- #1

AKG

Science Advisor

Homework Helper

- 2,565

- 4

## Main Question or Discussion Point

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:

Z

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

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

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

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).

The sentences in bold are the places where I have problems. Anybody have any suggestions as to how to solve these problems? Thanks.

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 case

**H 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.