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:(adsbygoogle = window.adsbygoogle || []).push({});

Z_{p}x Z_{pq}and Z_{qp²}

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 x_{ψ}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 x_{ψ}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 x_{ψ}H = K x H which is an abelian group, namely Z_{qp²}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 caseH is necessarily isomorphic to Z. I'd like to know why this is the case. I have found the order of Aut(Z_{p}x Z_{p}_{p}x Z_{p}) to be (p - 1)(p + 1)(p² + 1) given a theorem which states that Aut(Z_{p}x Z_{p}) is isomorphic to GL_{2}(F_{p}), 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 x_{ψ1}H and K x_{ψ2}H are isomorphic if ψ_{1}(K) and ψ_{2}(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 ψ_{2}(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, ψ_{2}(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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# Classifying Groups of Order pq²

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