Group theory question

[titaniumx3]

Homework Statement

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

Homework Equations

n/a

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! :(

Please help!

 

[Dick]

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

 

[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?

 

[titaniumx3]

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?

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.

 

[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?

 

[ircdan]

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

 

[titaniumx3]

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?

Cyclic group of order p^2?

 

[JasonRox]

Assume G is non-abelian and go by contradiction.

And you'll need ircdan's statement.

It's a nice question.

 

[Kreizhn]

$$C_p \times C_p$$

 

[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 $C_{p^2} \text{ or } C_p \times C_p$

 

[JasonRox]

This question pretty much does that.

 

[titaniumx3]

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 $C_{p^2} \text{ or } C_p \times C_p$

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

 

[ircdan]

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

 

[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.

 

[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 similar 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?

 

[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.

 

[ircdan]

you can produce a counting arguement

 

[titaniumx3]

you can produce a counting arguement

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

 

[Kreizhn]

I imagine you should exploit the class equation

 

[morphism]

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?

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.)

 

[Kummer]

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

Sylow theorems say G has one or three Sylow 2-subgroups or order 16. If there is one such group then the proof is complete. Otherwise we need to show that if it has three Sylow 2-subgroups or der 16 it is still normal. Let H an K be two of these subgroups. And consider $$H\cap K$$ how many elements does it have? Firstly it must divide 16 (which is order of H) by Lagranges theorem and it cannot be 16 because H!=K that means it can have either order 1 or order 4 or order 8. But since $$|HK||H\cap K|=|H||K|$$ it means it cannot be that $$|H\cap K|=1$$ or $$H\cap K|=4$$. So their intersection must share 8 elements. Now $$H\cap K$$ is normal in H and in K (because it is an index 2 subgroup) so the normalizer $$N(H\cap K)$$ contains both H and K. So this normalizer must have order > 16 and be a divisor of 48 and is a multiple of 16. So the only possibility is that $$H(H\cap K)= G$$ so $$H\cap K$$ is a normal subgroup.

 

[titaniumx3]

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.)

Since we have a homomorphism between G and Sym(X), it's kernel would be a normal subgroup in G. This would end the proof, but I'm not sure if it is a proper subgroup.

 

[morphism]

It's a nontrivial homomorphism, because we have 3 Sylow-2s.

 

[titaniumx3]

So say we denote this homomorphism by f,

Then using the fact that |G| = |ker(f)|*|im(f)|, can we say that |ker(f)| != |G| because im(f) is a non-trivial subgroup, hence ker(f) is a proper subgroup of G?

... or is it obvious (by some other notion) that ker(f) is a proper subgroup of G?

 

[JasonRox]

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

Do you have the textbook by Gallian?

Even my graduate textbook has two examples how to show this and I know Gallian has atleast two as well. (I own both.)

One is a counting argument like ircdan is speaking of and the other uses homomorphisms.

 

[titaniumx3]

No unfortunately I don't (though by the looks of it I will pretty soon).

Anyway, I understand the homomorphism based arguement, but I don't understand what exactly you mean by a counting argument (maybe I'm missing something?).

 

[titaniumx3]

I am also stuck on this related question:

From Sylows theorems (or Cauchy's theorem to be exact) we know that if p is prime and p divides |G|, G a finite group, there exists an element of order p contained in G.

How would you go about proving that there exists an element x of order p (as above) such that hcf([G: C_G(x)], p) = 1, i.e. [G: C_G(x)] and p are co-prime.

Note that C_G(x) is the centraliser of x in G and the index [G: C_G(x)] = |Cl(x)| which is the order of the conjugacy class of x.

I know I'm supposed to use Sylow's theorem but I'm not sure what can be said using the theorem.

 

[morphism]

No unfortunately I don't (though by the looks of it I will pretty soon).

Anyway, I understand the homomorphism based arguement, but I don't understand what exactly you mean by a counting argument (maybe I'm missing something?).

Kummer gave you a counting argument. Counting arguments rely on using the Sylow theorems and the class equation and counting orders.

As to why kerf isn't the entire group, it's because the conjugation action isn't trivial. Call the Sylow-2 subgroups H, K and J. Take an element g in G and look at how it acts on the Sylow-2s. If gHg^-1 = H, gKg^-1 = K and gJg^-1 = J for all g in G, that means they're each normal in G, and in fact that H=K=J. But we're assuming that n isn't 1. So some g_0 acts nontrivially on X, and thus kerf cannot be all of G (for instance g_0 isn't in the kernel).

 

[morphism]

I am also stuck on this related question:

From Sylows theorems (or Cauchy's theorem to be exact) we know that if p is prime and p divides |G|, G a finite group, there exists an element of order p contained in G.

How would you go about proving that there exists an element x of order p (as above) such that hcf([G: C_G(x)], p) = 1, i.e. [G: C_G(x)] and p are co-prime.

Note that C_G(x) is the centraliser of x in G and the index [G: C_G(x)] = |Cl(x)| which is the order of the conjugacy class of x.

I know I'm supposed to use Sylow's theorem but I'm not sure what can be said using the theorem.

Can you show us what you refer to as Sylow's theorem?

 

[titaniumx3]

Can you show us what you refer to as Sylow's theorem?

A part of Sylow's theorem states that if the order of G can be written as (p^n)*m (where p and m are coprime) then there exists a subgroup of order (p^n).

I corrollary based on this says that if p divides |G| then there exists and element x of G such that that the order of x is p.

It's pretty much what's given on the wikipedia page about http://en.wikipedia.org/wiki/Sylow_theorems" .

 

[titaniumx3]

I know that |G| = [G: C_G(x)] * |C_G(x)| = |Cl(x)| * |C_G(x)|. I guess if we show that |C_G(x)| is more than (p^n) then |Cl(x)| would be coprime to p. I'm not sure how you'd show something like that or if it is the right method at all.

 

[morphism]

Any element x of order p sits inside a Sylow p-subgroup P. So C(x) = N(x) >= N(P). Now think about how the third Sylow theorem (as stated in wikipedia) helps us.

 

[titaniumx3]

Does it show that [G:C(x)] <= np (where np is the number of Sylow p-subgroups)?

 

[Kreizhn]

I'm not sure if you guys are still going on about trying to show that a Group of order 48 is non-simple, since I really don't want to read all the way back, but like I said before in corroboration with ircdan's counting argument problem, we can solve this very simply by exploiting the class equation.

$$|G| = |Z(G)| + \displaystyle \sum\frac{|G|}{|C_G(a)|}$$

where the summation is over the order of the conjugacy classes of G with $$a$$ a member from each class. This is equivalent to the other definition where

$$\frac{|G|}{|C_G(a)|}$$ is substituted for [tex] [G_i][/tex]

Now, if $$|G| = 48$$ then we have three cases

1) $$\{ 1 \} \neq Z(G) \neq G$$ then G is a proper normal subgroup of G and so G is not simple

2) $$Z(G) = G$$ which implies that G is abelian and as such has a normal subgroup of order 2

3) $$Z(G) = \{1\}$$

For this case, the class equation tells us that

$$48 = 1 + \displaystyle \sum\frac{|G|}{|C_G(a)|}$$

which implies that

$$\sum\frac{|G|}{|C_G(a)|} = 47$$

However, 47 is a prime, and we know that

$$\frac{|G|}{|C_G(a)|}$$ cannot be a prime, thus case 3 cannot happen, and so in all cases, we get that G is simple

 

[titaniumx3]

Unbelievable, that is so much more simple, lol.