Automorphisms of Metacyclic p-Groups with Cyclic Maximal Subgroups

Rose-Hulman Undergraduate Mathematics Journal

Rose-Hulman Undergraduate Mathematics Journal

Volume 2

Issue 2

Article 4

Automorphisms of Metacyclic p-Groups with Cyclic Maximal

Automorphisms of Metacyclic p-Groups with Cyclic Maximal

Subgroups

Subgroups

Mark Schulte

St. Olaf College, mark.schulte@us.wmmercer.com

Follow this and additional works at: https://scholar.rose-hulman.edu/rhumj

Recommended Citation

Recommended Citation

Schulte, Mark (2001) "Automorphisms of Metacyclic p-Groups with Cyclic Maximal Subgroups," Rose-Hulman Undergraduate Mathematics Journal: Vol. 2 : Iss. 2 , Article 4.

Automorphisms of Metacyclic p-Groups With Cyclic Maximal Subgroups

Mark Schulte

Abstract

This paper deals with the determination of the automorphism group of the metacyclicp-groups,P(p, m), given by the presentation

P(p, m) =x, y|xpm = 1, yp= 1, yxy−1=xpm−1+1 (1) wherepis an odd prime number andm >1. We will show that Aut(P) has a unique Sylowp-subgroup,S_p, and that in fact

Aut(P)∼=SpZp−1.

A general metacyclicp-group has a presentation of the form

G=x, y|xpm = 1, ypn=xpq, yxy−1=xpl+1,

with certain restrictions on the parameters m, n, q, and l [4]. Thus, x is maximal in G if and only if |G| = pm+1 _{[3]. Checking the conditions on the}

parameters, we see that this forcesn= 1,q = 0, andl =m−1. This implies that the groupsP(p, m) described in the abstract are exactly those metacyclic p-groups with a cyclic maximal subgroup. Note that when m= 2, we get one of the extra-special groups of order p3. While we will restrict our study of automorphisms to metacyclic p-groups of odd order, it may be useful to note thatP(2,2) is the dihedral group of order 8.

We begin our examination of the group of automorphisms of P =P(p, m), Aut(P), with the fact that any automorphism is determined by its action on the generators of P(p, m). Also, the third relation in equation 1 implies that any element ofP(p, m) can be written uniquely in the formxa_yb_{, where 0≤a≤p}m_,

and 0≤b≤p. We therefore define any automorphismφ on the groupP(p, m) byφ(x) =xi_yj _{andφ(y) =x}r_ys_.

Throughout this paper we will represent an automorphismφby the matrix

i r j s

and will writeφ=

i r j s

. This does not imply thatφis a matrix. Rather, we are merely using matrix notation to organize and represent the action ofφupon the generators ofP(p, m).

Proposition 1 Ifφ=

i r j s

∈Aut(P), thenr≡0 modpm−1_{, j ∈Z} p, s≡

1 modp, andi∈U(pm_{), where U(p}m_{) is the group of integers under}

Proof. From the first and second relations in equation 1 we conclude that i and r are integers considered modpm while j and s are integers considered modp. We complete the proof of this proposition by using the results of the following three claims.

Claim 2 If ris as in Proposition 1, thenr≡0 modpm−1.

Proof. We begin by noting that for p > 2, P(p, m) is a regular group [1]. This implies that for anyg₁, g₂∈P(p, m) andpa prime number, we have that (g1g2)pk =gp₁kg₂pkzpk where z ∈ [P, P], the commutator subgroup of P(p, m). From the third relation in the presentation of P(p, m) we see that [P, P] = xpm−1_{. Thus, for any k ≥ 1, we have that (g}

1g2)pk = g₁pkg₂pk. As a result, an application of an automorphism in Aut(P) to the second relation in the presentation ofP(p, m) yields

1 =φ(1) =φ(yp) =φ(y)p= (xrys)p=xrpysp=xrp. It follows thatrp≡0 modpm_{, which implies thatr≡0 modp}m−1_.

✷ Claim 3 If iis as in Proposition 1, then i∈U(pm_).

Proof. Since P(p, m) is a finite p-group, its Frattini subgroup is Φ(P) = Pp[P, P] [5]. It follows that Φ(P) =xpandP/Φ(P)∼=Zp×Zp. Also, Φ(P) is

characteristic so any automorphismαin Aut(P) restricts to an automorphism of Φ(P).

Let Θ : Aut(P) −→ Aut (P/Φ(P)) be the natural homomorphism defined by Θ(α) (xΦ(P)) =α(x)Φ(P), whereα∈Aut(P) andxΦ(P)∈P/Φ(P).

Now, Aut (P/Φ(P))∼= Aut(Zp×Zp) =GL2(Zp), the group of 2×2 invertible

matrices overZ_p. Thus, Θ(α) = Θ i r j s = i r j s

, where a “bar” signifies that an

entry is considered modp. However,

i r j s

∈GL₂(Zp) implies thatis−rj ≡

0 modp, while Claim 2 states thatr≡0 modp. It follows thati≡0 modp. Using this result and the fact that values of i are considered mod pm_{, we}

arrive at the conclusion that we can consideri∈U(pm_).

✷ We next prove a lemma which will be used extensively in Claim 5.

Lemma 4 Let x, y be the generators of P(p, m),α=pm−1_{+ 1, and a, b >0.}

Proof. The third relation in equation 1 states that yxy−1 = xα, which is equivalent toyx=xαy.By applying this relation toybxa we have

ybxa = yb−1(yx)xa−1 = yb−1(xαy)xa−1 .. . = xαb_yb_xa−1 .. . = x2αbybxa−2 .. . = xaαb_yb_. ✷ Claim 5 If sis as in Proposition 1, then s≡1 mod p.

Proof. We apply an automorphism to both sides of the third relation in equation 1 and then use the result of the previous lemma. By applyingφto the left hand side of the third relation we have

φ(yxy−1) = (xrys)(xiyj)(xrys)−1=xr+iαs−rαjyj. Similarly, an application ofφto the right hand side yields

φ

xpm−1+1

= (xiyj)pm−1(xiyj) =xi(pm−1+1)yj. This implies that

xr+iαs−rαj_yj _{= x}iα_yj

⇔ r+iαs_−rαj _{≡ iαmodp}m

⇔ r1−αj _{≡ i(α−α}s_{) modp}m_.

Now, α = (pm−1_{+ 1) implies that α ≡ 1 modp, so it follows that α}j _≡

1 modp, and hence 1−αj _{≡0 modp. Likewise, r≡0 modp}m−1 _{implies that}

r(1−αj_{)≡ 0 modp}m_{. Applying this to our result above, we must have that}

i(α−αs_{)≡0 modp}m_{. However, i≡0 modpimplies thati≡0 modp}m_.Thus,

i(α−αs_{)≡0 modp}m _{if and only ifα≡α}s _modpm_.

Sinceαis relatively prime topmwe can conclude thatα∈U(pm) andα−1 exists mod pm. As a result we have that αs−1 ≡ 1 modpm, so |α| | (s−1). Furthermore, a binomial expansion ofαpproduces

αp _{= p}m−1_{+ 1}p = p(m−1)p₊ p 1 p(m−1)(p−1)_{+· · ·+} p p−1 pm−1_{+ 1.}

Using the fact thatp|

p k

for 1≤k≤p−1, we see thatαp _{≡1 modp}m_.

✷ Finally, analysis of the relations ofP(p, m) yields no limits onjexcept that its values are evaluated modp. This implies that we can considerj∈Z_p, which completes the proof of Proposition 1.

✷ Our next task is to use our previous results to determine the structure of Aut(P). We begin with a definition and several lemmas which will be used extensively in the remainder of this paper.

Definition 6 Let α=pm−1_{+ 1 anda, b >0. Then we defineΛ(a, b)by}

Λ(a, b) = 1 +αa+α2a+· · ·+α(b−1)a. Lemma 7 Λ(a, b)≡bmodpk _{for all1≤k≤m−1.}

Proof. Let 1≤k≤m−1.Then we have the following: Λ(a, b) = 1 +αa+α2a+· · ·+α(b−1)a

= 1 + (pm−1_{+ 1)}a_{+ (p}m−1_{+ 1)}2a_{+· · ·+ (p}m−1_{+ 1)}(b−1)a

≡ 1 + 1a_{+ 1}2a_{+· · ·+ 1}(b−1)a _modpk

≡ bmodpk_.

✷ Lemma 8 If x and y are the generators of P(p, m) and a, b, c > 0, then

(xayb)c=xaΛ(b,c)ybc.

Proof. We expand (xa_yb₎c _{as follows:}

(xa_yb₎c _{= x}a_(yb_xa_)(yb_xa_{)∗ · · · ∗x}a_yb = xa_xaαb_y2b_xa_{∗ · · · ∗x}a_yb = xaxaαbxaα2by3bxa∗ · · · ∗xayb .. . = xa(1+αb+···+α(c−1)b)ybc = xaΛ(b,c)_ybc_. ✷ Lemma 9 Ifα=pm−1+ 1, thenαpk≡1 modpm for allk≥1.

Proof. The result follows from binomially expandingαpk as shown below: αpk _{= p}m−1_{+ 1}pk = pm−1pk + pk 1 pm−1 _pk₋₁ +· · ·+ pk pk₋₁ pm−1_{+ 1} ≡ 1 modpm _{for allk≥1.}

✷ Proposition 10 There exists a one-to-one relationship between the set of ma-trices of the form given in Proposition 1, and the automorphisms in Aut(P). Proof. We know from Proposition 1 and Claim 5 that every automorphism φin Aut(P) can be represented by a matrix of the form

i r j 1 , where i∈ U(pm_{), r ≡ 0 modp}m−1_{, and j ∈ Z}

p. In order to show the converse is true,

we will prove that the homomorphism induced by φ=

i r j 1

is indeed an automorphism by determining its unique inverse.

Letψ=

a d c 1

.We will determinea, c,andd, then show that they satisfy the conditions of Proposition 1. In order forψ to be an inverse map forφ, we needx= (ψ◦φ)(x) andy= (ψ◦φ)(y). Hence we must have that

x= (ψ◦φ)(x) = (xa_yc₎i_(xd_y)j _{= x}aΛ(c,i)_yci_xdΛ(1,j)_yj

= xaΛ(c,i)+dΛ(1,j)αciyj+ci. and

y= (ψ◦φ)(y) = (xayc)r(xdy) = xaΛ(c,r)ycrxdy = xaΛ(c,r)+dαcr_ycr+1_.

This implies we need the following relationships to hold for some unique a, d∈Zpm andc∈Zp as discussed in Proposition 1.

i) j+ci≡0 modp

ii) aΛ(c, i) +dΛ(1, j)αci≡1 modpm iii) aΛ(c, r) +dαcr≡0 modpm

iv) cr+ 1≡1 modp

From (i) it can be concluded that c ≡ −ji−1modp, which exists since i ≡ 0 modp. However, −ji−1 may range through all values of Zp, so this does

not impose additional restrictions on the value of c. Analysis of (iv) yields no further information onc since r≡0 modpm−1_{, and we will later see that (ii)} and (iii) also do not limit the value of c. As far as determining the value ofa, we first note that (iii) implies

aΛ(c, r) +dαcr≡aΛ(c, r) +dmodpm

sincer≡0 modpm−1_{. It follows thatd≡ −aΛ(c, r) modp}m_{.Substituting this}

result into (ii) yields

a[Λ(c, i)−Λ(c, r)∗Λ(1, j)∗αci]≡1 modpm.

Now, [Λ(c, i)−Λ(c, r)∗Λ(1, j)∗αci]≡i−rjmodpby Lemma 7. Furthermore, i−rj ≡0 modpbecausei ≡0 modpand r ≡0 modp, so it follows that we can choosea≡ Λ(c, i)−Λ(c, r)∗Λ(1, j)αci−1 _modpm_.

Finally, using this information we have that d = −aΛ(c, r)

= − Λ(c, i)−Λ(c, r)∗Λ(1, j)αci−1Λ(c, r).

In order to show thata, c,anddsatisfy the conditions of Proposition 1, we first note that Λ(c, r)≡0 modpm−1 _{implies thatd≡0 modp}m−1_{. Similarly,c}

can be considered modp, and we have already seen that a∈U(pm).Hence we have determined values fora,c, anddsuch thatψ◦φ= 1, soψ=φ−1. It follows that every matrix

i r j s

whose entries satisfy the conditions of Proposition 1 yields an element of Aut(P). This implies there exists a one-to-one relationship between the set of matrices of this form and the group Aut(P).

✷ Using this information, we are now able to infer the order of the automor-phism group ofP(p, m).

Theorem 11 If P =P(p, m), then|Aut(P)|=pm+1_(p−1).

Proof. From the previous proposition, we know thati,j, andrcompletely and uniquely determine each matrix representation of an automorphism in Aut(P). This implies that the order of Aut(P) is completely determined by the possible combinations ofi,j, andr. There arepchoices forj sincej∈Zp,pm−1(p−1)

possibilities forisincei∈U(pm) and|U(pm)|=pm−1(p−1), andpchoices for r. It follows that|Aut(P)|=pm+1(p−1).

✷ Now that the order of Aut(P) has been determined, we next factor the matrix representation of an element of this group into the product of more basic matrices. Since pis odd,U(pm_{) is cyclic [2]. Hence, we can find a generator g}

such thatgp−1_{≡1 modp. Leta=g}p−1 _andc=gpm−1_{, and note thataandc}

have orderpm−1_{andp−1 respectively. Moreover,a∩c={1}and|ac|=} (p−1)pm−1 _{implies thatU(p}m₎∼_=Z

pm−1×Zp−1. LetV =a ∼=Zpm−1 and

B = c ∼= Zp−1. Then there exists unique l1, l2 ∈ Z such that g = al1cl2.

Furthermore, letl ∈Zand note thatgl_=all1_cll2_{. It follows that a}ll1 ≡_{1 mod} psincea≡1 modp. Now, leti∈U(pm_{). Then we have shown that there exists}

uniquev∈V andb∈B such thati=vbwith v≡1 modp. Claim 12 Let φ =

i r j 1

∈ Aut(P). Then there exist ν ∈V, β ∈B, and

j∈Zp such thatφ=θ1◦θ2, whereθ1=

ν r j 1 andθ2= β 0 0 1 .

Proof. Leti=vb, where v∈V and b∈B as in the discussion above. Also, let j = jb−1 and consider Λ(j, b). As an element of U(pm_{), we can write}

Λ(j, b) =wbwherew∈V andw= 1. Now let ν=vw−1 andβ =b, and note that ν ∈ V since v, w−1 ∈ V. It follows that νΛ(j, b) = vw−1wb = vb = i. Now, consider (θ1◦θ2)(x) = ν r j 1 ◦ β 0 0 1 (x) = (xνyj)β = xνΛ(j,β)_yjβ = xνΛ(j,b)_yjb = xi_yj = φ(x). Similarly, (θ1◦θ2)(y) = ν r j 1 ◦ β 0 0 1 (y) = xry =φ(y). This implies thatφ=θ₁◦θ₂. ✷ Proposition 13 Let L =    v r j 1 v∈V r≡0 modpm−1 j∈Zp  

 and let L be the corresponding subset of Aut(P). ThenL=Sp, the unique Sylow p-subgroup of

Aut(P).

Proof. From Sylow theory we know that|Aut(P)|=pm+1_{(p−1) implies that} Aut(P) has a unique Sylowp-subgroup of orderpm+1_{. In order to proveL=S}

p

we may show thatL ≤Aut(P) and|L|=pm+1_.

It can be proved thatL ≤Aut(P) by showing thatLsatisfies the axioms of closure and existence of inverse.

a) Closure: Letl₁= v₁ r₁ j₁ 1 and l₂= v₂ r₂ j₂ 1

∈ L. We must show that (l₁◦l₂)∈ L. Consider (l1◦l2)(x) = l1 xv2_yj2 = xv1_yj1 v2 (xr1_y)j2 = xv1Λ(j1,v2)+r1Λ(1,j2)αj 1v2 yj1v2+j2_. Hence we need the following relations to hold.

i) j₁v₂+j₂ ∈Z_p

ii) v1Λ(j₁, v2) +r1Λ(1, j₂)αj1v2 ∈_V.

By definition, (i) is true. In order to show thatv1Λ(j₁, v2)+r1Λ(1, j₂)αj1v2∈ V, we must prove that

v₁Λ(j₁, v₂) +r₁Λ(1, j₂)αj1v2

pk

≡1 modpm_{, for some}

1≤k≤m−1. The binomial expansion of this polynomial is equal to [v₁Λ(j₁, v₂)]pk+ pk 1 [v₁Λ(j₁, v₂)]pk−1 r₁Λ(1, j₂)αj1v2₊· · ·

· · ·+ pk pk₋₁ [v₁Λ(j₁, v₂)] r₁Λ(1, j₂)αj1v2p k₋₁ + r₁Λ(1, j₂)αj1v2p k . Now, notice that each term in the binomial expansion which contains a factor of r₁Λ(1, j₂)αj1v2 _{is equivalent to 0 modp}m_sincer

1 ≡0 modpm−1 and each of these terms will have a coefficient which is divisible by p. The only exception is the final term of the expansion, but this is raised to a power of p so it is equivalent to 0 modpm _{also. Therefore we must simply check to see if}

[v1Λ(j₁, v2)]pk ≡1 modpm.

First note that Λ(j₁, v2)≡v2 modpm−1. Thus, Λ(j₁, v2) = v2+cpm−1 for some constant c∈Z. This implies that |Λ(j₁, v₂)| =|v₂+cpm−1_{|. Now, using}

binomial expansion we see that|v₂+cpm−1_{| ≤ |v}

2|=pl, for some 1≤l≤m−1. It follows that|Λ(j₁, v₂)|is equal to some power ofp, so Λ(j₁, v₂)∈V.

Thus, v₁Λ(j₁, v₂)∈V ⇒ v1Λ(j₁, v2) +r1Λ(1, j₂)αj1v2 pk ≡1 modpm, for 1≤k≤m−1 ⇒ v₁Λ(j₁, v₂) +r₁Λ(1, j₂)αj1v2 ∈_V. Next we consider (l₁◦l₂)(y) = l₁(xr2_y) = (xv1_yj1₎r2_(xr1_y) = xv1Λ(j1,r2)+r1αj1r2 yj1r2+1_. Hence we need the following relations to hold.

i) j₁r₂+ 1≡1 modp

ii) v₁Λ(j₁, r₂) +r₁αj1r2 ≡_{0 modp}m−1

Both of these items follow immediately from the fact thatr₂≡0 mod pm−1_. b) Inverse: From Proposition 10, we already know that for anyl =

v r j 1

there exists a uniquel−1=

a d c 1 such thatl−1◦l= 1 0 0 1 . The values of a, c, and d are derived as a = [Λ(c, v)−Λ(c, r)Λ(1, j)αcv]−1, c =−jv−1, and d = −[Λ(c, v)−Λ(c, r)Λ(1, j)αcv_]−1_{Λ(c, r). Our next task is to show if}

l−1=

a d c 1

, thenl−1∈Lby verifying the following. i) a∈V

ii) c∈Zp

iii) d≡0 modpm−1

From the proof of Proposition 10 we have already shown thatl−1∈Aut(P) implies thata, c, anddsatisfy the conditions of Proposition 1. Thus, (ii) and (iii) are accounted for. In order to prove a ∈ V, we must show that apk _≡

1 modpm _{for some 1 ≤ k ≤ m−1. For ease of computation, we use the}

fact that|a|=|a−1|and instead show that a−1pk ≡1 modpm_{by expanding}

a−1pk = [Λ(c, v)−Λ(c, r)Λ(1, j)αcv_]pk = Λ(c, v)pk− pk 1 [Λ(c, v)]pk−1[Λ(c, r)Λ(1, j)αcv] +· · ·+ pk pk−1 Λ(c, v) [Λ(c, r)Λ(1, j)αcv]pk−1−[Λ(c, r)Λ(1, j)αcv]pk. Now, notice that Λ(c, r) ≡ 0 modpm−1 implies that each term in the bi-nomial expansion which includes a factor of Λ(c, r) is equivalent to 0 modpm since each of these terms will either be raised to a power ofpor be multiplied by a coefficient which is divisible byp. Therefore, we may simply check to see if [Λ(c, v)]pk≡1 modpm_.

First note that Λ(c, v)≡vmodpm−1_{. Thus, Λ(c, v) =v+kp}m−1 _{for some}

k∈Z.This implies that|Λ(c, v)|=|v+kpm−1_{|.Now, using binomial expansion}

we see that|v+kpm−1_{| ≤ |v|=p}l_{, for some 1≤l≤m−1.It follows that|Λ(c, v)|}

is equal to some power ofp. Thus,apk _{≡1 modp}m _{for some 1≤k ≤m−1,}

which implies thata∈V.It follows thatL ≤Aut(P).

We now show that |L|=pm+1_{. First note that v, j, andr completely and}

uniquely determine each matrix representation of an automorphism inL. This implies that the order ofLis completely determined by the possible combina-tions ofv, j,andr. There arepchoices forj sincej∈Zp,pm−1 choices forv

sincev∈V, andpchoices forr. It follows that|L|=pm−1_∗p∗p=pm+1_=|S p|.

Combining this with the fact that Aut(P) has a unique p-Sylow subgroup of orderpm+1, the conclusion can be made thatL=Sp.

✷ Theorem 14 Aut(P)∼=S_pZ_p−1.

Proof. Recall from Claim 12 that any automorphismφin Aut(P) can be rep-resented asφ= i r j 1 = v r j 1 ◦ b 0 0 1 wherev∈V,r≡0 modpm−1_,

j=jb−1∈Zp, andb∈B, the subgroup ofU(pm) that is isomorphic toZp−1. Let ˆB =

b 0 0 1

|b∈B∼=Zp−1} and let ˆB be the corresponding

sub-group of Aut(P). We now check to see whether Aut(P)∼=LBˆby confirming the following items.

i) L ∩Bˆ=e ii) LBˆ= Aut(P) iii) L'Aut(P)

Again by Claim 12 we see that (ii) is true, while (i) is confirmed by inspec-tion. Similarly,L=Sp implies thatL'Aut(P), so (iii) holds also.

✷ The author first determined the structure of the automorphism group of some metacyclic 2-groups while a student in an Abstract Algebra II course at

St. Olaf College. The study of automorphism groups progressed to include metacyclicp-groups of odd order as an independent research project under the direction of Professor Jill Dietz.

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