A finite groupGis ap-groupif the order ofGa power of a primep.
Each abelianp-group is a direct productG=Cpn1 ×Cpn2 × · · · ×Cpnk of cyclicp-groups, there being
one isomorphism class of such groups for every set of positive integers{n1, . . . , nk}. When allni = 1,
the groupCk
p =Cp×Cp× · · · ×Cpis calledelementary abelian of rankk. The dihedral groupsD2n
and generalized quaternion groupsQ2n are examples of nonabelian2-groups.
One cannot hope to classify allp-groups, except those whose orders are small powers ofp.
Proposition 3.11 Letpbe a prime and letGbe ap-group. 1. If|G|=pthenG'Cp.
2. If |G| = p2 then Gis abelian. We have G ' C
p2 if Gis cyclic and G ' Cp ×Cp if Gis not
cyclic.
3. If |G| = p3 then either G is one of two nonabelian groups or G is one of Cp3, Cp × Cp2 or
Cp×Cp×Cp.
Proof: We already noted that part 1 is a consequence of Lagrange’s theorem. We will prove part 2 here, and postpone the proof and a more detailed statement of part 3.
Assume|G| = p2. We have seen in Prop. 2.15 that every p-group has a nontrivial centerZ(G). By
Lagrange’s theorem, we have|Z(G)| = por p2. If|Z(G)| = p then G/Z(G) has order p, hence is cyclic, soGis abelian, contradictingZ(G)6=G. HenceGis abelian.
The order of every element ofGalso dividesp2. IfGhas an element of orderp2thenG'C
p2. Assume
Ghas no element of orderp2. Then every nonidentity element ofGhas orderp. Chooseh, k ∈Gwith
h 6= 1andk /∈ hhi. The subgroupsH = hhiand K = hkihaver orderpand are both normal in the abelian groupG. NowHK is a subgroup ofGproperly containing H. Since [G: H] = p, it follows thatHK =G. Likewise,H∩K is a proper subgroup ofK, which has orderp, soH∩K ={1}. Now
by Prop.2.10we haveG'H×K 'Cp×Cp.
Prop. 2.15can be extended to prove the converse of Lagrange’s theorem forp-groups. First we need a lemma.
Lemma 3.12 IfA is a finite abelian group whose order is divisible by a primepthenA contains an element of orderp.
Proof: By induction, we may assume the result is true for groups of smaller order. Letb ∈ A have orderm >1, and letB =hbi. Ifp|mthenbm/phas orderp. Assumep
-m. Thenpdivides|A/B|and |A/B|< |A|, soA/B has an element of orderp, by the induction hypothesis. This element is aBfor somea∈Asuch thata /∈B, butap ∈B. Thereforeap =brfor some integerr. Sincegcd(p, m) = 1,
we can writer =kp+`mfor integersr, `. The elementc=ab−kdoes not belong toBsincea /∈B, but sinceAis abelian we have
cp =apb−kp =br−kp =b`m = 1.
Hencec∈Ahas orderp.
As we will see in the next result, the lemma is true without the assumption thatA is abelian, but the proof is not as constructive.
Proposition 3.13 Let G be a finite group of order pr, where p is a prime. Then G has a chain of subgroups
1 =G0 < G1 < G2 <· · ·< Gr−1 < Gr =G such that for all0≤i < rwe have
1. |Gi|=pi;
2. Gi is a normal subgroup ofGandGi+1/Gi 'Cp; 3. Gi+1/Gi is contained in the center ofG/Gi.
Proof: We argue by induction on r. By Prop. 2.15, the centerZ(G)is a nontrivial abelianp-group. By Lemma3.12, there exists a subgroup G1 < Z(G) of orderp. Since G1 is central in G we have
G1/ G. The groupG=G/G1has orderpr−1. Applying the induction hypothesis toG, there is a chain
of subgroups
1 =G0 < G1 < G2 <· · ·< Gr−2 < Gr−1 =G
such that for all0≤i < r−1we have|Gi|=pi andGi/ GandGi+1/Giis contained in the center of
G/Gi.
By the Correspondence Theorem applied toG/G1there are normal subgroupsGi EGsuch that
Gi =Gi/G1.
Moreover, the canonical projectionG→Ginduces isomorphisms
G/Gi
∼
−→G/Gi
which restrict to isomorphisms
Gi+1/Gi
∼
−→Gi+1/Gi
for each0≤i < r. It follows thatGi+1/Giis contained in the center ofG/Gi, as claimed.
Thus, everyp-group has a tower of normal subgroups whose quotients are cyclic of orderp. Despite this apparent simplicity, the number of isomorphism classes of groups of orderpr grows rapidly with
|G| number of groups 2 1 22 2 23 5 24 15 25 51 26 267 27 2 328 28 56 092 29 10 494 213 210 49 487 365 422
It has been determined1that the total number of all groups of order≤2000is49 910 529 484, so over
99%of these groups have order210.