} =
x_n.

Let H = < x, x_2, x_3,...>

It is now easy to verify that H is nontrivial and infinite, so H=G. It
is now also easy to given an isomorphism from Z_{p^{infty}} to G: sent
1/p + Q/Z to x, and 1/p^n +Q/Z to x_n, n>1.

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Jun 28, 2003