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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


    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


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    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


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    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


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    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


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    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


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    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


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    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


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    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.
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