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Ametacyclic groupis a groupGhaving a cyclic normal subgroupA with cyclic quotientB = G/A. Any cyclic groupC is metacyclic. Indeed, we can takeA to be any subgroup ofC. ThenAis cyclic and so isB =C/A.

The classification of metacyclic groups is simpler if we specify the groupsA andB in advance. For this we use the language of extensions: Ametacyclic extensionis an extension

1−→A−→ι G−→π B −→1. (47) whereAandB are cyclic. In this section we classify metacyclic extensions withGfinite. So we fix cyclic groupsA'CmandB 'Cn, as well as generatorsα, βofAandB, respectively.

Proposition 9.11 Let (G, ι, π) be a metacyclic extension of B by A. Let a = ι(α) and choose an elementb∈Gsuch thatπ(b) =β. Then

1. Every element ofGcan be written uniquely as asai_bj_{, fori∈}

Z/mZand some integer0≤j < n. 26

2. The groupGhas the presentation

G' ha, b|am =e, bab−1 =aq, bn=ari (48)

for some elementsq, r in_Z/m_Zsuch that

qn= 1, and qr =r. (49)

26_{Note thatbneed not have ordern, so we cannot writej∈}

3. Let (G1, ι1, π1)be another extension ofB by A, let a1 = ι1(α)and choose b1 ∈ G1 such that

π1(b1) = β. Letq1, r1 be as in part 2, fora1, b1. Then the extensions (G1, ι1, π1) and(G, ι, π)

are equivalent if and only ifq1 =qandr1 =r+ (1 +q+· · ·+qn−1)k, for somek∈Z/mZ.

Proof: For part 1, letx ∈ Gis an arbitrary element, we haveπ(x) = βj _{for some j ∈}

Z, and also π(bj) = βj, so xb−j ∈ kerπ = hai, which means that x = aibj for some i ∈ _Z/m_Z and integerj. Sinceπ(bn_{) = π(b)}n _=βn _{= 1,we haveb}n _{∈kerπ =hai, sob}n _=ar_{, for somer ∈}

Z/mZ. Hence in

the expressionx=ai_bj _{we may replacej by its remainder when divided byn.}

We are given thatα has orderm and β has order n. Since ι is injective this means a has order m. And sincehai _EGwe havebab−1 = aq_{for someq ∈(}

Z/mZ)×. We have proved thatGsatisfies the

relations (48). These calculations also show that the groupha, b | am _{=e, bab}−1 _=aq_{, b}n _{= a}r_i

has order at mostnm, hence is isomorphic toG.

To see thatq, r satisfy (49), note thatbn _{= a}r _{commutes witha. Hencea = b}n_ab−n _{= a}(qn₎

, which impliesqn _{= 1. Also b commutes withb}n _{= a}r_{, so we havea}r _{= ba}r_b−1 _{= (bab}−1₎r _{= a}qr_{,so that}

r=qr, proving (49).

For part 3, let(G1, ι1, π1)be another extension ofBbyA, setι1(α) = a1and chooseb1 ∈G1such that

π1(b1) = β. Applying part 2 to this extension, we get relations analogous to (48) and (49), namely

am₁ =e, b1a1b−11 =a q1 1 , b n 1 =a r1 1 , (50) along with q₁n = 1, and q1r1 =r1. (51)

Suppose the extensions(G, ι, π)and(G1, ι1, π1)are equivalent. This means there is an isomorphism

f : G1 → Gsuch that f ι1 = ι and πf = π1. The former relation means that f(a1) = awhile the

second implies thatf(b1)is another lift ofβ inG. Hence we have

f(b1) =akb,

for somek ∈_Z/m_Z. And sincef(a1) =a, we have

ar1 _=f(ar1

1 ) = f(b1)n= (akb)n=a(1+q+···+q

n−1_)k+r

,

where the last equality follows by induction from the relations bab−1 = aq _{and b}n _{= a}r_{. Hence we}

haver1 = (1 +q+· · ·+qn−1)kinZ/mZ, as claimed.

Finally, we have

aq1 _=f(aq1

1 ) = f(b1a1b−11 ) =f(b1)·a·f(b1)−1 =akb·a·b−1a−k =ak·aq·a−k=aq,

soq1 =qinZ/mZ.

Conversely, if r1 = (1 + q +· · · +qn−1)k in Z/mZ then the above calculations show that a and

ak_{b satisfy the relations of a}

1 andb1, so there is a surjective homomorphismf : G1 → Gsuch that

f(a1) = a and f(b1) = akb. These equations imply that f ι1 = ι and πf = π1. Finally, f is an

We next prove that such extensions exist, whenever the conditions (49) are satisfied.

Proposition 9.12 Supposeq, rare elements of_Z/m_Zsatisfyingqn _{= 1andqr =r. Then there exists} an extension

1−→A−→ι G−→π B −→1

and an elementb∈Gwithπ(b) =β, such that

bab−1 =aq, and bn=ar.

Proof: We will constructGas a quotient of a split extension. LetC be a cyclic group of order mn, choose a surjective homomorphismψ :C →B, and letγbe a generator ofCsuch thatf(γ) = β. Sinceqn _{= 1, we haveq}mn_{= 1, so there is a homomorphism}

ϕ:C →(_Z/m_Z)× = Aut(A), such that ϕ(γ) = q.

Form the semidirect productGe = Ao_ϕC, where we have the relationγαγ−1 =αq. We observe that

γnαγ−n = α(qn) =αandγαrγ−1 =αqr =αr. Thus, αcommutes withγnandγ commutes withαr. SinceGe is generated byα andγ, this implies that the cyclic subgroupH_r =hα−rγniis contained in the center ofGe. In particular,H_rEGe, so we can form the quotient group

Gr:=G/He _r.

Let ιr : A → Gr be the projection of A ⊂ Ge to G_r. This is injective since A ∩H_r = {1}. Set

a=ιr(α) =αHrandb=γHr, both elements ofGr. Thenaandbsatisfy the relations

am = 1, bab−1 =aq, bn=ar.

Letπ˜ :Ge→B be the composition

˜

π:Ge

p

−→C −→ψ B,

wherepis the natural projectionG=A_oC →C. We haveπ˜(αi_γj_{) = β}j_{. In particular,π(α}−r_γn_{) =}

βn _{= 1. Thus, π˜ induces a surjective map π}

r : Gr → B such that πr(aibj) = βj. We see that

πr(b) =β and thatιr(A)≤ kerπr. On the other hand, ifπr(aibj) = 1thenn| j, sayj =nk, and we

haveai_bj _=ai_ark _∈ι

r(A). This shows thatkerπr =ιr(A)and completes the proof that(Gr, ιr, πr)is

the desired extension.

The conditions on r and q can be understood more simply if we regard “q” as the endomorphism of _Z/m_Z given by multiplication by q. Likewise we view q − 1 and qn := 1 + q +· · · + qn−1

as endomorphisms of _Z/m_Z. The condition rq = r means that r ∈ ker(q − 1). The condition

qn = 1means thatimqn ⊂ ker(q−1). Part 3 of the proposition means that the equivalence class of

the extension (G, π)depends only on the class of r inker(q−1)/imNq. In this language, the two

Theorem 9.13 Fix generatorsαofA ' Cm andβ ofB ' Cn and letq ∈ (Z/mZ)×satisfyqn = 1.

Then there is a bijection from the subquotient ker(q −1)/imqn of Z/mZ to the set of equivalence classes of extensions(G, ι, π)ofB byAsuch that any lift ofβinGacts onι(A)by the powerq. To the class ofr ∈ker(q−1)/imqncorresponds the equivalence class of the extension(Gr, ιr, πr)where

Gr =ha, b|am =e, bab−1 =aq, bn=ari,

ιr(α) =aandπr(b) = β. This extension splits iff the class ofrinker(q−1)/imqnis zero.

We have now classified all metacyclic extensions, and have shown that every metacyclic group is isomorphic to one of the groups

G(m, n, q, r) =ha, b|an=e, bab−1 =aq, bn=ari,

whereq, r ∈_Z/msatisfyqn = 1and(q−1)r = 0. However, the same group can appear in different extensions. For example, ifgcd(j, n) = 1thenker(qj_{−1) = ker(q−1)on}

Z/mZ G(m, n, qj, r)'G(m, n, q, r)

viaa 7→ a, b7→ bj. Hence the groupG(m, n, q, r)depends only on the subgroup of(_Z/m_Z)×gener- ated byq.