1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Group theory question

  1. Nov 14, 2007 #1
    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! :(

    Please help!
     
  2. jcsd
  3. Nov 14, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Consider an element q that is not in Z(G). How big is the subgroup generated by q and Z(G)?
     
  4. Nov 14, 2007 #3
    In the case of your argument, if [itex] |Z(G)| = p [/itex] then we have that

    [itex]\frac{|G|}{|Z(G)|} = p[/itex] which we can't have.

    So on the other hand, we know that if [itex] \exists a \in G[/itex] such that [itex] o(a) = p^2[/itex] where [itex] o(a) [/itex] is the order of a, then [itex] G = C_{p^2} [/itex]. 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 [itex]<a>[/itex]. Such an element is guaranteed to exist since [itex]o(a) = p \Rightarrow |<a>|\neq|G| [/itex]. Consider the subgroup generated by [itex]b[/itex].

    What can we say about the order of [itex] <a> [/itex] and the order of [itex] <b> [/itex] ?. What can we say about their intersection? What can we say about the order of their product?
     
  5. Nov 14, 2007 #4
    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.
     
  6. Nov 14, 2007 #5
    That's precisely correct. They're intersection is empty, and so

    [tex] |<a> \times <b>| = \frac{|<a>||<b>|}{|<a>\cap<b>|} = |<a>||<b>| = p^2 [/tex]

    Thus, since [itex] a \in G, \; b\in G, \; \text{ and } |<a> \times <b>|=|G| [/itex] then

    [tex] <a> \times <b> = G [/tex]

    Now the question is, what is [itex] <a> \times <b> [/itex] isomorphic to?
     
  7. Nov 14, 2007 #6
    for any group G, if G/Z(G) has prime order, then G/Z(G) is cyclic, hence G is abelian
     
  8. Nov 14, 2007 #7
    Cyclic group of order p^2?
     
  9. Nov 14, 2007 #8

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Assume G is non-abelian and go by contradiction.

    And you'll need ircdan's statement.

    It's a nice question.
     
  10. Nov 14, 2007 #9
    [tex] C_p \times C_p [/tex]
     
  11. Nov 14, 2007 #10
    Via this method we've actually proved something stronger than the actual question. Namely that every group of order [tex] p^2 [/tex] is isomorphic to either [itex]C_{p^2} \text{ or } C_p \times C_p [/itex]
     
  12. Nov 14, 2007 #11

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    This question pretty much does that.
     
  13. Nov 14, 2007 #12
    Is [tex]C_p \times C_p[/tex] also cyclic? How do you know it is abelian?
     
  14. Nov 14, 2007 #13
    C_p x C_p is not cyclic but it is abelian since each factor is
     
  15. Nov 14, 2007 #14

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  16. Nov 14, 2007 #15
    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?
     
  17. Nov 14, 2007 #16
    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.
     
  18. Nov 15, 2007 #17
    you can produce a counting arguement
     
  19. Nov 15, 2007 #18
    Can you elaborate on that? (I assume you are talking about |G| = 48 not being simple)
     
  20. Nov 15, 2007 #19
    I imagine you should exploit the class equation
     
  21. Nov 15, 2007 #20

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    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
  22. Nov 15, 2007 #21
    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 [tex]H\cap K[/tex] 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 [tex]|HK||H\cap K|=|H||K|[/tex] it means it cannot be that [tex]|H\cap K|=1[/tex] or [tex]H\cap K|=4[/tex]. So their intersection must share 8 elements. Now [tex]H\cap K[/tex] is normal in H and in K (because it is an index 2 subgroup) so the normalizer [tex]N(H\cap K)[/tex] 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 [tex]H(H\cap K)= G[/tex] so [tex]H\cap K[/tex] is a normal subgroup.
     
  23. Nov 15, 2007 #22
    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.
     
  24. Nov 15, 2007 #23

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    It's a nontrivial homomorphism, because we have 3 Sylow-2s.
     
  25. Nov 15, 2007 #24
    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?
     
  26. Nov 15, 2007 #25

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook