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Classifying Groups of Order pq²

  1. Aug 24, 2005 #1

    AKG

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    Let p and q be primes with q < p, and q ł (p-1). If G is a group with |G| = qp², then there are two possibilities if G is abelian by the fundamental theorem for finitely generated abelian groups:

    Zp x Zpq and Zqp²

    G has a normal Sylow-p subgroup (of order p²) and if we call it H, and call any q-subgroup K, G is isomorphic to HK, which is isomorphic to K xψ H where ψ sends any k in K to left-conjugation by k. So to classify the remaining groups of order qp², we need simply classify the groups of the form K xψ H where |H| = p², |K| = q and ψ : K → Aut(H) is a homomorphism. If H has an element of order p², it is cyclic, and |Aut(H)| = φ(p²) = p(p - 1), where φ is Euler's φ-function. p is a prime number, so q does not divide p, and q does not divide (p-1) by assumption, so |Aut(H)| has no subgroup of order q. This means that ψ maps every element to the identity, and K xψ H = K x H which is an abelian group, namely Zqp² and has already been accounted for.

    If H contains no element of order p², then every nonidentity element has order p, by Lagrange's theorem. My book suggests that in this case H is necessarily isomorphic to Zp x Zp. I'd like to know why this is the case. I have found the order of Aut(Zp x Zp) to be (p - 1)(p + 1)(p² + 1) given a theorem which states that Aut(Zp x Zp) is isomorphic to GL2(Fp), the general linear group of 2 x 2 matrices with elements from the finite field of order p.

    I have proven that if we have a cyclic group K (and our K is indeed cyclic) and an arbitrary group H, that semi-direct products K xψ1 H and K xψ2 H are isomorphic if ψ1(K) and ψ2(K) are conjugate. Now if q | |Aut(H)| but q² ł |Aut(H)|, then subgroups of Aut(H) of order q will by Sylow subgroups, hence they would all be conjugate. Again, if ψ2(K) = {I}, where I is the identity automorphism, then this semi-direct product is again just a direct product, giving an already-accounted-for abelian group G. Otherwise, ψ2(K) is a subgroup of Aut(H) of order q. So if I can prove that:

    q | |Aut(H)| but q² ł |Aut(H)|

    then there will be a unique isomorphism type for non-abelian G, and I'll be done my classification. Recall |Aut(H)| = (p - 1)(p + 1)(p² + 1) and by assumption q ł (p - 1) so it remains to see how q divides (p + 1)(p² + 1). Notice that:

    (p² + 1) - (p + 1) = p(p - 1)

    so if q | (p + 1) and q | (p² + 1), then clearly q | [(p² + 1) - (p + 1)], which is a contradiction, so q only divides one of (p + 1) and (p² + 1). So to complete the classification, we have to show existence by showing that either q | (p + 1) or q | (p² + 1), and show uniqueness by showing that both q² ł (p + 1) and q² ł (p² + 1).

    The sentences in bold are the places where I have problems. Anybody have any suggestions as to how to solve these problems? Thanks.
     
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  3. Aug 24, 2005 #2

    matt grime

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    |H| is p^2 and is therefore abelian it is not cyclic since there is no element of oder p^2 so it must be as stated in the first bold sentence.
     
  4. Aug 24, 2005 #3

    AKG

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    How do I know H is abelian if it's order is p²? Also, would you be able to shed any light on the second problem (regarding q dividing (p + 1), etc.)? Thanks.
     
  5. Aug 24, 2005 #4

    matt grime

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    because it is an elemenatary problem that is done in any group theory course? (this of course means that if i am asked to prove it then i'll fail miserably, but it should be straight forward. it certainly has nontrivial centre, so wither it is abelian or it is a non abelian extension of Z_p by Z_p and they don't exist as we must be able to show easily if we tried)
     
  6. Aug 24, 2005 #5

    AKG

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    Yeah, it seems that it should have a non-trivial center, but I'm not sure why. But it's center, if it is non-trivial, has order p and is normal. Any x not in Z(G) will generate <x>, a subgroup of order p intersecting the center trivially, so G = <x>Z(G). Therefore, G is isomorphic to <x> xf Z(G) where f : <x> --> Aut(Z(G)) is a homomorphism. Z(G) is isomorphic to Zp, so Aut(Z(G)) is isomorphic to Zp-1, and since p-1 and p are co-prime, f maps every element of <x> to the identity automorphism, leaving us with a direct product, and thus an abelian group (and in fact this gives us a contradiction because Z(G) would not have order p, it would have order p², i.e. Z(G) = G). But isn't it possible that Z(G) = {e}? Well, I guess not, but how do we know that this case will also lead to a contradiction?
     
  7. Aug 24, 2005 #6

    matt grime

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    no, the venter is abso,utely non-trivial, and that is easily deduced from the class equation (or just consider the gruop as the union of the conjugacy classes each has order dividing p^2, none has order p^2 and at least 1 has order 1, that containing the identity, the others have order p, but then p^2=|Z(G)| +p*{no of conjugacy classes with p elements} and as we noted Z(G) contains at least the identiy. it must contain other elements otherise the RHS is 1 mod p but the LHS is 0 mod p. indeed this shows that the center of any p group is nontririval and by induction that all p groups are solvable.
     
  8. Aug 24, 2005 #7

    AKG

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    I just noticed that my book actually has a proof that a group of order p² is either (Zp)² or Z, but it was in a section that was supposedly covered by the other book I read so I missed it (it was given in this book simply as a corollary to the class equation). Thanks again for your help, now to figure out the second problem
     
  9. Aug 25, 2005 #8

    AKG

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    Actually, the second problem is going to be more problematic than I thought:

    5 doesn't divide 7-1, and 5 doesn't divide 7+1, but not only does 5 divide 7²+1, 5² divides 7²+1.

    11 doesn't divide 13-1, 11 doesn't divide 13+1 and 11 doesn't divide 13²+1.

    So we have a case where we have primes satisfying the hypothesis, but the smaller prime doesn't divide either of those other factors of |Aut(H)|, so we'll only have abelian groups (2 of them), and we have a case where the smaller prime does divide the order of |Aut(H)|, but that smaller prime has a power that divides |Aut(H)|, so we could have numerous subgroups of Aut(H) of order q that aren't conjugate (since they wouldn't be Sylow subgroups, so we aren't guaranteed that they're conjugate) and thus semi-direct products might gives us more than just one more non-abelian isomorphism type.

    I guess the question that arises next (although it seems like it would be too much to ask) is whether we can predict in general when q will divide |Aut(H)| or not, and in those cases when it does, where we can predict if q will not divide |Aut(H)|/q, making subgroups of order q of Aut(H) Sylow subgroup, and hence conjugate, and hence ensuring that there is only one more isomorphism type (which will be non-abelian). We could also ask whether we can do any sort of classification even if q² divides Aut(H), but that seems like an even bigger task.

    On the other hand, my book says, after asking me to classify groups of order 3*5², that the classification of groups of order qp² is similar. Maybe someone knows a nice way to show that cases where q doesn't divide |Aut(H)| at all and cases when q² divides |Aut(H)| are rare, or easily predictable, and that cases where qn divides |Aut(H)| are still easily classifiable. It seems like a long shot, but I guess it doesn't hurt to ask.
     
  10. Aug 25, 2005 #9

    mathwonk

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    the more elementary parts of your question, groups of order p^2, are discussed on page 17 of the free class notes 843.1, on the webpage below:

    http://www.math.uga.edu/~roy/
     
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