Abstract Algebra - no Sylow allowed

1. Oct 7, 2007

nebbish

Abstract Algebra -- no Sylow allowed

Please note Sylow's theorem(s) may not be used.

Using Theorem 1 as a tool, prove that if $$o(G)=p^{n}$$, p a prime number,
then G has a subgroup of order $$p^m$$ for all $$0\leq m\leq n$$.

Theorem 1:
If $$o(G)=p^{n}$$, p a prime number, then $$Z(G)\neq (e)$$.

Theorem 1 uses the class equation to do the proof and I imagine that we are expected to use the class equation to do this exercise as well.

I can use the class equation to set up a proof by induction but my method fails because the class equation does not break G into smaller pieces if G is abelian.

Here's how my attempted solution goes (I basically pretend I'm only dealing with the non-abelian case):

Assume that the theorem is true for n = 1 and proceed by induction on n (it is trivial for
n=1).
In the following, $$p^{n_a}$$ is the order of the normalizer of conjugacy class representative a (when the normalizer is smaller than G) and z is the order of the center:

The class equation is: $$p^n=z+ \Sigma \frac{p^n}{p^{n_a}}$$

Since G is non-abelian the largest value of $$p^{n_a}$$ is $$p^j$$ where
$$1 \leq j \leq n-1$$.

Then by the class equation, $$p^{n-j} \mid z$$. Then $$z = p^{n-j+k}$$ where k < j because G is non-abelian. By induction, the normalizer of a, N(a), has
subgroup H with $$o(H)=p^{j-k-1}$$, and Z(G)H is a group
with $$o(Z(G)H)= o(Z(G))o(H) = p^{n-1}$$.

By the inductive hypothesis, Z(G)H has subgroups of the desired orders, i.e. Z(G)H has a subgroup of order $$p^m$$ for all $$0\leq m\leq n-1$$.

Thus G has a subgroup of order $$p^m$$ for all $$0\leq m\leq n$$.

The problem that I see of course is that I have not addressed the case where G is abelian and that throws off the whole induction.

For the abelian case, there's a little Cauchy Theorem I could use saying that G has a subgroup H of order p (normal because G is abelian), and from there I could infer that $$o(G/H)=p^{n-1}$$. However, I failed to find a bijection between G/H and some subgroup of G.

Last edited: Oct 7, 2007
2. Oct 7, 2007

matt grime

The abelian case is trivial, though. I didn't read your post in detail, but if you don't see that the abelian case is trivial, then something is worrying.

3. Oct 7, 2007

nebbish

Still, a hint would be appreciated. A pointed reference to a theorem or equation that gives the answer away would be fine with me. I've spent tremendous time and effort on this problem and the course in general. I'm disappointed you didn't evaluate the work that I did for the non-abelian case. It suggests my presentation was poor. Perhaps it merely reflects the quirks of my particular textbook, in which case the professor should be able to evaluate the work ok.

I just noticed that in the latter part of my proof I used z when I meant Z(G). That is, I used the order of the center when I meant to reference the center itself. It was just a notational error but I imagine it could cause some confusion.

Last edited: Oct 7, 2007
4. Oct 8, 2007

matt grime

You have a structure theorem for abelian groups, you know precisely what abelian groups look like. Or perhaps you've not been taught it, which would be odd since it is more elementary than Sylow or the class equation.

I didn't read your post in detail because it was late, I will do so later today. But I thought it only fair to warn you you weren't going to get a detailed response for a reason.

5. Oct 8, 2007

nebbish

The text has a theorem stating that every finite abelian group is the direct product of cyclic subgroups, and indeed I think I could use this to prove the abelian case. However, the theorem appears later in the book than the Sylow material, and indeed Sylow material is used in the proof.

6. Oct 9, 2007

matt grime

Really? that seems very odd indeed. It is way more elementary than that, I think.

Anyway, the abelian case: if G is cyclic, there is nothing to show. If G is not cyclic then pick any element h, and let H be the subgroup it generates. By induction H has a subgruop of every correct order, and so does G/H and we're done.

Last edited: Oct 9, 2007
7. Oct 9, 2007

nebbish

If H happens to have order $$p^{n-1}$$ then by induction we are done. However, if H had order of just p, the induction really tells us nothing. It is critical that we find a subgroup of order $$p^{n-1}$$, not just any order less than $$p^n$$.

We could attempt to generate a subgroup of order $$p^{n-1}$$ by taking the product of cyclic subgroups of order less than $$p^n$$. If these subgroups happen to be disjoint then we can easily generate a group of order $$p^{n-1}$$ as such a product. If we were in danger of generating a group the size of all of G we would simply use induction to find a smaller sugroup in one of the subgroups (and of just the right size!) so we end up with a subgroup of order $$p^{n-1}$$. However, with the material I currently have available, it does not seem fair to assume that such disjoint cyclic subgroups exist. They would have to be constructed.

Otherwise, if we simply start taking the product of cycles we have the problem of keeping track of the overlap between the cycles. We need to be sure not only that we generate a group sufficiently large (of order $$p^{n-1}$$) but we must also prove that the final product group we end up with is smaller than all of G (order exactly $$p^{n-1}$$).

By the way, I was very careful in the non-abelian case to construct a subgroup of order exactly $$p^{n-1}$$, no more and no less. (Ok, probably too strong to say I constructed anything. Just showed it exists. Except I still need the abelian case to make it fly!)

P.S. I think I have an idea on how to do the construction now...

Last edited: Oct 9, 2007
8. Oct 9, 2007

nebbish

My original non-abelian proof now seems fatally flawed. The normalizer always contains the center, I can't hope to generate a larger subgroup by the product of the center with some subgroup of a normalizer group.

9. Oct 10, 2007

nebbish

According to my Professor, the key to the problem is to recognize that Z(G) must have an element of order p, say element a. Then "lift up" G/(a) to find a larger proper subgroup within G itself as necessary. Anybody have an idea what "lift up" means? I asked him about it. The phrase seems to be related to the various isomorphism theorems as they relate to factor groups, but I still don't understand what he's saying. Any ideas?

10. Oct 11, 2007

matt grime

Let H=G/(a), this is a group of order p^{n-1}, so inductively there is a sequence of groups

e<H_1 <H_2<..<H_{n-1}=H

with H_i of order p^i.

These are all quotients of subgroups of G by <a>. The preimage of H_i is the lift of H_i.

11. Oct 11, 2007

nebbish

Does this mean there exists a subgroup H of G such that $$o(H/(a)) = p^{n-2}$$
and therefore $$o(H) = p^{n-1}$$ since $$o(H/(a))= \frac{o(H)}{o(a)}$$?

(Since the n =1 case is trivial I am assuming n > 1. Also note that the H I am using is a subgroup of G whereas the H you were using above is a subgroup of G/(a). I'm referring to the pre-image here.)

If so then H would complete the induction for G.

If not then I'm still completely lost.

Also, which isomorphism theorem tells us that subgroups of G/(a) are the quotient group of subgroups of G by (a)?

Last edited: Oct 11, 2007