ON THE ORDER OF THE SCHUR MULTIPLIER OF p-GROUPS

arXiv:1705.02520v1 [] 6 May 2017

PRADEEP K. RAI
Abstract. We give a bound on the order of the Schur multiplier of p-groups
refining earlier bounds. As an application we complete the classification of
groups having Schur multiplier of maximum order. Finally we prove that the
order of the Schur multiplier of a finite p-group of maximal class and order pn
n
is at most p 2 .

1. Introduction
Let G be a group. The center and the commutator subgroup of G are denoted
by Z(G), and γ2 (G) respectively. By d(G) we denote the minimal no of generators
of G. We write γi (G) and Zi (G) for the i-th term in the lower and upper central
series of G respectively. Finally, the abelianization of the group G, i.e. G/γ2 (G), is
denoted by Gab .
Let G be finite p-group of order pn and M (G) be the Schur multiplier of G.
1
In 1956 Green proved that |M (G)|≤ p 2 n(n−1) [5]. Since then, Green’s bound has
been refined by many mathematicians. For a detailed account we refer the reader
to a note recently written by the author [12]. There has also been interest in the
classification of finite p-groups given the order of their Schur multiplier. By Green’s
1
result note that |M (G)|= p 2 n(n−1)−t for some non-negative integer t. The groups
with t = 0, 1, 2, 3, 4, 5 have been classified, see [1, 3, 14, 15, 11]. The first such result
1
was proved by Berkovich [1]. He proved that |M (G)|= p 2 n(n−1) if and only if G is
elementary abelian.
Let G be a non-abelian p-group of order pn with |γ2 (G)|= pk and d(G) = d.
Niroomand proved in [10] that
(1.1)

1

|M (G)|≤ p 2 (n−k−1)(n+k−2)+1 .

The author noted in [12] that a bound of Ellis and Weigold is actually better
than this bound and derived from their bound that
1

1

|M (G)|≤ p 2 (d−1)(n+k−2)+1 (= p 2 (d−1)(n+k)−(d−2) ).
Niroomand also classified finite p-groups such that k = 1 and bound (1.1) is
attained. Note that if k = 1, then the group is of nilpotency class 2. The author
further classified the groups of nilpotency class 2 such that this bound is attained
[13]. Recently Hatui proved that there are no p-groups, for p ≥ 5, of nilpotency
class 3 or more attaining the bound [6]. She also gave an example of 3-group of
nilpotency class 3 such that the bound is attained.
2010 Mathematics Subject Classification. 20J99, 20D15.
Key words and phrases. Schur multiplier, finite p-group, maximal class.
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P. K. RAI

A natural question is: whether this bound can be improved for groups of higher
nilpotency classes. We offer the following theorem which improves the bound for
the groups with large min(d(G), c), where c is the nilpotency class of G.
Theorem 1.1. Let G be a non-abelian p-group of order pn and nilpotency class c
with |γ2 (G)|= pk and d(G) = d. Then
Pmin(d,c)
1
d−i
|M (G)|≤ p 2 (d−1)(n+k)− i=2
.
In the following theorem we give a more direct proof of Hatui’s result and classify
3-groups such that bound (1.1) is attained completing the classification of such
groups.
Theorem 1.2. Let G be a finite p-group of order pn with |γ2 (G)|= pk . Then
1
|M (G)|= p 2 (n−k−1)(n+k−2)+1 if and only if G is one of the following group.
(n−3)

1. G1 = Ep × Cp
, where Ep is the extraspecial p-group of order p3 and
exponent p for an odd prime p,
(4)
2. G2 = Cp ⋊ Cp for an odd prime p,
D
3. G3 = α1 , β1 , α2 , β2 , α3 , β3 | [α1 , α2 ] = β3 , [α2 , α3 ] = β1 , [α3 , α1 ] = β2 ,
E
[αi , βj ] = 1, αpi = βip = 1 (p an odd prime) (i, j = 1, 2, 3) .
D
4. G4 = α1 , β1 , α2 , β2 , α3 , β3 , γ | [α1 , α2 ] = β3 , [α2 , α3 ] = β1 , [α3 , α1 ] = β2 ,
E
[βi , αi ] = γ, [αi , βj ] = 1, α3i = βi3 = 1 (i = 1, 2, 3, j = 2, 3) .
A finite p-group of order pn is said to be of maximal class if its nilpotency class is
n − 1. Let G be finite p-group of maximal class and order pn . Since G is generated
by 2 elements, it follows by a result of Gaschütz [4] that |M (G)|≤ pn−1 . Moravec
p+1 n−1
proved for n > p + 1, that |M (G)|≤ p 2 ⌈ p−1 ⌉ [9]. Improving his result we prove
the following theorem.
Theorem 1.3. Let G be a finite p-group of maximal class and of order pn for an
n
odd prime p and n ≥ 4. Then |M (G)|≤ p 2 .
2. Prerequisites
Let G be a finte p-group of nilpotency class c and G be the factor group G/Z(G).
Define homomorphism
ab

Ψ2 : G

ab

⊗G

ab

⊗G

7→

γ2 (G)
ab
⊗G
γ3 (G)

by Ψ(x1 ⊗ x2 ⊗ x3 ) = [x1 , x2 ] ⊗ x3 + [x2 , x3 ] ⊗ x1 + [x3 , x1 ] ⊗ x2 .
For 3 ≤ i ≤ c define homomorphisms
ab

Ψi : G
|

γi (G)
ab
ab
ab
7→
⊗G
⊗ G ··· ⊗ G
{z
}
γi+1 (G)
i+1 times

ON THE ORDER OF THE SCHUR MULTIPLIER OF p-GROUPS

3

by
Ψi (x1 ⊗ x2 ⊗ · · · ⊗ xi+1 )

= [x1 , x2 , · · · , xi ]l ⊗ xi+1 + [xi+1 , [x1 , x2 , · · · xi−1 ]l ] ⊗ xi
+[[xi , xi+1 ]r , [x1 , · · · , xi−2 ]l ] ⊗ xi−1
+[[xi−1 , xi , xi+1 ]r , [x1 , x2 , · · · , xi−3 ]l ] ⊗ xi−2
+ · · · + [x2 , · · · , xi+1 ]r ⊗ x1

where
[x1 , x2 , · · · xi ]r = [x1 , [· · · [xi−2 , [xi−1 , xi ]] . . .]
and
[x1 , x2 , · · · xi ]l = [. . . [[x1 , x2 ], x3 ], · · · , xi ].
The following proposition was given by Ellis and Weigold [2, Proposition 1 and
the comments on page 192 following the proof of Theorem 2].
Proposition 2.1. Let G be a finite p-group and G be the factor group G/Z(G).
Then
c
c
Y
Y
γi (G)
ab
ImΨi ≤ M (Gab )
M (G) γ2 (G)
⊗G .
γ
(G)
i+1
i=2
i=2
The following Lemma is from [12].
Lemma 2.2. [12, Lemma 2.1] Let G be an abelian p-group of order pn such that
G = Cpα1 × Cpα2 × · · · × Cpαd (α1 ≥ α2 ≥ · · · ≥ αd ) and |G|= pn , then |M (G)|≤
1
p 2 (d(G)−1)(n−(α1 −αd )) .
The following Lemma is from [8].
Lemma 2.3. [8, Lemma 3.6 (c)] Let G be a group such that G = hM i. Then γk (G)
is generated by simple commutators of weight ≥ k in the elements m± , m ∈ M .
3. Proofs of Theorems
Proof of Theorem 1.1 Let Ψi be as defined in Section 2 and d(G/Z(G)) = δ.
Following Proposition 2.1 we have that
c
Y
|M (G)||γ2 (G)| |ImΨi |≤ |M (Gab )|pkδ .
i=2

Applying Lemma 2.2 this gives
c
Y
1
|M (G)| |ImΨi |≤ |p 2 (d−1)(n−k)+k(δ−1) .
i=2

so that
(3.1)

|M (G)|

c
Y

1

|ImΨi |≤ |p 2 (d−1)(n+k)−k(d−δ) .

i=2

Choose a subset S = {x1 , x2 , . . . , xδ } of G such that {x1 , x2 , · · · , xδ } be a minimal
generating set for G/Z(G). Fix i ≤ min(δ, c). Since i ≤ c, γi (G)/γi+1 (G) is a
non-trivial group. Using Lemma 2.3 we can choose a commutator [y1 , y2 , · · · , yi ]
of weight i such that [y1 , y2 , · · · , yi ] ∈
/ γi+1 (G) and y1 , . . . , yi ∈ S . Since i ≤ δ,

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P. K. RAI

S\{y1 , y2 , · · · , yi } contains at least δ − i elements. Choose any δ − i elements
z1 , z2 , . . . , zδ−i from S\{y1, y2 , · · · , yi }. Since [y1 , y2 , · · · , yi ] ∈
/ γi+1 (G) and zj ∈
/
{y1 , y2 , · · · , yi }, Ψi (y1 , . . . , yi , zj ) 6= 1. Notice that the set {Ψi (y1 , . . . , yi , zj ) | 1 ≤
j ≤ δ − i} is a minimal generating set for h{Ψi (y1 , . . . , yi , zj ) | 1 ≤ j ≤ δ − i}i
because {x1 , x2 , · · · , xδ } is a minimal generating set for G/Z(G). It follows that
|ImΨi |≥ pδ−i . Putting this in Equation 3.1 we get the required result.
1

Proof of Theorem 1.2 Let |M (G)|= p 2 (n−k−1)(n+k−2)+1 . In view of [13, Theorem 1.1] suppose that the nilpotency class of G is at least 3. From the following
exact sequence [7, Corollary 3.2.4 (ii)]
1 7→ X 7→ G/γ2 (G) ⊗ γc (G) 7→ M (G) 7→ M (G/γc (G)) 7→ γc (G) 7→ 1,
it follows that the Bound 1.1 is attained for the group G/γc (G). Applying induction
the bound 1.1 is attained for G/γ3 (G). But G/γ3 (G) is of nilpotency class 2,
therefore by [13, Theorem 1.1], p 6= 2. Also, it follows from Theorem 1.1 that
d(G) ≤ 3. Now using [10, Theorem 2.2] we see that d(G) = 3. Let Ψi be the maps
as given in Section 2. By simplyfying notations
Ψ3 (x1 ⊗ x2 ⊗ x3 ⊗ x4 ) =

[[x1 , x2 ], x3 ] ⊗ x4 + [x4 , [x1 , x2 ]] ⊗ x3 + [[x3 , x4 ], x1 ] ⊗
x2 + [x2 , [x3 , x4 ]] ⊗ x1 .

As in the proof of Theorem 1.1 we have |ImΨ2 |≥ pδ−2 . Applying Equation 3.1
we get that d = δ. It follows from Equation 3.1 that
(3.2)

|M (G)|

c
Y

1

|ImΨi |≤ p 2 (d−1)(n+k−2)+1 .

i=3

Therefore ImΨ3 = {1}. Let G be generated by α1 , α2 , α3 . Then for i 6= j
Ψ3 (αi ⊗ αj ⊗ αi ⊗ αj ) = 2([αi , αj , αi ] ⊗ αj ) + 2([αj , [αi , αj ]] ⊗ αi ).
This shows that [αi , αj , αi ] ∈ γ4 (G) because p 6= 2. Now for i 6= j 6= k 6= i, consider
Ψ3 (αi ⊗ αj ⊗ αk ⊗ αi ) = [αi , αj , αk ] ⊗ αi + [αj , [αk , αi ]] ⊗ αi .
Therefore
[αi , αj , αk ]γ4 (G) = [αk , αi , αj ]γ4 (G).
Putting (i, j, k) = (1, 2, 3) and (2, 3, 1) gives
[α1 , α2 , α3 ]γ4 (G) = [α3 , α1 , α2 ]γ4 (G).
and
[α2 , α3 , α1 ]γ4 (G) = [α1 , α2 , α3 ]γ4 (G).
respecticely.
Applying Hall-Witt identity we see that [α2 , α3 , α1 ]3 ∈ γ4 (G). Since [αi , αj , αi ] ∈
γ4 (G) and γ3 (G)/γ4 (G) is non-trivial, we have [α1 , α2 , α3 ] ∈
/ γ4 (G). It follows that
p = 3. Let G be a group of nilpotency class at least 4. Consider the map Ψ4 . By
simplyfying notations

ON THE ORDER OF THE SCHUR MULTIPLIER OF p-GROUPS

Ψ4 (x1 ⊗ x2 ⊗ x3 ⊗ x4 ⊗ x5 ) =

5

[x1 , x2 , x3 , x4 ] ⊗ x5 + [x5 , [x1 , x2 , x3 ]] ⊗ x4
+[[x4 , x5 ], [x1 , x2 ]] ⊗ x3 + [[x3 , [x4 , x5 ]], x1 ] ⊗ x2 .
+[x2 , [x3 , [x4 , x5 ]]] ⊗ x1 .

Since γ4 (G)/γ5 (G) is non-trivial, one of the elements [α1 , α2 , α3 , αi ], i = 1, 2, 3
does not belong to γ5 (G). Suppose [α1 , α2 , α3 , α1 ] ∈
/ γ5 (G). Then Ψ4 (α1 ⊗ α2 ⊗
α1 ⊗ α2 ⊗ α3 ) is non-identity so that ImΨ4 is non-trivial. Similarly supposing
[α1 , α2 , α3 , α2 ] ∈
/ γ5 (G), the element Ψ4 (α1 ⊗ α2 ⊗ α2 ⊗ α1 ⊗ α3 ), while supposing
[α1 , α2 , α3 , α3 ] ∈
/ γ5 (G), the element Ψ4 (α1 ⊗ α2 ⊗ α3 ⊗ α3 ⊗ α1 ) give that ImΨ4 is
non-trivial. This, in view of Equation 3.2, gives a contradiction. Therefore G is a
3-group of nilpotency class 3. Hence we have
[α1 , α2 , α3 ] = [α3 , α1 , α2 ] = [α2 , α3 , α1 ].
Since [α1 , α2 , α3 ] 6= 1 we get that [αi , αj ] ∈
/ γ3 (G) for i, j = 1, 2, 3. Also, since
[αi , αj , αi ] = 1, it follows that [αi , αj ]γ3 (G) can not be generated by {[αk , αl ]γ3 (G) |
(k, l) 6= (i, j) or (j, i)}. This shows that γ2 (G)/γ3 (G) is generated by 3 elements.
Using Equation 3.2 G/γ2 (G) is elementary abelian. So that γ2 (G)/γ3 (G) is elementary abelian. Hence |γ2 (G)/γ3 (G)|= 33 . Therefore |G|= 37 . Now it can be checked
using GAP that the bound is attained if and only if G = G4 . This completes the
proof.
Proof of Theorem 1.3 Let P1 = CG (γ2 (G)/γ4 (G)). Choose arbitrary elements
s ∈ G\P1 ∪ CG (γn−2 (G)) and s1 ∈ P1 \γ2 (G). Then s and s1 generate G. If we
define si = [si−1 , s] for i ≥ 2, then si ∈ γi (G)\γi+1 (G). Let Ψi , i ≥ 3, be the map
as defined in Section 2. Then
Ψi (s1 ⊗ s ⊗ s ⊗ · · · ⊗ s ⊗ s1 ) =

[s1 , s, s, · · · , s]l [s, s, · · · , s, s1 ]r ⊗ s1 + t ⊗ s

for some t ∈ G.
Notice, for an odd i, that
[s1 , s, s, · · · , s]l = [s, s, · · · , s, s1 ]r .
Since p 6= 2, it follows that Ψi (s1 ⊗ s ⊗ s ⊗ · · · ⊗ s ⊗ s1 ) is non-identity so that ImΨi
is non-trivial. Using this fact Equation 3.1 gives the required result.
Acknowledgements: I am very grateful to my post-doctoral superviser Prof.
Boris Kunyavskiı̆ for his encouragement and support. This research was supported
by Israel Council for Higher Education’s fellowship program and by ISF grant
1623/16.
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P. K. RAI

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(Pradeep K. Rai) Department of Mathematics, Bar-Ilan University Ramat Gan, Israel
E-mail address: raipradeepiitb@gmail.com

