Schur multipliers of special 𝑝-groups of rank 2

Theorem 1.1.

Let G be a special p-group of rank 2 with d=d⁢(G) and Gp=G′. Then the following assertions hold:

1. Either Z*⁢(G)=Z⁡(G) or G is capable.

2. M⁡(G) is elementary abelian of order p12⁢d⁢(d-1)-2 iff Z*⁢(G)=Z⁡(G).

3. M⁡(G) is of order p12⁢d⁢(d-1)-1 and exponent at most p2 iff G is capable.

4. For every central subgroup Z of order p, G/Z is isomorphic to

   ESp2(p2⁢m+1)×ℤpd-2⁢m (m≥2),ESp2⁢(p3)×ℤpd-2 𝑜𝑟 Q8×ℤ2d-2.

Corollary 1.2.

If G is a special p-group of rank 2 of order pn (n≥8) with Gp=G′, then M⁡(G) is elementary abelian of order p12⁢(n-2)⁢(n-3)-2.

Theorem 1.3.

If G is a special p-group of rank 2 with d=d⁢(G) and Gp≅Zp, then the following assertions hold:

1. G is not capable, and either Z*⁢(G)=Z⁡(G) or Z*⁢(G)=Gp.

2. M⁡(G) is elementary abelian.

3. The following are equivalent:

   1. M⁡(G) is of order p12⁢d⁢(d-1)-2.

   2. Z*⁢(G)=Z⁡(G).

   3. G/Gp≅ESp⁢(p2⁢m+1)×ℤpd-2⁢m, m≥2.

4. The following are equivalent:

   1. M⁡(G) is of order p12⁢d⁢(d-1).

   2. Z*⁢(G)=Gp.

   3. G/Gp≅ESp⁢(p3)×ℤpd-2.

Theorem 1.4.

If G is a special p-group of rank 2 with d=d⁢(G) and Gp=1, p odd, then the following assertions hold:

1. M⁡(G) is elementary abelian.

2. p12⁢d⁢(d-1)-2≤|M⁡(G)|≤p12⁢d⁢(d-1)+3.

3. G is capable if and only if G is isomorphic to one of the following groups:

   Φ4⁢(15)=〈α,α1,α2,β1,β2∣⁢[αi,α]=βi,αp=αip=βip=1(i=1,2)〉,

   Φ12⁢(16)=ESp⁢(p3)×ESp⁢(p3),

   Φ13⁢(16)=〈α1,α2,α3,α4,β1,β2∣⁢[αi,αi+1]=βi,[α2,α4]=β2,αip=α3p=α4p=βip=1(i=1,2)〉,

   Φ15⁢(16)=〈α1,α2,α3,α4,β1,β2∣⁢[αi,αi+1]=βi,[α3,α4]=β1,

     [α2,α4]=β2g,αip=α3p=α4p=βip=1(i=1,2)〉,

     ⁢𝑤ℎ𝑒𝑟𝑒⁢g⁢is non-quadratic residue modulo⁢p,

   T=〈x1,…,x5,c1,c2∣⁢[x2,x1]=[x5,x3]=c1,[x3,x1]=[x5,x4]=c2,xip=cjp=1,1≤i≤5, 1≤j≤2〉.

4. |M⁡(G)|=p12⁢d⁢(d-1)+3 if and only if G is isomorphic to Φ4⁢(15).

5. |M⁡(G)|=p12⁢d⁢(d-1)+2 if and only if G is isomorphic to Φ12⁢(16), Φ13⁢(16) or Φ15⁢(16).

6. |M⁡(G)|=p12⁢d⁢(d-1)-1 if and only if G is isomorphic to T.

7. M⁡(G) is of order p12⁢d⁢(d-1)-2 if and only if Z*⁢(G)=Z⁡(G). In this case, G/Z≅ESp⁢(p2⁢m+1)×ℤpd-2⁢m, m≥2, for every central subgroup Z of order p.

8. M⁡(G) is of order p12⁢d⁢(d-1) if and only if Z*⁢(G)≅ℤp. In this case,

   G/Z*⁢(G)≅ESp⁢(p3)×ℤpd-2.

Theorem 1.5.

Let G be a special 2-group of rank 2. Then G2=G′ holds.

Theorem 2.1 ([3, Theorem 3.1]).

M⁡(G)≅M.

Note.

It is easy to observe that X1 is generated by the set

where S is a set of generators of G and x¯ is the image of x in G/G′.

Theorem 2.2 ([4]).

Let Z be a central subgroup of a finite group G. Then the following sequence is exact:

Theorem 2.3 ([2]).

Let Z be a central subgroup of a finite group G. Then we have Z⊆Z*⁢(G) if and only if (G/G′)⊗Z=Ker⁡λZ.

Lemma 2.4.

Let Z be a central subgroup of a group G of nilpotency class 2. Then Z⊆Z*⁢(G) if and only if (G/G′)⊗Z is contained in X.

Proposition 2.5 ([12, Proposition 3.3]).

Let G be a special p-group, d=d⁢(G), and Gp of order r. Then |X2|=pr⁢d-12⁢r⁢(r-1).

Theorem 2.6 ([9, Main Theorem]).

Let G be a p-group of order pn. Then we have |M⁡(G)|=p12⁢(n-1)⁢(n-2)+1 if and only if G≅ESp⁢(p3)×Zpn-3.

Theorem 2.7 ([10, Theorem 21]).

Let G be a p-group of order pn. Then we have |M⁡(G)|=p12⁢(n-1)⁢(n-2) if and only if G is isomorphic to one of the following groups:

1. G≅ℤp2×ℤpn-2,

2. G≅D8×ℤ2n-3,

3. G≅ℤp4⋊ℤp (p≠2).

Theorem 2.8 ([11, Theorem 11]).

Let G be a group of order pn such that G/G′ is elementary abelian of order pn-1. Then |M⁡(G)|=p12⁢(n-1)⁢(n-2)-1 if and only if G is isomorphic to one of the following groups:

1. ESp2⁢(p3)×ℤpn-3,

2. Q8×ℤ2n-3,

3. ES⁢(p2⁢m+1)×ℤpn-2⁢m-1 (m≥2).

Proof of Theorem 1.1.

Consider Gp=G′≅ℤp×ℤp. Now, by Proposition 2.5, |X2|=p2⁢d-1. Let v1,v2∈V such that {f⁢(v1),f⁢(v2)} is a basis of W. Observe that X2 is generated by the set

Hence p2⁢d≥|X|≥p2⁢d-1, and by (2.2),

(a) Observe that |X|=p2⁢d, i.e., the set {v1⊗f⁢(v2),v2⊗f⁢(v1)} is contained in X if and only if Z*⁢(G)=Z⁡(G), follows by Lemma 2.4. Another possibility is |X|=p2⁢d-1, i.e., X1⊆X2 if and only if v1⊗f⁢(v2),v2⊗f⁢(v1) are not in X. Hence, by Lemma 2.4, Z*⁢(G)=1. So G is capable.

(b) By (2.2), |X|=p2⁢d if and only if

By Theorem 2.3 and Theorem 2.2, taking Z=Z⁡(G), we see that M⁡(G) embeds in M⁡(G/Z⁡(G)) which is elementary abelian. So, in this case, M⁡(G) is elementary abelian.

(c) For x∈M⁡(G), xp∈(V⊗W)/X and V⊗W/X is elementary abelian, so xp2=1.

By (2.2), X1⊆X2 i.e., |X|=p2⁢d-1 if and only if, by (2.2),

Hence, by Lemma 2.4, Z*⁢(G)=1, so G is capable. The converse follows from (b).

(d) Observe that, in both the cases, |M⁡(G/Z)|=p12⁢d⁢(d-1)-1 follows from (2.1), taking Z a central subgroup of order p. Therefore, by Theorem 2.8,

Proof of Theorem 1.3.

(a) Assume Gp is cyclic of order p. Now, by Proposition 2.5, |X2|=pd. Let v1∈V such that Gp=〈f⁢(v1)〉. Observe that X2 is generated by the set

Hence, by Lemma 2.4, Gp⊆Z*⁢(G). Hence G is not capable with Z*⁢(G)=Gp or Z*⁢(G)=Z⁡(G).

Using Theorem 2.3 and taking Z=Gp in (2.1), we have

Now let v1,v2∈V such that (v1,v2)∈G′∖Gp. Then the set

is linearly independent in X1, so |X1|≥pd-2. The set M′∪N′ is linearly independent in X and M′∩N′=∅. Thus

Hence, by (2.2),

(c) Now similarly, as described in the proof of Theorem 1.1, Z*⁢(G)=Z⁡(G) if and only if |M⁡(G)|=p12⁢d⁢(d-1)-2, i.e., by (3.1),

which happens if and only if

follows from Theorem 2.8.

(d) By Theorem 2.7, it follows that there is no G/Gp such that

Thus, by (3.1), |M⁡(G)| cannot be of order p12⁢d⁢(d-1)-1. Hence

if and only if Z*⁢(G)=Gp. By (3.1),

which happens if and only if

follows by Theorem 2.6.

(b) By (c) and (d), it follows that p must be odd. The group G/Gp is of exponent p and p odd, so the homomorphism σ, described in Section 2, is the trivial map, and therefore σ⁢ξ⁢(x)=xp=1 for x∈M⁡(G/Gp). Thus M⁡(G/Gp) is elementary abelian. Since Gp⊆Z*⁢(G), by Theorem 2.3 and Theorem 2.2, M⁡(G) embeds in M⁡(G/Gp). Therefore, M⁡(G) is also elementary abelian. The proof is complete now. ∎

Proof of Theorem 1.4.

(a) Since p is odd and Gp=1, the homomorphism σ, described in Section 2, is the trivial map, and therefore σ⁢ξ⁢(x)=xp=1. Thus M⁡(G) is elementary abelian.

Let z,z′ be the generators of Z⁡(G), and let x1,x2,…,xd be the generators of G such that [x1,x2]∈〈z〉 is non-trivial. Then the set

consists of d-2 linearly independent elements of X1.

Now if, for some xk∈{3,4,…,d}, [x1,xk]∈〈z′〉 is non-trivial, then the set

consists of d-3 linearly independent elements of X1. Thus A∪B consists of (2⁢d-5) linearly independent elements of X1.

If we have 1≠[x2,xk]∈〈z′〉 for some k∈{3,4,…,d}, then a similar conclusion holds. Suppose then that [x1,xk] and [x2,xk] are all trivial or in 〈z〉 for all k∈{3,4,…,d}. Say, [x3,x4]∈〈z′〉. In this case,

consists of d-2 independent elements of X1. Thus A∪B1 consists of 2⁢d-4 linearly independent elements of X1.

Hence, in both cases, 2⁢d≥|X|=|X1|≥p2⁢d-5 holds. By (2.2), it follows that

which proves (b).

Now if G is not capable, then, by (2.1) and [9, Main Theorem], we have

By Theorem 2.7, there is no G and central subgroup Z such that

Hence

Assume then that G is capable. By [5, Proposition 3], p5≤|G|≤p7, with p odd. If |G|=p5, then, looking through the list of groups given in [7], it follows that G≅Φ4⁢(15). Since |X|=p, it follows by (2.2) that

If |G|=p6, then, looking through the list of groups given in [7], it follows that G≅Φ12⁢(16), Φ13⁢(16) or Φ15⁢(16). Since |X|=p4, it follows by (2.2) that |M⁡(G)|=p8=p12⁢d⁢(d-1)+2.

Now consider groups of order p7 of exponent p. By [6], it follows that there is only one capable group

up to isomorphism. By (2.2), |M⁡(G)|=p9=p12⁢d⁢(d-1)-1.

Now (c), (d), (e), (f) follow by (3.2).

(g) By (2.1), we have Z*⁢(G)≅Z⁡(G) if and only if |M⁡(G)|=p12⁢d⁢(d-1)-2, taking Z=Z⁡(G). Hence, by (2.1), we have that, for every central subgroup Z of order p, |M⁡(G/Z)|=p12⁢d⁢(d-1)-1, and thus

follows from Theorem 2.8.

(h) Suppose Z*⁢(G)≅ℤp. By (3.2) and (g), it follows that |M⁡(G)|=p12⁢d⁢(d-1).

Conversely, suppose |M⁡(G)|=p12⁢d⁢(d-1). From the previous cases, it follows that G is not capable since Z*⁢(G)≅ℤp.

In this case, we have

By Theorem 2.6, it follows that |M⁡(G/Z*⁢(G))|=p12⁢d⁢(d-1)+1 if and only if

The proof is complete now. ∎

Proof of Theorem 1.5.

By Theorem 1.3, it follows that there is no special 2-group of rank 2 with G2≅ℤ2.

Assume that p=2. As in the proof of Theorem 1.4, we conclude that

Let Z be a central subgroup of order 2. If G is not capable, then, by (2.1),

By Theorems 2.6, 2.7 and 2.8, there is no group G and central subgroup Z such that G/Z of exponent 2 and 212⁢d⁢(d-1)-1≤|M⁡(G/Z)|≤212⁢d⁢(d-1)+1. Hence G must be capable, and |M⁡(G/Z)|≤212⁢d⁢(d-1)-2.

Suppose |M⁡(G)|=212⁢d⁢(d-1)-2+k, 0≤k≤5. Then, by (2.2),

Hence 1≤|Im⁡λZ|=2m≤2k. By (2.1), it follows that

Hence k-m≤-1, which is not possible. So there is no special 2-group of rank 2 with G2=1. The proof is complete now. ∎