Metabelian thin Beauville p-groups

Theorem A.

Let G be a metabelian thin p-group of class p or p+1 for p≥5. Then there are four cases in which there is a Beauville structure:

1. cl⁡(G)=p and |γp⁢(G)|=p2.

2. cl⁡(G)=p+1.

3. cl⁡(G)=p, |γp⁢(G)|=p and Gp=γp-1⁢(G).

4. cl⁡(G)=p, |γp⁢(G)|=p, Gp=γp⁢(G) and G has at least three maximal subgroups of exponent p.

Notation.

We use standard notation in group theory. If G is a group, H,K≤G and N⊴G, then H≡K⁢(mod⁡N) means H⁢N/N=K⁢N/N. If p is a prime, then we write Gpi for the subgroup generated by all powers gpi as g runs over G, and Ωi⁢(G) for the subgroup generated by the elements of G of order at most pi. The exponent of G, denoted by exp⁡G, is the maximum of the orders of all elements of G.

Lemma 2.1.

Let G be a metabelian thin p-group, and let l be the largest integer such that Gp≤γl⁢(G). Then 3≤l≤p, γl+1⁢(G) is cyclic, γl+2⁢(G)=1 and γ2⁢(G)p≤γl+1⁢(G).

Proof.

See [6, proofs of Lemmas 1.2, 1.3 and 3.3]. ∎

Corollary 2.2.

Let G be a metabelian thin p-group. Then |Gp|≤p3.

Lemma 2.3.

Let G be a metabelian thin p-group such that its lattice of normal subgroups ends with a chain, i.e. monolithic. Then the order of Gp cannot be p2.

Proof.

If G is of class p+1, then Gp=γp⁢(G), and hence |Gp|=p3. Thus we assume that cl⁡(G)=c≤p. Next observe that if M is a maximal subgroup of G, then cl⁡(M)<c≤p, and so M is regular.

Now suppose, on the contrary, that |Gp|=p2. Consider the quotient group G¯=G/γc⁢(G), which is regular. Then |G¯p|=p, and hence |G¯:Ω1(G¯)|=p. Write Ω1⁢(G¯)=M/γc⁢(G) for some maximal subgroup M of G. Since G¯ is regular, it follows that exp⁡Ω1⁢(G¯)=p, and hence Mp≤γc⁢(G). This implies that |Mp|=|M:Ω1(M)|≤p as M is regular. It then follows that G′≤Ω1⁢(M), and thus exp⁡G′=p.

On the other hand, if M is an arbitrary maximal subgroup of G, we have G′≤M and since exp⁡G′=p, we obtain G′≤Ω1⁢(M)≤M. It follows that |Mp|=|M:Ω1(M)|≤p. Since G is thin, this implies that Mp≤γc⁢(G) for any maximal subgroup M. But Gp=〈Mp∣M maximal in G〉≤γc⁢(G). Thus |Gp|≤p, which is a contradiction. ∎

Theorem 2.4.

A metabelian thin 3-group is a Beauville group if and only if it is one of SmallGroup⁢(35,3), SmallGroup⁢(36,34), or SmallGroup⁢(36,37).

Proof.

Let G be a metabelian thin 3-group. Then |G|≤36. Note that the smallest Beauville 3-group S=𝚂𝚖𝚊𝚕𝚕𝙶𝚛𝚘𝚞𝚙⁢(35,3) is of order 35, and it is the only Beauville 3-group of that order [1]. Furthermore, by using the computer algebra system GAP [21], it can be seen that S is metabelian thin. Thus, if |G|=35, then G is a Beauville group if and only if G≅S. We next assume that |G|=36. It has been shown in [1] that there are only three Beauville 3-groups of order 36, namely S=𝚂𝚖𝚊𝚕𝚕𝙶𝚛𝚘𝚞𝚙⁢(36,n) for n=34,37,40. However, if n=40, then S is not thin since |Z⁢(S)|=9 and thus Z⁢(S)≠γ4⁢(S). On the other hand, if n=34 or 37, then again by using the computer algebra system GAP, one can show that S is a metabelian thin 3-group. Consequently, G is a Beauville group if and only if G≅S for n=34 or 37. ∎

Lemma 2.5 ([19, Lemma 6]).

Let G be a metabelian p-group and x,y∈G. Set σ1=y and σi=[σi-1,x] for i≥2. Then

where

and

Lemma 2.6.

Let G be a metabelian thin p-group such that γ2⁢(G)p≤γp+1⁢(G). If x and y are the generators of G satisfying (2.3), then for all 0≤t≤p-1,

Proof.

By Lemma 2.5, we have

Since γ2⁢(G)p≤γp+1⁢(G), it then follows that

Note that for i+j>0, C⁢(i,j) is the coefficient of ui⁢vj in ∑k=0p-1(1+u)k⁢(1+v)k, where

In the previous expression the monomials of total degree less than p appear only in (u+v)p-1≡∑r=0p-1(-1)r⁢ur⁢vp-r-1⁢(mod⁡p), and hence

Thus the condition γ2⁢(G)p≤γp+1⁢(G) implies that

On the other hand, notice that for 1≤t≤p-1,

Since G is metabelian, for every a,b∈G and c∈G′ we have [c,a,b]=[c,b,a], and this, together with (2.5), yields that

and hence (modulo γp+1⁢(G))

Note that since h is a quadratic non-residue, we have

Consequently, we get

for 0≤t≤p-1, as desired. ∎

Lemma 2.7.

Let G be a metabelian thin p-group such that |γp⁢(G)|≥p2 and let x and y be the generators of G satisfying (2.3). Then for every t0∈{0,1⁢…,p-1} there exist at most three distinct t∈{0,1⁢…,p-1} such that

Proof.

Since |γp⁢(G)|≥p2, it follows that Gp=γp⁢(G) and γ2⁢(G)p≤γp+1⁢(G), by Lemma 2.1. Also note that |γp(G):γp+1(G)|=p2. Notice that, as a consequence of (2.2), we have that l=[y,x,p-2y] and m=[y,x,p-3y,x] are linearly independent modulo γp+1⁢(G). Thus (l,m) is a basis of γp⁢(G) modulo γp+1⁢(G). If we set xp≡lα⁢mβ⁢(mod⁡γp+1⁢(G)) and yp≡lγ⁢mδ⁢(mod⁡γp+1⁢(G)) for some α,β,γ,δ∈𝔽p, then by (2.4), we have

Observe that as rational functions in t, neither

nor

are zero.

We now fix t0∈{0,1⁢…,p-1}. Then (2.6) holds if and only if there exists λ∈𝔽p* such that

If f⁢(t0)=0 or g⁢(t0)=0, then we have f⁢(t)=0 or g⁢(t)=0, that is,

or

Otherwise, we have f⁢(t)f⁢(t0)=g⁢(t)g⁢(t0). Then g⁢(t0)⁢f⁢(t)-f⁢(t0)⁢g⁢(t)=0, that is,

which is a polynomial in t of degree ≤3. Thus in every case, there are at most three distinct t∈{0,1⁢…,p-1} such that 〈(xt0⁢y)p〉≡〈(xt⁢y)p〉⁢(mod⁡γp+1⁢(G)). ∎

Lemma 2.8.

Let G be a metabelian thin p-group such that γ2⁢(G)p≤γp+1⁢(G). If M is a maximal subgroup of G and a,b∈M∖G′, then

Proof.

If we write b=ai⁢c for some c∈G′ and for some integer i not divisible by p, then by the Hall–Petrescu collection formula (see [16, III.9.4]), we have

where cj∈γj⁢(〈a,c〉)≤γj+1⁢(G). Thus we have (ai⁢c)p≡ap⁢i⁢(mod⁡γp+1⁢(G)), and hence 〈a〉p≡〈b〉p⁢(mod⁡γp+1⁢(G)). ∎

Remark 2.9.

If we replace x with x*, where x*∈G∖G′ is not a power of x, there exists a corresponding y* satisfying (2.3). Then x∈〈(x*)t0⁢y*,G′〉∖G′ for some 0≤t0≤p-1, and according to Lemma 2.8, we have

It follows from Lemma 2.7 that there exist at most three distinct t∈{0,1⁢…,p-1} such that 〈xp〉≡〈(xt⁢y)p〉⁢(mod⁡γp+1⁢(G)).

Corollary 2.10.

Let G be a metabelian thin p-group such that |γp⁢(G)|≥p2. If M is a maximal subgroup of G, then there exist at most two maximal subgroups M1, M2 different from M such that Mp≡M1p≡M2p⁢(mod⁡γp+1⁢(G)).

Remark 2.11.

Let G be a finite 2-generator p-group. Then we can always find elements x,y∈G∖Φ⁢(G) such that x,y and xy fall into the given three maximal subgroups of G. Let M1, M2 and M3 be three maximal subgroups of G. Choose the elements x∈M1∖Φ⁢(G) and y∈M2∖Φ⁢(G). Since each element in the set {x⁢yj∣1≤j≤p-1} falls into different maximal subgroups, there exists 1≤j≤p-1 such that x⁢yj∈M3∖Φ⁢(G). Thus if we put x*=x and y*=yj, then elements in the triple {x*,y*,x*⁢y*} fall into the given three maximal subgroups.

Theorem 2.12.

Let G be a metabelian thin p-group with cl⁡(G)=p such that |γp⁢(G)|=p2, where p≥5. Then G has a Beauville structure in which one of the two triples has all elements of order p2.

Proof.

We divide our proof into three cases depending on the number of maximal subgroups whose pth powers coincide, and in every case, we take into account Corollary 2.10 and Remark 2.11. First of all, note that since G has at most three maximal subgroups of exponent p and p≥5, there are at least three maximal subgroups of exponent p2.

Case 1. Assume that there is a one-to-one correspondence between maximal subgroups Mi of exponent p2 and Mip. Choose a set of generators {x1,y1} such that o⁢(x1)=o⁢(y1)=o⁢(x1⁢y1)=p2.

Case 2. Assume that there exist three maximal subgroups of exponent p2 such that their pth power subgroups coincide. Then choose a set of generators {x1,y1} such that x1,y1 and x1⁢y1 fall into those maximal subgroups.

In both Case 1 and Case 2, since p≥5, we can choose another set of generators {x2,y2} so that each pair of elements in {xi,yi,xi⁢yi∣i=1,2} is linearly independent modulo G′ by Remark 2.11.

Case 3. Assume that we are not in the first two cases. Then there exist two maximal subgroups M1, M2 of exponent p2 such that M1p=M2p and Mp≠M1p for all other maximal subgroups M.

Let us first deal with p≥7. We start by choosing a set of generators {x1,y1}, where x1∈M1 and y1∈M2 are such that o⁢(x1⁢y1)=p2, say x1⁢y1∈M3. Then there might be a maximal subgroup M4 such that M3p=M4p (note that there is no other i≠3,4 satisfying Mip=M3p, otherwise we are in Case 2). Since p≥7, we can choose another set of generators {x2,y2} so that x2,y2,x2⁢y2∉M4 and each pair of elements in {xi,yi,xi⁢yi∣i=1,2} is linearly independent modulo G′, by Remark 2.11.

If p=5, then by using the construction of metabelian thin p-groups in [6] and the computer algebra system GAP, one can show that there is no metabelian thin 5-group of class 5 such that |γ5⁢(G)|=52 and fifth powers of maximal subgroups coincide in pairs. Thus in Case 3, there exists a maximal subgroup, say M3, of exponent 52, where M5≠M35 for all maximal subgroups M with M≠M3. Then choose sets of generators {x1,y1} and {x2,y2} so that x1∈M1, y1∈M2 and x1⁢y1∈M3 and each pair of elements in {xi,yi,xi⁢yi∣i=1,2} is linearly independent modulo G′.

We claim that, in every case, {x1,y1} and {x2,y2} form a Beauville structure for G. If A={x1,y1,x1⁢y1} and B={x2,y2,x2⁢y2}, then we need to show that

for all a∈A, b∈B and g,h∈G. Note that o⁢(a)=p2 for every a∈A. Assume first that o⁢(b)=p. If 〈ag〉∩〈bh〉≠1 for some g,h∈G, then 〈bh〉⊆〈ag〉, and hence 〈a⁢G′〉=〈b⁢G′〉, which is a contradiction, since a and b are linearly independent modulo G′. Thus we assume that o⁢(b)=p2. If (2.7) does not hold, then 〈(ag)p〉=〈(bh)p〉, which contradicts the choice of b. ∎

Lemma 2.13 ([13, Lemma 4.2]).

Let G be a finite group and let {x1,y1} and {x2,y2} be two sets of generators of G. Assume that, for a given N⁢⊴⁢G, the following hold:

1. {x1⁢N,y1⁢N} and {x2⁢N,y2⁢N} is a Beauville structure for G/N,

2. o⁢(u)=o⁢(u⁢N) for every u∈{x1,y1,x1⁢y1}.

Then {x1,y1} and {x2,y2} is a Beauville structure for G.

Theorem 2.14.

Let G be a metabelian thin p-group with cl⁡(G)=p+1, where p≥5. Then G has a Beauville structure.

Proof.

By Theorem 2.12, G¯=G/γp+1⁢(G) has a Beauville structure in which one of the two triples has all elements of order p2, i.e. they have the same order in both G and G¯. Then we can apply Lemma 2.13 and thus G is a Beauville group. ∎

Theorem 2.15.

Let G be a group in case (i). Then G has a Beauville structure.

Proof.

First of all, notice that there exists a pair of generators a and b of G such that ap and bp are linearly independent modulo γp⁢(G). By the Hall–Petrescu formula, we have

where cj∈γj⁢(〈at,b〉). Since γ2⁢(G)p≤γp⁢(G), by Lemma 2.1, we get

for 1≤t≤p-1. Observe that, similarly to Lemma 2.8, for every maximal subgroup M, m∈M and c∈G′ , we have (m⁢c)p≡mp⁢(mod⁡γp⁢(G)). It then follows that the power subgroups Mp are all distinct modulo γp⁢(G).

On the other hand, since G¯=G/γp⁢(G) is of class p-1, it is a regular p-group such that |G¯p|=p2. According to [11, Corollary 2.6], G¯ is a Beauville group since p≥5. From the observation above, all elements outside G′ are of order p2 in both G and G¯. Then we can apply Lemma 2.13 to conclude that G is a Beauville group. ∎

Theorem 2.16.

Let G be a group in case (ii). Then G has a Beauville structure if and only if it has at least three maximal subgroups of exponent p.

Proof.

If the number of maximal subgroups of exponent p is less than three, then Ω1⁢(G) is contained in the union of at most two maximal subgroups. Since |Gp|=p, it then follows from [11, Proposition 2.4] that G has no Beauville structure.

On the other hand, if at least three maximal subgroups have exponent p, then we choose a triple in which all elements have order p. Since p≥5, we can choose another triple such that each pair of elements in the union of the two triples is linearly independent modulo G′. Then G has a Beauville structure. ∎