# Homework Help: Group theory question

1. Nov 14, 2007

### titaniumx3

1. The problem statement, all variables and given/known data

If p is a prime and G is group of order p^2, then show that G is abelian.

2. Relevant equations

n/a

3. The attempt at a solution

I first consider Z(G), the centre of G. Since it is a normal subgroup of G, then by Lagrange's Theorem, |Z(G)| divides |G|. Hence |Z(G)| = 1, p or p^2. We know that Z(G) not the trivial subgroup (proof already given) hence it must be of order p^2 or p.

If |Z(G)| = p^2, then Z(G) = G and hence by definition it is abelian.

If |Z(G)| = p, then .... well this is where I am stuck! :(

2. Nov 14, 2007

### Dick

Consider an element q that is not in Z(G). How big is the subgroup generated by q and Z(G)?

3. Nov 14, 2007

### Kreizhn

In the case of your argument, if $|Z(G)| = p$ then we have that

$\frac{|G|}{|Z(G)|} = p$ which we can't have.

So on the other hand, we know that if $\exists a \in G$ such that $o(a) = p^2$ where $o(a)$ is the order of a, then $G = C_{p^2}$. Thus we can assume that every non-identity element has order p, since the order of the elements must divide the order of the group.

Thus consider a non-identity element of G, say a, and the subgroup it generates. Furthemore, consider another non-identity element, say b, that is not in $<a>$. Such an element is guaranteed to exist since $o(a) = p \Rightarrow |<a>|\neq|G|$. Consider the subgroup generated by $b$.

What can we say about the order of $<a>$ and the order of $<b>$ ?. What can we say about their intersection? What can we say about the order of their product?

4. Nov 14, 2007

### titaniumx3

If we assume that every non-identity element has order p, then <a>, <b> would have order p also. Wouldn't their intersection be the empty set if b is defined as an element not in <a>? Sorry I'm not sure where I'm supposed to go with this.

5. Nov 14, 2007

### Kreizhn

That's precisely correct. They're intersection is empty, and so

$$|<a> \times <b>| = \frac{|<a>||<b>|}{|<a>\cap<b>|} = |<a>||<b>| = p^2$$

Thus, since $a \in G, \; b\in G, \; \text{ and } |<a> \times <b>|=|G|$ then

$$<a> \times <b> = G$$

Now the question is, what is $<a> \times <b>$ isomorphic to?

6. Nov 14, 2007

### ircdan

for any group G, if G/Z(G) has prime order, then G/Z(G) is cyclic, hence G is abelian

7. Nov 14, 2007

### titaniumx3

Cyclic group of order p^2?

8. Nov 14, 2007

### JasonRox

Assume G is non-abelian and go by contradiction.

And you'll need ircdan's statement.

It's a nice question.

9. Nov 14, 2007

### Kreizhn

$$C_p \times C_p$$

10. Nov 14, 2007

### Kreizhn

Via this method we've actually proved something stronger than the actual question. Namely that every group of order $$p^2$$ is isomorphic to either $C_{p^2} \text{ or } C_p \times C_p$

11. Nov 14, 2007

### JasonRox

This question pretty much does that.

12. Nov 14, 2007

### titaniumx3

Is $$C_p \times C_p$$ also cyclic? How do you know it is abelian?

13. Nov 14, 2007

### ircdan

C_p x C_p is not cyclic but it is abelian since each factor is

14. Nov 14, 2007

### JasonRox

This question tells us that all groups of order p^2 are abelian.

What kind of abelian groups are there of order p^2?

Well, it can be cyclic if it has an element of order p^2 if not, then all the elements must be of order p (Lagrange's Theorem). And you work from there.

15. Nov 14, 2007

### titaniumx3

Thanks for all the help! I've written the proof both ways and they seem to be pretty much equivalent.

I'm not sure if I should open a new thread for this but I would appreciate some help on another related question:

"If G is a group of order 48 then show that it is not simple"

Now |G| = 2^4 * 3. I'm thinking it would be a similiar argument to how you show groups of order (p^2 * q) are not simple, but in all honesty I don't even understand the proof for that very well.

I guess if you apply Sylow's theorem, then clearly there are Sylow p-subgroups of order 2^4. Are these p-subgroups normal? If so, how would you go about showing this?

16. Nov 14, 2007

### Kummer

A theorem by Burnside states that the center of a finite p-group is non-trivial. So if |G|=p^2 and Z(G)!=G choose x in G so that x not in Z(G). We know by divisibility that Z(G) >= p (Burnside). But that means the centralizer C(x) must be G a contradiction. So Z(G)=G.

17. Nov 15, 2007

### ircdan

you can produce a counting arguement

18. Nov 15, 2007

### titaniumx3

Can you elaborate on that? (I assume you are talking about |G| = 48 not being simple)

19. Nov 15, 2007

### Kreizhn

I imagine you should exploit the class equation

20. Nov 15, 2007

### morphism

How many of these can we have? Let n be the number of Sylow-2's. Then n=1(mod2) and n|3. This leaves us with n=1 or n=3. If n=1, we're done, because this implies the unique Sylow-2 is normal in G (why?). So suppose n=3. One possible way you can proceed from here is via group actions. Let G act on the set X of 3 Sylow-2's by conjugation. This induces a nontrivial homomorphism from G into Sym(X). What is its kernel? (Don't think too hard about what the kernel actually is; think about what kernels are, and how they could be relevant to proving the non-simplicity of G.)

Last edited: Nov 15, 2007