# A special basis for the image of an alternating map and its applications to
the size of the Schur multiplier of $p$-groups

Sathasivam Kalithasan<EMAIL_ADDRESS>Tony Nixon Mavely
<EMAIL_ADDRESS>Viji Zachariah Thomas  School of Mathematics,
Indian Institute of Science Education and Research Thiruvananthapuram,
695551 Kerala, India<EMAIL_ADDRESS>
###### Abstract

We consider two vector spaces $U,\ V$ over a field $F$ and an alternating
bilinear map $A:U\times U\to V$ such that the $\text{Im }{(A)}$ spans $V$. We
construct a special basis for $V$ consisting of elements of $\text{Im }{(A)}$,
and then study several properties of this basis. As an application, we obtain
bounds for the size of the Schur multiplier of $p$-groups of nilpotency class
$c$, which improves existing bounds.

###### keywords:

Alternating bilinear map , Schur multiplier , Coclass , Special $p$-groups ,
$p$-groups , Nilpotent groups

###### MSC:

[2020] 15A63 , 15A69 , 11E39 , 20D15 , 20F18 , 20J05 , 20J06

## 1 Introduction

The study of Schur multiplier of a $p$-group and bounding the size of Schur
multiplier of a $p$-group has attracted many researchers over the years [2, 9,
17, 4, 3, 16, 7, 8, 13, 14, 15, 11, 12, 10, 5, 6]. The aim of the present
paper is to give a bound for the Schur multiplier of $p$-groups which improves
existing bounds for $p$-groups of nilpotency class $c$. One of the main tools
used to bound the size of the Schur multiplier of a $p$-group in recent years
is the following beautiful inequality of Ellis and Wiegold [4, Proposition 1],

$\displaystyle|M(G)||\gamma_{2}G||\text{Im
}{(\Psi_{2})}|\leq|M(G^{ab})|\prod^{c}_{i=2}|\gamma_{i}G/\gamma_{i+1}G\otimes\overline{G}^{ab}|.$
(1.0.1)

In recent years, the study of the size of the Schur multiplier for $p$-groups
of maximal class [12, 16, 15] and special $p$-groups of different ranks [2,
14, 8, 11] have received considerable interest. In [16, Corollary B] and [15,
Theorem 1.7], both the authors prove that if $G$ is a $p$-group of maximal
class and order $p^{n}$, then $|M(G)|\leq p^{\lceil\frac{n}{2}\rceil}$.

Berkovich and Janko asked to find the Schur multiplier of special groups of
rank $2$ and rank $\binom{d}{2}$ [1, Problem 2027, Problem 1729]. These
questions were addressed in [8] and [14]. It is natural to ask about the size
of the Schur multiplier of intermediate ranks. This was achieved for all ranks
$(2\leq k\leq\binom{d}{2})$ simultaneously in [11, Theorem 3.2]. For special
$p$-groups with $|\gamma_{2}G|=p^{\frac{d(d-1)}{2}}$, Rai proves the following
theorem:

######  Theorem 1.1.

([14, Theorem 1.1]) Let $G$ be a special $p$-group with $d(G)=d\geq 3$ and
$|\gamma_{2}G|=p^{\frac{1}{2}d(d-1)}$. Then

$\displaystyle|M(G)|\leq p^{\frac{1}{3}(d-1)d(d+1)}$ (1.1.1)

and the equality holds if and only if $G$ is of exponent $p$.

More recently, Rai proves the following theorem using a modification of
(1.0.1).

######  Theorem 1.2.

([15, Theorem 1.4]) Let $G$ be a non-abelian $p$-group of nilpotency class $c$
with $|\gamma_{2}G|=p^{k}$, $d(\gamma_{2}G/\gamma_{3}G)=k^{\prime}$ and
$d=d(G)$. The following inequality holds:

$\displaystyle|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-\sum\limits_{i=2}^{\min(d,k^{\prime}+1)}d-i}.$
(1.2.1)

Moreover, in the proof of the above theorem, the author obtains the bound

$\displaystyle|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)-\sum\limits_{i=2}^{\min(\delta,k^{\prime}+1)}\delta-i},$
(1.2.2)

which is better than (1.2.1).

It is easy to see that when Theorem 1.2 is restricted to the class of special
$p$-groups with $|\gamma_{2}G|=p^{\frac{1}{2}d(d-1)}$, one does not obtain the
bound given in Theorem 1.1. In fact, the bound in Theorem 1.1 is better than
the one given in Theorem 1.2. So in this sense, Theorem 1.2 is not a
generalization of Theorem 1.1.

One of the main aims of this paper is to remedy such a drawback. We would like
to give a bound for $p$-groups of nilpotency class $c$ such that when this
bound is restricted to special $p$-groups, it is still the best general bound
for special $p$-groups. Our next Theorem achieves this and improves the bound
given in Theorem 1.2 (cf. Lemma 3.5), and it also matches the bound given in
Theorem 1.1 when restricted to the class of special $p$-groups with
$|\gamma_{2}G|=p^{\frac{1}{2}d(d-1)}$. Furthermore, it is worth noting that
when Theorem 3.4 is restricted to the class of special $p$-groups, it gives
the exact bound obtained in [11, Theorem 3.2]. Hence, it is a generalization
of the bounds given in [15, 11, 8, 14, 13].

######  Theorem 3.4.

Let $G$ be a $p$-group of order $p^{n}$ and nilpotency class $c\geq 2$. Assume
that $d(G)=d$, $d(G/Z)=\delta$, $d(\gamma_{2}G/\gamma_{3}G)=k^{\prime}$ and
$|\gamma_{2}G|=p^{k}$. Let $r\in\mathbb{N}$ be such that ${r\choose
2}\leq{\delta\choose 2}-k^{\prime}<{r+1\choose 2}$. Then,

$|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)-\binom{\delta}{3}+\binom{r}{3}+\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}}.$

If $\delta\geq k^{\prime}+1$, then Theorem 1.2 and Theorem 3.4 when restricted
to special $p$-groups give the same bound. The aim of the next theorem is to
improve the bounds given in Theorems 1.2 and 3.4 for $\delta\geq
k^{\prime}+1$. Note that $d=\delta$, and $k=k^{\prime}$ for special
$p$-groups.

######  Theorem 3.7.

Let $G$ be a special $p$-group of order $p^{n}$, and that $d(G)=d$,
$d(\gamma_{2}G)=k>2$. If $d>k+1$, then

$|M(G)|\leq p^{\frac{1}{2}(d-1)(n+k)-\sum\limits_{i=2}^{k+1}(d-i)-(k-2)}$

The table below shows that as $k^{\prime}-\delta$ increases, our bound is much
better than the recent bound given in [15].

Table 1: Comparison of bounds on size of Schur multipliers of $p$-groups

Order | d | $\delta$ | k | k’ | Bounds on $|M(G)|$
---|---|---|---|---|---
Theorem 3.4 | [15, Theorem 1.4]
$5^{21}$ | 7 | 6 | 14 | 14 | $5^{71}$ | $5^{81}$
$5^{24}$ | 7 | 7 | 17 | 17 | $5^{89}$ | $5^{108}$
$3^{39}$ | 8 | 8 | 29 | 28 | $3^{182}$ | $3^{217}$
$3^{50}$ | 10 | 10 | 40 | 40 | $3^{287}$ | $3^{369}$

Outline of the paper: In Section 2, we consider two vector spaces $U,\ V$ over
a field $F$ and an alternating bilinear map $A:U\times U\to V$ such that the
$\text{Im }{(A)}$ spans $V$. We construct a special basis for $V$ consisting
of elements of $\text{Im }{(A)}$, and then study several properties of this .
Using the map $A$, we construct a trilinear alternating map $\Psi:U\times
U\times U\to V\otimes U$, which induces a linear map $\Psi:U\otimes U\otimes
U\to V\otimes U$. We then estimate a lower bound for the dimension of the
$\text{Im }{(\Psi)}$ using graph theory and some basic counting arguments. In
Section 3, using this lower bound for $\dim\text{Im }{(\Psi)}$ in (1.0.1), we
obtain a bound on the size of the Schur multiplier given in Theorem 3.4. In
the rest of Section 3, we turn our attention to special $p$-groups, and
improve the bounds given in Theorem 3.4 for $d>k+1$.

$\textbf{Notation}:$ We use fairly standard notations. We will denote the
commutator subgroup of $G$ as $\gamma_{2}G$; $\gamma_{i}G$ will denote the
$i^{th}$ term of the lower central series of $G$; the nilpotency class of $G$
will be denoted by $c$; $d(G)$ will denote the cardinality of a minimal
generating set for $G$; $G^{ab}$ will denote the abelianization of $G$;
$\Phi(G)$ will denote the Frattini subgroup of $G$; $Z(G)=Z$ will denote the
center of the group. The Schur multiplier of $G$ is denoted by $M(G)$.

## 2 Constructing a special basis for $p$-groups

Let $U$ and $V$ be finite-dimensional $F$ vector spaces and $A:U\times U\to V$
be an alternating bilinear map such that $\text{Im }{(A)}$ spans $V$. Let
$X=\\{u_{1},\cdots,u_{n}\\}$ be a basis for $U$. We have that
$\\{A(u_{i},u_{j})\mid u_{i},u_{j}\in X\\}$ spans $V$.

### 2.1 A spanning set of $V$ and a construction of a basis

Let $u_{i_{1}}<u_{i_{2}}<\dots<u_{i_{n}}$ be an order relation on X. Clearly
$Y=\\{A(u_{i},u_{j})\mid 1\leq i,j\leq n,u_{i}<u_{j}\\}$ spans $V$. We define
a set $\mathcal{Y}=\\{\\{i,j\\}\mid 1\leq i,j\leq n,i\neq j\\}$. The total
order on $X$ induces a total order on the set $\mathcal{Y}$ defined as
$\\{i,j\\}<\\{k,l\\}$, whenever $\max\\{u_{i},u_{j}\\}<\max\\{u_{k},u_{l}\\}$
or $\max\\{u_{i},u_{j}\\}=\max\\{u_{k},u_{l}\\}=u_{a}$ and
$\\{u_{i},u_{j}\\}\setminus\\{u_{a}\\}<\\{u_{k},u_{l}\\}\setminus\\{u_{a}\\}$.
The order relation on the set $\mathcal{Y}$ can be visualized as follows:

$\begin{array}[]{cccccc}&\\{i_{1},i_{2}\\}\\\
&\\{i_{1},i_{3}\\}&\\{i_{2},i_{3}\\}\\\
&\\{i_{1},i_{4}\\}&\\{i_{2},i_{4}\\}&\\{i_{3},i_{4}\\}\\\ &\cdots\\\
&\\{i_{1},i_{n}\\}&\\{i_{2},i_{n}\\}&\\{i_{3},i_{n}\\}&\cdots&\\{i_{n-1},i_{n}\\}.\\\
\end{array}$

Note that $\\{i_{a},i_{b}\\}<\\{i_{s},i_{t}\\}$ if $\\{i_{a},i_{b}\\}$ is in
one of the rows above $\\{i_{s},i_{t}\\}$ or $\\{i_{a},i_{b}\\}$ appears to
the left of $\\{i_{s},i_{t}\\}$ in the same row. Now we construct
$\mathcal{B}\subseteq\mathcal{Y}$ with very important properties. We begin the
construction of $\mathcal{B}$ by choosing the least element
$\\{i_{1}^{1},i_{2}^{1}\\}$ of $\mathcal{Y}$ such that
$A(u_{i_{1}^{1}},u_{i_{2}^{1}})$ is non zero. Either
$A(u_{i_{1}^{1}},u_{i_{2}^{1}})\in Y$ or $A(u_{i_{2}^{1}},u_{i_{1}^{1}})\in Y$
and we choose the one in $Y$. Without loss of generality assume
$A(u_{i_{1}^{1}},u_{i_{2}^{1}})\in Y$. Consider the sets
$\mathcal{B}_{1}=\\{\\{{i_{1}^{1}},{i_{2}^{1}}\\}\\}$ and
$B_{1}=\\{A(u_{i_{1}^{1}},u_{i_{2}^{1}})\\}$. We define $\mathcal{B}_{k+1}$
and $B_{k+1}$ inductively as follows: Let $\\{i_{1}^{k+1},i_{2}^{k+1}\\}$ be
the least element of $\mathcal{Y}$ such that
$A(u_{i_{1}^{k+1}},u_{i_{2}^{k+1}})$ is not in the span of $B_{k}$. Again,
without loss of generality we can assume that
$A(u_{i_{1}^{k+1}},u_{i_{2}^{k+1}})\in Y$. Define
$\mathcal{B}_{k+1}:=\mathcal{B}_{k}\cup\\{\\{{i_{1}^{k+1}},i_{2}^{k+1}\\}\\}$
and $B_{k+1}:=B_{k}\cup\\{A(u_{i_{1}^{k+1}},u_{i_{2}^{k+1}})\\}$. This process
will stop at $m=\dim_{F}V$. We will ignore the subscript and call the set
$\mathcal{B}_{m}$ as $\mathcal{B}$. Note that the construction of
$\mathcal{B}$ depends on $X$ and the order on $X$.

In the next lemma we study some properties of the set $\mathcal{B}$.

######  Lemma 2.1.

The set $\mathcal{B}\subseteq\mathcal{Y}$ constructed above satisfies the
following properties:

  1. (i)

The set $B=\\{A(u_{i},u_{j})\in Y\mid\\{i,j\\}\in\mathcal{B}\\}$ is a basis of
$V$.

  2. (ii)

If $\\{r,s\\}\in\mathcal{Y}$, then $A(u_{r},u_{s})\in\text{span
}(\\{A(u_{i},u_{j})\in B\mid\
\\{i,j\\}\in\mathcal{B},\\{i,j\\}\leq\\{r,s\\}\\})$.

  3. (iii)

If $\\{r,s\\}\in\mathcal{Y}$, then $\text{span }(\\{A(u_{i},u_{j})\in
Y\mid\\{i,j\\}\leq\\{r,s\\}\\})=\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}\leq\\{r,s\\}\\})=\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}\in\mathcal{B},\\{i,j\\}\leq\\{r,s\\}\\})$.

  4. (iv)

If $\\{r,s\\}\in\mathcal{Y}$, then set
$\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}$ is either trivial or
$\mathcal{B}_{k}$ for some $1\leq k\leq m$

  5. (v)

If $\\{r,s\\}\in\mathcal{B}$, then $A(u_{r},u_{s})\notin\text{span
}(\\{A(u_{i},u_{j})\in B\mid\\{i,j\\}<\\{r,s\\}\\})$. (Note that for some
$\\{a,b\\}\in\mathcal{Y}\setminus\mathcal{B}$ such that $\\{r,s\\}<\\{a,b\\}$
and $A(u_{a},u_{b})=A(u_{r},u_{s})\in B$, we have that
$A(u_{a},u_{b})\in\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}<\\{a,b\\}\\})$.)

###### Proof.

  1. (i)

By construction, the set $B_{m}$ is linearly independent and hence a basis of
$V$. Note that $B=B_{m}$ and hence $\mathcal{B}$ satisfies the first property.

  2. (ii)

For all $\\{r,s\\}\in\mathcal{Y}$, either $\\{r,s\\}\in\mathcal{B}$ or
$\\{r,s\\}\notin\mathcal{B}$. If $\\{r,s\\}\in\mathcal{B}$, then, clearly,
$A(u_{r},u_{s})\in\text{span }(\\{A(u_{i},u_{j})\in B\mid\
\\{i,j\\}\leq\\{r,s\\}\\})$. Now assume that $\\{r,s\\}\notin\mathcal{B}$. If
$A(u_{r},u_{s})$ is trivial, then the result holds trivially. Thus we can
assume that $A(u_{r},u_{s})$ is non-zero. As $\\{i_{1}^{1},i_{2}^{1}\\}$ is
the least element in $\mathcal{Y}$ such that $A(u_{i^{1}_{1}},u_{i_{2}^{1}})$
is non-zero, we have that $\\{i_{1}^{1},i_{2}^{1}\\}<\\{r,s\\}$. If
$A(u_{r},u_{s})\in\text{span }(\\{A(u_{i^{1}_{1}},u_{i_{2}^{1}})\\})$, we are
done. Thus assume that $A(u_{r},u_{s})\notin\text{span }(B_{1})$. We know that
$A(u_{r},u_{s})\in\text{span }(B_{m})$. Therefore, there exists
$j\in\mathbb{N}$ such that $A(u_{r},u_{s})\notin\text{span }(B_{j})$ and
$A(u_{r},u_{s})\in\text{span }(B_{j+1})$. By the construction of the basis,
$\\{i_{1}^{j+1},i_{2}^{j+1}\\}$ is the least element in $\mathcal{Y}$ such
that $A(u_{i_{1}^{j+1}},u_{i_{2}^{j+1}})\notin\text{span }(B_{j})$. Thus we
have that $\\{i_{1}^{j+1},i_{2}^{j+1}\\}<\\{r,s\\}$. Similarly,
$\\{i_{1}^{l},i_{2}^{l}\\}<\\{r,s\\}$ for all $l\leq j+1$. As
$A(u_{r},u_{s})\in\text{span }(B_{j+1})$, we have that
$A(u_{r},u_{s})\in\text{span }(\\{A(u_{i},u_{j})\in B\mid\
\\{i,j\\}\in\mathcal{B},\\{i,j\\}\leq\\{r,s\\}\\})$, and hence the proof.

  3. (iii)

Follows from $\textit{(ii)}.$

  4. (iv)

Suppose the set $\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}$ is non
trivial, then we claim it is $\mathcal{B}_{k}$ for some $1\leq k\leq m$. Let
$\mathcal{B}_{1}=\\{i_{1}^{1},i_{2}^{1}\\}$ and for $2\leq l<m$, let
$\mathcal{B}_{l}\setminus\mathcal{B}_{l-1}=\\{i_{1}^{l},i_{2}^{l}\\}.$ By
construction,
$\\{i_{1}^{1},i_{2}^{1}\\}<\cdots<\\{i_{1}^{1},i_{2}^{1}\\}<\cdots<\\{i_{1}^{m},i_{2}^{m}\\}.$
Observe that for $1\leq t\leq m$,
$\mathcal{B}_{t}=\\{\\{i_{1}^{l},i_{2}^{l}\\}\mid l\leq t\\}$. Let $k$ be the
largest integer such that $\\{i_{1}^{k},i_{2}^{k}\\}<\\{r,s\\}$. Clearly
$\mathcal{B}_{k}=\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}$.

  5. (v)

We will prove that if $A(u_{r},u_{s})\in\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}<\\{r,s\\}\\})$, then $\\{r,s\\}\notin\mathcal{B}$. By (iv), the
set $\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}$ is either trivial
or $\mathcal{B}_{k}$ for some $1\leq k\leq m$. If
$\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}$ is trivial, then the
proof follows. Thus we assume
$\\{\\{i,j\\}\in\mathcal{B}\mid\\{i,j\\}<\\{r,s\\}\\}=\mathcal{B}_{k}$ for
some $1\leq k\leq m$, which implies $\\{r,s\\}\notin\mathcal{B}_{k}.$ By
(iii), $\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}<\\{r,s\\}\\})=\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}\in\mathcal{B},\\{i,j\\}<\\{r,s\\}\\})$, and by the definition
of $\mathcal{B}_{k}$, the latter is equal to $\text{span
}(\\{A(u_{i},u_{j})\in B\mid\\{i,j\\}\in\mathcal{B}_{k}\\})$. Thus, if
$A(u_{r},u_{s})\in\text{span }(\\{A(u_{i},u_{j})\in
B\mid\\{i,j\\}<\\{r,s\\}\\})=\text{span }(B_{k})$, then
$\\{r,s\\}\notin\mathcal{B}_{l}$ for any $k<l\leq m$. Thus
$\\{r,s\\}\notin\mathcal{B}$, a contradiction.

∎

Throughout this section, when a basis of $V$ is mentioned, it is assumed to be
the basis constructed above. We give an example to show how $\mathcal{B}$ is
constructed.

######  Example 1 (Construction of $\mathcal{B}$).

Consider the $\mathbb{F}_{p}$ vector spaces $U$ and $V$. Let
$X=\\{u_{1},u_{2},u_{3},u_{4},u_{5}\\}$ be a basis of $U$ and
$\\{v_{1},v_{2},v_{3},v_{4}\\}$ be a basis of $V$. Assume that $A:U\times U\to
V$ is an alternating bilinear map defined as follows,

$\displaystyle A(u_{1},u_{2})$ $\displaystyle=v_{1}$ $\displaystyle
A(u_{1},u_{3})$ $\displaystyle=v_{2}$ $\displaystyle A(u_{2},u_{3})$
$\displaystyle=0$ $\displaystyle A(u_{1},u_{4})$ $\displaystyle=0$
$\displaystyle A(u_{2},u_{4})$ $\displaystyle=v_{3}$ $\displaystyle
A(u_{3},u_{4})=0$ $\displaystyle A(u_{1},u_{5})$
$\displaystyle=-(v_{1}+v_{2})$ $\displaystyle A(u_{2},u_{5})$
$\displaystyle=v_{4}$ $\displaystyle A(u_{3},u_{5})=0$ $\displaystyle
A(u_{4},u_{5})=v_{3}.$

Let $X$ be ordered as $u_{1}<u_{2}<\cdots<u_{5}$. Observe that $\\{1,2\\}$ is
the least element of $\mathcal{Y}$ such that $A(u_{1},u_{2})$ is non-zero.
Thus $\mathcal{B}_{1}=\\{\\{1,2\\}\\}$ and $B_{1}=\\{A(u_{1},u_{2})\\}$. Next,
observe that $A(u_{1},u_{3})\notin\text{span }(B_{1})$ and $\\{1,3\\}$ is the
smallest such element in $\mathcal{Y}$. Thus
$\mathcal{B}_{2}=\\{\\{1,2\\},\\{1,3\\}\\}$ and
$B_{2}=\\{A(u_{1},u_{2}),A(u_{1},u_{3})\\}$. Now,
$A(u_{2},u_{4})\notin\text{span }(B_{2})$ and $\\{2,4\\}$ is the smallest such
element ($\\{4,5\\}$ is another element such that
$A(u_{4},u_{5})\notin\text{span }(B_{2})$ but $\\{4,5\\}$ is not the smallest
such element.) Thus $\mathcal{B}_{3}=\\{\\{1,2\\},\\{1,3\\},\\{2,4\\}\\}$ and
$B_{3}=\\{A(u_{1},u_{2}),A(u_{1},u_{3}),A(u_{2},u_{5})\\}$. Lastly,
$A(u_{2},u_{5})\notin\text{span }(B_{3})$ and $\\{2,5\\}$ is the smallest such
element. Thus $\mathcal{B}_{4}=\\{\\{1,2\\},\\{1,3\\},\\{2,4\\},\\{2,5\\}\\}$
and $B_{4}=\\{A(u_{1},u_{2}),A(u_{1},u_{3}),A(u_{2},u_{4}),A(u_{2},u_{5})\\}$.
As the dimension of $V$ over $\mathbb{F}_{p}$ is $4$, $B=B_{4}$ and
$\mathcal{B}=\mathcal{B}_{4}$.

######  Definition 2.2.

The set $\mathcal{B}\subseteq\mathcal{Y}$ constructed in section 2.1 is called
a tree of height one if there exists $i\in\\{1,\dots,n\\}$ such that
$i\in\\{k,l\\}$ for every $\\{k,l\\}\in\mathcal{B}.$

For any $\mathcal{S}\subseteq\mathcal{Y}$, let $\mathcal{G}(\mathcal{S})$ be
the graph whose vertex set is $\\{1,\dots,n\\}$, and there exists edge between
$i$ and $j$ if $\\{i,j\\}\in\mathcal{S}$. The set $\mathcal{B}$ is a tree of
height one if and only if graph $\mathcal{G}(\mathcal{B})$ is a tree with
height one. Such a set $\mathcal{B}$ can be visualized as follows:

$i$ $i$

######  Proposition 2.3.

Let $U$ and $V$ be finite dimensional vector spaces over $\mathbb{F}$ with
dimension $m$ and $n$ respectively, and let $A:U\times U\to V$ be a
alternating map such that image of $A$ spans $V$. If $m>2$ and $n>m+1$, there
exists a basis $\\{u_{1},\ldots,u_{n}\\}$ of $U$ with the order
$u_{1}<\cdots<u_{n}$ such that the set $\mathcal{B}$ constructed using the
algorithm given in Section 2.1 is not a tree of height one or satisfies the
following properties:

$\displaystyle\mathcal{B}=\\{\\{1,2\\},\dots,\\{1,{m+1}\\}\\}.$ (2.3.1)
$\displaystyle A(u_{1},u_{i})=0\text{ for all }m+2\leq i.$ (2.3.2)
$\displaystyle A(u_{i},u_{j})=0\text{ for all }i,j\text{ with }m+1<i<j.$
(2.3.3) $\displaystyle A(u_{i},u_{j})\in\text{span }\\{A(u_{1},u_{i})\\}\text{
for all }i,j\text{ with }2\leq i\leq m+1<j\leq n.$ (2.3.4)

###### Proof.

Since the proof is long, we will divide the proof into 4 steps. In i-th step,
we will show that the set $\mathcal{B}$ is not a tree of height one or
$\mathcal{B}$ satisfies properties (2.3.1),…, (2.3.i).

Step 1

Let $X=\\{u_{1},\ldots,u_{n}\\}$ be any basis of $U$ with order
$u_{1}<\dots<u_{n}$. Corresponding to this order, let $\mathcal{B}$ and $B$ be
as constructed in Section 2.1. If $\mathcal{B}_{W}$ is not a tree of height
one, then Step 1 is complete. So we can assume $\mathcal{B}$ is a tree of
height one. Thus there exists an $i_{1}\in\\{1,\dots,n\\}$ such that
$\mathcal{B}=\\{\\{i_{1},i_{2}\\},\dots,\\{i_{1},i_{m+1}\\}\\}$. let
$X\setminus\\{u_{i_{1}},u_{i_{2}},\dots,u_{i_{m+1}}\\}=\\{w_{m+2},\dots,w_{n}\\}$,
set $w_{1}=u_{i_{1}},w_{2}=u_{i_{2}},\ldots,w_{m+1}=u_{i_{m+1}}$. The set
$X_{1}=\\{v_{1},\dots,v_{n}\\}$ is a basis of $V$. Let $v_{1}<\cdots<v_{n}$ be
a total order on $X_{1}.$ Corresponding to this order, consider the order on
set $\mathcal{Y}$ defined in Section 2.1; namely $\\{i,j\\}<\\{s,t\\}$,
whenever $\max\\{w_{i},w_{j}\\}<\max\\{w_{s},w_{t}\\}$ or
$\max\\{w_{i},w_{j}\\}=\max\\{w_{s},w_{t}\\}=w_{a}$ and
$\\{w_{i},w_{j}\\}\setminus\\{w_{a}\\}<\\{w_{s},w_{t}\\}\setminus\\{w_{a}\\}$.
The order on $\mathcal{Y}$ can be visualized as follows:

$\begin{array}[]{cccccc}&\\{{1},{2}\\}\\\ &\\{{1},{3}\\}&\\{{2},{3}\\}\\\
&\vdots\\\
&\\{{1},{n}\\}&\\{{2},{n}\\}&\\{{3},{n}\\}&\cdots&\\{{n-1},{n}\\}.\\\
\end{array}$

Now we construct the set $\mathcal{B}^{1}$ using the algorithm given in
Section 2.1. Observe that $A(w_{1},w_{2})=A(u_{i_{1}},u_{i_{2}})$ and
$A(w_{1},w_{3})=A(u_{i_{1}},u_{i_{3}})$ are linearly independent and hence
$\\{1,2\\},\\{1,3\\}\in\mathcal{B}^{1}$. If $\mathcal{B}^{1}$ is not a tree of
height one, then Step 1 is complete. If $\mathcal{B}^{1}$ is a tree of height
one, then $\mathcal{B}^{1}=\\{\\{1,2\\},\dots,\\{1,m\\}\\}$ and hence Step 1
is complete.

Step 2

We will start with a basis $X_{1}=\\{u_{1},\ldots,u_{n}\\}$ of $U$ with order
$u_{1}<\dots<u_{n}$, that we obtained in Step 1. Corresponding to this order,
let $\mathcal{B}^{1}$ and the basis $B^{1}$ of $V$ be as constructed in
Section 2.1. If $\mathcal{B}^{1}$ is not a tree of height one, then Step 1 is
complete. So we can assume $\mathcal{B}^{1}$ satisfies property (2.3.1).
Consider the linear map

$\displaystyle L:U$ $\displaystyle\to V$ $\displaystyle u$
$\displaystyle\mapsto A(u_{1},u)$

Since $\\{A(u_{1},u_{i})\mid 2\leq i\leq m+1\\}$ is a basis of $V$, the map
$L$ is surjective. Note that dim Ker $(L)=n-m$. Let
$\\{u_{1},w_{m+2},\dots,w_{n}\\}$ be a basis of Ker $(L)$. Set $w_{i}=u_{i}$
for all $1\leq i\leq m+1$. Now note that $X_{2}=\\{w_{1},\dots,w_{n}\\}$ is a
basis of $V$. Let $w_{1}<\dots<w_{\delta}$ be a total order on $X_{2}$.
Corresponding to this order, we construct $\mathcal{B}^{2}$ and the basis
$B^{2}$ of $V$ using the algorithm given in Section 2.1. It is easy to see
that $\mathcal{B}^{2}=\\{\\{1,2\\},\dots,\\{1,i_{m+1}\\}\\}$ and
$A(w_{1},w_{r})=0$ for every $r\geq m+2.$ and hence the Step 2 is complete.

Step 3

We will start with a basis $X_{2}=\\{u_{1},\ldots,u_{n}\\}$ of $U$ with order
$u_{1}<\dots<u_{n}$, that we obtained in Step 2. Let $\mathcal{B}^{2}$ and the
basis $B^{2}$ of $V$ be as constructed in Section 2.1. Suppose
$A(u_{r},u_{s})\neq 0$ for some $r,s$ with $\delta\geq s>r>m+1$, we will show
that there exist a basis $X_{3}=\\{w_{1},\ldots,w_{n}\\}$ of $V$ with order
$w_{1}<\dots<w_{n}$ such that corresponding $\mathcal{B}^{3}$ is not a tree of
height one. Let $j\in\\{2,\dots,m+1\\}$ be such that $A(u_{r},u_{s})$ and
$A(u_{1},u_{j})$ are linearly independent, and let
$X_{2}\setminus\\{u_{r},u_{s},u_{1},u_{j}\\}=\\{w_{5},\dots,w_{n}\\}$. Set
$w_{1}=u_{1},w_{2}=u_{r},w_{3}=u_{s},w_{4}=u_{j}$. Let
$X_{3}=\\{w_{1},\ldots,w_{n}\\},$ and let $w_{1}<\cdots<w_{n}$ be a total
order on $X_{3}$. Corresponding to this order, consider the order on set
$\mathcal{Y}$ defined in Section 2.1; namely $\\{i,j\\}<\\{s,t\\}$, whenever
$\max\\{w_{i},w_{j}\\}<\max\\{w_{s},w_{t}\\}$ or
$\max\\{w_{i},w_{j}\\}=\max\\{w_{s},w_{t}\\}=w_{a}$ and
$\\{w_{i},w_{j}\\}\setminus\\{w_{a}\\}<\\{w_{s},w_{t}\\}\setminus\\{w_{a}\\}$.
Now we construct the set $\mathcal{B}^{3}$ using the algorithm given in
Section 2.1. We have $A(w_{1},w_{2})=A(u_{1},u_{r})=0$ and
$A(w_{1},w_{3})=A(u_{1},u_{s})=0$. Moreover $A(w_{2},w_{3})=A(u_{r},u_{s})$
and $A(w_{1},w_{4})=A(u_{1},u_{j})$ are linearly independent. Thus
$\\{2,3\\},\\{1,4\\}\in\mathcal{B}^{3}$. Thus the basis $B_{3}$ is not a tree
of height one and Step 3 is complete.

Step 4

We will start with a basis $X_{3}=\\{u_{1},\ldots,u_{n}\\}$ of $U$ with order
$u_{1}<\dots<u_{n}$, that we obtained in Step 3. We will show if
$A(u_{r},u_{s})\notin\text{span }\\{A(u_{1},u_{r})\\}$ for all $r,s\text{ with
}2\leq r\leq m+1<s\leq n$, then there exists a basis
$X_{4}=\\{w_{1},\ldots,w_{n}\\}$ of $U$ with order $w_{1}<\dots<w_{n}$ such
that corresponding $\mathcal{B}^{4}$ is not a tree of height one. Suppose
there exists $r,s\text{ with }2\leq r\leq m+1<s\leq\delta$ such that
$A(u_{r},u_{s})\notin\text{span }\\{A(u_{1},u_{r})\\}$. Since $\mathcal{B}$
satisfies (2.3.1) and $m\geq 3$, there exists $2\leq j\leq m+1$ with $j\neq r$
such that $A(u_{1},u_{r}),A(u_{r},u_{s})$ and $A(u_{1},u_{j})$ are linearly
independent. Let
$X_{3}\setminus\\{u_{1},u_{r},u_{s},u_{j}\\}=\\{w_{5},\dots,w_{n}\\}$. Set
$w_{1}=u_{1},w_{2}=u_{r},w_{3}=u_{s},w_{4}=u_{j}$, and let
$X_{4}=\\{w_{1},\ldots,w_{\delta}\\},$ and let $w_{1}<\cdots<w_{\delta}$ be a
total order on $X_{4}$. Corresponding to this order, let $\mathcal{B}^{4}$ and
the basis $B^{4}$ of $V$ be as constructed in Section 2.1. Since $s>m+1$, we
have $A(w_{1},w_{3})=A(u_{1},u_{s})=0$, by (2.3.2). Moreover,
$A(w_{1},w_{2})=A(u_{1},u_{r})$, $A(w_{2},w_{3})=A(u_{r},u_{s})$ and
$A(w_{1},w_{4})=A(u_{1},u_{j})$ are linearly independent. Thus we have
$\\{1,2\\},\\{2,3\\},\\{1,4\\}\in\mathcal{B}^{4}$, and hence $\mathcal{B}^{4}$
is not a tree of height one, which completes Step 4. ∎

### 2.2 A set of linear independent elements in the image of $\Psi$

Consider the following map,

$\displaystyle\Psi$ $\displaystyle:U\times U\times U\to V\otimes U$
$\displaystyle(x,y,z)$ $\displaystyle\mapsto A(x,y)\otimes z+A(y,z)\otimes
x+A(z,x)\otimes y.$

The following lemma lists some properties of $\Psi$ which can be easily
verified. We omit the proof of this lemma.

######  Lemma 2.4.

Let $U$ and $V$ be finite-dimensional $F$ vector spaces with an alternating
bilinear map $A:U\times U\to V$. Let $\sigma$ be a permutation on
$\\{x,y,z\\}$. We have the following properties for $\Psi$,

  1. (i)

$\Psi(x,x,y)=\Psi(x,y,x)=\Psi(y,x,x)=0$,

  2. (ii)

$\Psi(\sigma(x),\sigma(y),\sigma(z))=sgn(\sigma)\Psi(x,y,z)$,

  3. (iii)

$\Psi(x_{1}x_{2},y,z)=\Psi(x_{1},y,z)+\Psi(x_{2},y,z)$.

Observe that this trilinear map induces the following linear map,

$\displaystyle\Psi$ $\displaystyle:U\otimes U\otimes U\to V\otimes U.$

For a basis $B$ of $V$ and a basis $X$ for $U$, it is well known that the
basis of $V\otimes U$ is given by $B\otimes
X=\\{A(u_{i_{1}^{j}},u_{i_{2}^{j}})\otimes u_{l}\mid
A(u_{i_{1}^{j}},u_{i_{2}^{j}})\in B,u_{l}\in X\\}$. Thus we have the
decomposition, $V\otimes U=\bigoplus\limits_{A(u_{i},u_{j})\in
B}\bigoplus\limits_{k=1}^{n}\langle A(u_{i},u_{j})\otimes u_{k}\rangle$. Hence
we have the projection maps $P_{i,j,k}:V\otimes U\to\langle
A(u_{i},u_{j})\otimes y_{k}\rangle$. Consider the following set,

$\displaystyle\mathcal{W}(\mathcal{B})$ $\displaystyle=\\{\\{a,b,c\\}\mid
1\leq a,b,c\leq n,\\{a,b\\}\in\mathcal{B},c\notin\\{a,b\\}\\}.$ (2.4.1)

######  Remark 1.

Note that $\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$ if and only if $1\leq
a,b,c\leq n$, $\\{a,b\\}\in\mathcal{B}$ or $\\{b,c\\}\in\mathcal{B}$ or
$\\{c,a\\}\in\mathcal{B}$ and $a,b,c$ are pairwise distinct.

We define $f:\mathcal{W}(\mathcal{B})\to V\otimes U$ given by
$f(\\{a,b,c\\})=\Psi(u_{i}\otimes u_{j}\otimes u_{k})$ where
$u_{k}=\max\\{u_{a},u_{b},u_{c}\\}$, $u_{i}=\min\\{u_{a},u_{b},u_{c}\\}$ and
$u_{j}=\\{u_{a},u_{b},u_{c}\\}\setminus\\{u_{i},u_{k}\\}$. The following set
which depends on $\mathcal{B}$ will play an important role in the rest of the
paper:

$\displaystyle W(\mathcal{B})$ $\displaystyle=f(\mathcal{W}(\mathcal{B}))$
(2.4.2)

An element of $W(\mathcal{B})$ is of the form $A(u_{a},u_{b})\otimes
u_{c}+A(u_{b},u_{c})\otimes u_{a}+A(u_{c},u_{a})\otimes u_{b}$ where
$u_{a}<u_{b}<u_{c}$. We will prove that $W(\mathcal{B})$ is linearly
independent in $V\otimes U$. We impose a total order on
$\mathcal{W}(\mathcal{B})$ defined by $\\{a,b,c\\}<\\{i,j,k\\}$ if
$\max\\{u_{a},u_{b},u_{c}\\}<\max\\{u_{i},u_{j},u_{k}\\}$ or
$\max\\{u_{a},u_{b},u_{c}\\}=\max\\{u_{i},u_{j},u_{k}\\}=u_{l}$ and
$\\{a,b,c\\}\setminus\\{l\\}<\\{i,j,k\\}\setminus\\{l\\}$ where the comparison
on sets of order two is as defined in Section 2.1.

######  Example 2.

(Importance of $\mathcal{B}$) The use of $\mathcal{B}$ in the definition of
$W(\mathcal{B})$ is crucial in showing the linear independence of
$W(\mathcal{B})$ in Proposition 2.7.

One may wonder whether the set

$\displaystyle W(B)=\\{\Psi_{2}(g_{a}\otimes g_{b}\otimes g_{c})\mid$
$\displaystyle 1\leq a,b,c\leq\delta,g_{a}<g_{b}<g_{c},A(u_{a},u_{b})\in
B\text{ or }A(u_{b},u_{c})\in B$ $\displaystyle\text{ or }A(u_{a},u_{c})\in
B\\}$

is linearly independent. We give an example where $W(B)$ fails to be linearly
independent. From the definitions, it is clear that $W(\mathcal{B})\subseteq
W(B)$. We will give an example for which $W(\mathcal{B})\subsetneq W(B)$. This
example also shows that $W(B)$ is not necessarily linearly independent.
Consider the $\mathbb{F}_{p}$ vector spaces $U$ and $V$. Let
$X=\\{u_{1},u_{2},u_{3},u_{4},u_{5}\\}$ be a basis of $U$ and
$\\{v_{1},v_{2},v_{3},v_{4},v_{5}\\}$ be a basis of $V$. Assume that
$A:U\times U\to V$ is an alternating bilinear map defined as follows,

$\displaystyle A(u_{1},u_{2})$ $\displaystyle=v_{1}$ $\displaystyle
A(u_{1},u_{3})$ $\displaystyle=v_{2}$ $\displaystyle A(u_{2},u_{3})$
$\displaystyle=0$ $\displaystyle A(u_{1},u_{4})$ $\displaystyle=0$
$\displaystyle A(u_{2},u_{4})$ $\displaystyle=v_{3}$ $\displaystyle
A(u_{3},u_{4})=0$ $\displaystyle A(u_{1},u_{5})$
$\displaystyle=-(v_{1}+v_{2})$ $\displaystyle A(u_{2},u_{5})$
$\displaystyle=v_{4}$ $\displaystyle A(u_{3},u_{5})=0$ $\displaystyle
A(u_{4},u_{5})=v_{3}.$

Let $X=\\{u_{1},\ldots,u_{5}\\}$ be given the order $u_{1}<\cdots<u_{5}$. For
this group, we already observed that
$B=\\{A(u_{1},u_{2}),A(u_{1},u_{3}),A(u_{2},u_{4}),A(u_{2},u_{5})\\}$ and
$\mathcal{B}=\\{\\{1,2\\},\\{1,3\\},\\{2,4\\},\\{2,5\\}\\}$. By definition,

$\displaystyle W(B)=$ $\displaystyle\\{\Psi_{2}(g_{1}\otimes g_{2}\otimes
g_{3}),\Psi_{2}(g_{1}\otimes g_{2}\otimes g_{4}),\Psi_{2}(g_{1}\otimes
g_{2}\otimes g_{5}),\Psi_{2}(g_{1}\otimes g_{3}\otimes
g_{4}),\Psi_{2}(g_{1}\otimes g_{3}\otimes g_{5}),$
$\displaystyle\Psi_{2}(g_{2}\otimes g_{3}\otimes g_{4}),\Psi_{2}(g_{2}\otimes
g_{4}\otimes g_{5}),\Psi_{2}(g_{2}\otimes g_{3}\otimes
g_{5}),\Psi_{2}(g_{1}\otimes g_{4}\otimes g_{5}),\Psi_{2}(g_{3}\otimes
g_{4}\otimes g_{5})\\},$ $\displaystyle W(\mathcal{B})=$
$\displaystyle\\{\Psi_{2}(g_{1}\otimes g_{2}\otimes
g_{3}),\Psi_{2}(g_{1}\otimes g_{2}\otimes g_{4}),\Psi_{2}(g_{1}\otimes
g_{2}\otimes g_{5}),\Psi_{2}(g_{1}\otimes g_{3}\otimes
g_{4}),\Psi_{2}(g_{1}\otimes g_{3}\otimes g_{5}),$
$\displaystyle\Psi_{2}(g_{2}\otimes g_{3}\otimes g_{4}),\Psi_{2}(g_{2}\otimes
g_{4}\otimes g_{5}),\Psi_{2}(g_{2}\otimes g_{3}\otimes g_{5})\\}.$

Observe that $\Psi_{2}(g_{1}\otimes g_{4}\otimes g_{5})\in W(B)$ while
$\Psi_{2}(g_{1}\otimes g_{4}\otimes g_{5})\notin W(\mathcal{B})$, and hence
$W(\mathcal{B})\subsetneq W(B)$. Note that $\Psi_{2}(g_{1}\otimes g_{4}\otimes
g_{5})\in W(B)$ because $A(u_{4},u_{5})=A(u_{2},u_{4})\in B$ while
$\Psi_{2}(g_{1}\otimes g_{4}\otimes g_{5})\notin W(\mathcal{B})$ as
$\\{1,4\\},\\{4,5\\},\\{1,5\\}\notin\mathcal{B}$. We observe that

$\displaystyle\Psi_{2}(g_{1}\otimes g_{2}\otimes g_{4})+\Psi_{2}(g_{1}\otimes
g_{3}\otimes g_{4})=$ $\displaystyle A(u_{1},u_{2})\otimes
u_{4}+A(u_{2},u_{4})\otimes u_{1}+A(u_{4},u_{1})\otimes u_{2}+$ $\displaystyle
A(u_{1},u_{3})\otimes u_{4}+A(u_{3},u_{4})\otimes u_{1}+A(u_{4},u_{1})\otimes
u_{3}$ $\displaystyle=$ $\displaystyle(A(u_{1},u_{2})+A(u_{1},u_{3}))\otimes
u_{4}+A(u_{2},u_{4})\otimes u_{1}$ $\displaystyle=$ $\displaystyle
A(u_{5},u_{1})\otimes u_{4}+A(u_{4},u_{5})\otimes u_{1}+A(u_{1},u_{4})\otimes
u_{5}$ $\displaystyle=$ $\displaystyle\Psi_{2}(g_{1}\otimes g_{4}\otimes
g_{5}).$

Hence $W(B)$ is not linearly independent.

In the next example, we will show that given an arbitrary
$\mathcal{S}\subseteq\mathcal{Y}$ such that $S=\\{A(u_{i},u_{j})\in
Y\mid\\{i,j\\}\in\mathcal{S}\\}$ is a basis of $V$, we may not be able to
prove that $W(\mathcal{S})$ is linearly independent.

######  Example 3.

(Importance of method of construction of $\mathcal{B}$)The construction of
$\mathcal{B}$ is also crucial to the linear independence of $W(\mathcal{B})$
in Proposition 2.7.

Consider the $\mathbb{F}_{p}$ vector spaces $U$ and $V$. Let
$X=\\{u_{1},u_{2},u_{3},u_{4},u_{5}\\}$ be a basis of $U$ and
$\\{v_{1},v_{2},v_{3},v_{4},v_{5}\\}$ be a basis of $V$. Assume that
$A:U\times U\to V$ is an alternating bilinear map defined as follows,

$\displaystyle A(u_{1},u_{2})$ $\displaystyle=v_{1}$ $\displaystyle
A(u_{1},u_{3})$ $\displaystyle=v_{2}$ $\displaystyle A(u_{2},u_{3})$
$\displaystyle=0$ $\displaystyle A(u_{1},u_{4})$ $\displaystyle=v_{1}+v_{2}$
$\displaystyle A(u_{2},u_{4})$ $\displaystyle=v_{3}$ $\displaystyle
A(u_{3},u_{4})=0$ $\displaystyle A(u_{1},u_{5})$ $\displaystyle=0$
$\displaystyle A(u_{2},u_{5})$ $\displaystyle=v_{5}-v_{4}$ $\displaystyle
A(u_{3},u_{5})=v_{4}$ $\displaystyle A(u_{4},u_{5})=v_{5}.$

Let $X=\\{u_{1},\ldots,u_{5}\\}$ be given the order $u_{1}<\cdots<u_{5}$. Note
$\mathcal{S}=\\{\\{1,2\\},\\{1,3\\},\\{2,4\\},\newline
\\{3,5\\},\\{4,5\\}\\}\subseteq\mathcal{Y}$ is such that
$S=\\{A(u_{1},u_{2}),A(u_{1},u_{3}),A(u_{2},u_{4}),A(u_{3},u_{5}),A(u_{4},u_{5})\\}$
is a basis of $V$. By definition, $\Psi_{2}(g_{1}\otimes g_{2}\otimes
g_{5}),\Psi_{2}(g_{1}\otimes g_{3}\otimes g_{5}),\Psi_{2}(g_{1}\otimes
g_{4}\otimes g_{5})\in W(\mathcal{S})$. We observe that

$\displaystyle\Psi_{2}(g_{1}\otimes g_{2}\otimes g_{5})+\Psi_{2}(g_{1}\otimes
g_{3}\otimes g_{5})=$ $\displaystyle A(u_{1},u_{2})\otimes
u_{5}+A(u_{2},u_{5})\otimes u_{1}+A(u_{5},u_{1})\otimes u_{2}+$ $\displaystyle
A(u_{1},u_{3})\otimes u_{5}+A(u_{3},u_{5})\otimes u_{1}+A(u_{5},u_{1})\otimes
u_{3}$ $\displaystyle=$ $\displaystyle(A(u_{1},u_{2})+A(u_{1},u_{3}))\otimes
u_{5}+(A(u_{2},u_{5})+A(u_{3},u_{5}))\otimes u_{1}$ $\displaystyle=$
$\displaystyle A(u_{1},u_{4})\otimes u_{5}+A(u_{4},u_{5})\otimes
u_{1}+A(u_{5},u_{1})\otimes u_{4}$ $\displaystyle=$
$\displaystyle\Psi_{2}(g_{1}\otimes g_{4}\otimes g_{5}).$

Hence $W(\mathcal{S})$ is not linearly independent. Note that our construction
would yield
$\mathcal{B}=\\{\\{1,2\\},\\{1,3\\},\\{2,4\\},\\{2,5\\},\\{3,5\\}\\}$ and the
special basis as $B=\\{A(u_{1},u_{2}),A(u_{1},u_{3}),A(u_{2},u_{4}),\newline
A(u_{2},u_{5}),A(u_{3},u_{5})\\}$.

######  Lemma 2.5.

Let $\\{a,b,c\\},\\{x,y,z\\}\in\mathcal{W}(\mathcal{B})$. If
$\\{a,b,c\\}<\\{x,y,z\\}$ and $l\in\\{a,b,c\\}\cap\\{x,y,z\\}$, then
$\\{a,b,c\\}\setminus\\{l\\}<\\{x,y,z\\}\setminus\\{l\\}$.

###### Proof.

The proof proceeds by cases:

Case 1: $\max\\{u_{a},u_{b},u_{c}\\}<\max\\{u_{x},u_{y},u_{z}\\}$

Suppose $\max\\{u_{x},u_{y},u_{z}\\}=u_{l}$. As
$u_{l}\in\\{u_{a},u_{b},u_{c}\\}$, we get
$u_{l}\leq\max\\{u_{a},u_{b},u_{c}\\}<\max\\{u_{x},u_{y},u_{z}\\}=u_{l}$,
which is a contradiction. Thus we can assume that
$u_{l}\neq\max\\{u_{x},u_{y},u_{z}\\}$. We have
$\max\\{u_{a},u_{b},u_{c}\\}\setminus\\{u_{l}\\}\leq\max\\{u_{a},u_{b},u_{c}\\}<\max\\{u_{x},u_{y},u_{z}\\}$.
As $u_{l}\neq\max\\{u_{x},u_{y},u_{z}\\}$,
$\max\\{u_{x},u_{y},u_{z}\\}=\max\\{u_{x},u_{y},u_{z}\\}\setminus\\{u_{l}\\}$.
Thus, we get
$\max\\{u_{a},u_{b},u_{c}\\}\setminus\\{u_{l}\\}<\max\\{u_{x},u_{y},u_{z}\\}\setminus\\{u_{l}\\}$.
Therefore, $\\{a,b,c\\}\setminus\\{l\\}<\\{x,y,z\\}\setminus\\{l\\}$.

Case 2: $\max\\{u_{a},u_{b},u_{c}\\}=\max\\{u_{x},u_{y},u_{z}\\}$

If $u_{l}=\max\\{u_{a},u_{b},u_{c}\\}=\max\\{u_{x},u_{y},u_{z}\\}$, then
$\\{a,b,c\\}<\\{x,y,z\\}$ implies
$\\{a,b,c\\}\setminus\\{l\\}<\\{x,y,z\\}\setminus\\{l\\}$. Assume
$u_{k}=\max\\{u_{a},u_{b},u_{c}\\}=\max\\{u_{x},u_{y},u_{z}\\}$ and $u_{l}\neq
u_{k}$. Let $\\{u_{a},u_{b},u_{c}\\}=\\{u_{l},u_{k},u_{a^{\prime}}\\}$ and
$\\{u_{x},u_{y},u_{z}\\}=\\{u_{l},u_{k},u_{x^{\prime}}\\}$. Since the maximum
of both sets are equal, we have that
$\\{u_{l},u_{a^{\prime}}\\}<\\{u_{l},u_{x^{\prime}}\\}$. From this, it is easy
to see that $u_{a^{\prime}}<u_{x^{\prime}}$. Then we have
$\\{u_{k},u_{a^{\prime}}\\}<\\{u_{k},u_{x^{\prime}}\\}$ which is the required
result.

∎

######  Lemma 2.6.

Let $U$ and $V$ be finite-dimensional $F$ vector spaces with an alternating
bilinear map $A:U\times U\to V$ such that $\text{Im }{(A)}$ spans V. Let $X$
be a basis of $U$ with a total order and let $B$ be the basis of $V$ defined
in Lemma 2.1. Suppose $B\otimes X$ be the basis of $V\otimes U$ as defined in
Section 2.2. Assume that $\\{x,y\\}\in\mathcal{B}$, $A(u_{x},u_{y})\in B$ and
$A(u_{x},u_{y})\otimes u_{z}$ be a basis element of $V\otimes U$. Consider
$\\{a,b\\}\in\mathcal{Y}$. The following statements hold:

  1. (i)

If $u_{c}\neq u_{z}$, then $P_{x,y,z}(A(u_{a},u_{b})\otimes u_{c})=0$.

  2. (ii)

Suppose that $\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$. If
$P_{x,y,z}(\Psi(u_{a}\otimes u_{b}\otimes u_{c}))\neq 0$, then either
$u_{a}=u_{z}$ or $u_{b}=u_{z}$ or $u_{c}=u_{z}$.

  3. (iii)

Let $\\{a,b\\}\in\mathcal{Y}$. If $\\{a,b\\}<\\{x,y\\}$, then
$P_{x,y,z}(\Psi(u_{\sigma(a)}\otimes u_{\sigma(b)}\otimes u_{\sigma(z)}))=0$,
where $\sigma$ is a permutation of $\\{a,b,c\\}$.

###### Proof.

  1. (i)

Recall that $P_{x,y,z}:V\otimes U\to\langle A(u_{x},u_{y})\otimes
u_{z}\rangle$. We have $A(u_{a},u_{b})\otimes u_{c}=\sum_{i\in
I}\mu_{i}A(u_{s},u_{t})\otimes u_{c}$, where each $A(u_{s},u_{t})\otimes
u_{c}$ belong to the basis of $V\otimes U$ for some index $I$. Since
$u_{c}\neq u_{z}$, then $P_{x,y,z}(A(u_{a},u_{b})\otimes u_{c})=0$ as the term
$A(u_{x},u_{y})\otimes u_{z}$ does not appear in the linear combination.

  2. (ii)

Since $P_{x,y,z}(\Psi(u_{a}\otimes u_{b}\otimes u_{c}))\neq 0$, we have
$P_{x,y,z}(A(u_{a},u_{b})\otimes u_{c}+A(u_{b},u_{c})\otimes
u_{a}+A(u_{c},u_{a})\otimes u_{b})\neq 0$. Thus
$P_{x,y,z}(A(u_{a},u_{b})\otimes u_{c})+P_{x,y,z}(A(u_{b},u_{c})\otimes
u_{a})+P_{x,y,z}(A(u_{c},u_{a})\otimes u_{b})\neq 0$. Since
$A(u_{a},u_{b})=\sum_{i\in I}\mu_{i}A(u_{s},u_{t})$ for some index $I$, we
have $A(u_{a},u_{b})\otimes u_{c}=\sum_{i\in I}\mu_{i}A(u_{s},u_{t})\otimes
u_{c}$, for some index $I$. If $u_{c}\neq u_{z}$, then
$P_{x,y,z}(A(u_{a},u_{b})\otimes u_{c})=0$ by (i). Similarly, if $u_{b}\neq
u_{z}$ and $u_{a}\neq u_{z}$, then $P_{x,y,z}(A(u_{c},u_{a})\otimes u_{b})=0$
and $P_{x,y,z}(A(u_{b},u_{c})\otimes u_{a})=0$. But this is a contradiction as
the image under the projection is non-trivial.

  3. (iii)

Note that for $\sigma$, $\Psi(u_{\sigma(a)}\otimes u_{\sigma(b)}\otimes
u_{\sigma(z)})=\text{sgn}(\sigma)\Psi(u_{a}\otimes u_{b}\otimes u_{z})$. Hence
it is enough to prove that $P_{x,y,z}(\Psi(u_{a}\otimes u_{b}\otimes
u_{z}))=0$. As $\\{a,b\\}\in\mathcal{Y}$, by Lemma 2.1 (ii), we have
$A(u_{a},u_{b})\in\text{span }(\\{A(u_{s},u_{t})\in B\mid\
\\{s,t\\}\leq\\{a,b\\}\\})$. Thus $A(u_{a},u_{b})=\sum_{i\in
I}\mu_{i}A(u_{s},u_{t})$ for some index $I$ where $A(u_{s},u_{t})\in B$ and
$\\{s,t\\}\leq\\{a,b\\}$. As $\\{a,b\\}<\\{x,y\\}$, $\\{s,t\\}<\\{x,y\\}$.
Thus by Lemma 2.1 (v), $A(u_{x},u_{y})\notin\\{A(u_{s},u_{t})\in B\mid\
\\{s,t\\}\leq\\{a,b\\}\\}$, and hence $A(u_{x},u_{y})$ does not appear in the
linear combination of $A(u_{a},u_{b})$. This gives us that the linear
combination of $A(u_{a},u_{b})\otimes u_{z}$ does not contain the basis
element $A(u_{x},u_{y})\otimes u_{z}$. Thus $P_{x,y,z}(A(u_{a},u_{b})\otimes
u_{z}))=0$. By (i), $P_{x,y,z}(A(u_{b},u_{z})\otimes u_{a})=0$ and
$P_{x,y,z}(A(u_{z},u_{a})\otimes u_{b})=0$. Thus $P_{x,y,z}(\Psi(u_{a}\otimes
u_{b}\otimes u_{z}))=0$.

∎

In the next proposition, we will show that the set $W(\mathcal{B})$ is
linearly independent.

######  Proposition 2.7.

Let $U$ and $V$ be finite-dimensional $F$ vector spaces with an alternating
bilinear map $A:U\times U\to V$ such that $\text{Im }A$ spans V. Let $X$ be a
basis of $U$ with a total order and let $B$ be the basis of $V$ defined in
Lemma 2.1. Suppose $B\otimes X$ be the basis of $V\otimes U$ as defined in
Section 2.2. The following statements hold

  1. (i)

The elements $f(\\{a,b,c\\})$, where $\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$,
are linearly independent in $V\otimes U$.

  2. (ii)

The map $f:\mathcal{W}(\mathcal{B})\to W(\mathcal{B})$ is a bijection.

###### Proof.

  1. (i)

Suppose $\lambda_{\\{a,b,c\\}}\in F$ such that

$\sum\limits_{\\{a,b.c\\}\in\mathcal{W}(\mathcal{B})}\lambda_{\\{a,b,c\\}}f(\\{a,b,c\\})=0.$
(2.7.1)

Our aim is to prove that $\lambda_{\\{a,b,c\\}}=0$ for all
$\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$. If there is an
$\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$ with $\lambda_{\\{a,b,c\\}}\neq 0$,
then we can find the largest element $\\{x,y,z\\}$ in
$\mathcal{W}(\mathcal{B})$ with $\lambda_{\\{x,y,z\\}}\neq 0$. We will use
projection maps, $P_{i,j,k}:V\otimes U\to\langle A(u_{i},u_{j})\otimes
u_{k}\rangle$, to show that $\lambda_{\\{x,y,z\\}}=0$, thereby obtaining a
contradiction. As $\\{x,y,z\\}$ is the largest element in
$\mathcal{W}(\mathcal{B})$ such that $\lambda_{\\{x,y,z\\}}\neq 0$, we obtain
that $\lambda_{\\{i,j,k\\}}=0$ for all $\\{i,j,k\\}>\\{x,y,z\\}$. Let
$f({\\{x,y,z\\}})=\Psi(u_{a}\otimes u_{b}\otimes u_{c})$, where
$u_{c}=\max\\{u_{x},u_{y},u_{z}\\}$, $u_{a}=\min\\{u_{x},u_{y},u_{z}\\}$ and
$u_{b}=\\{u_{x},u_{y},u_{z}\\}\setminus\\{u_{a},u_{c}\\}$. Since
$\\{x,y,z\\}\in\mathcal{W}(\mathcal{B})$, we have that
$\\{a,b\\}\in\mathcal{B}$ or $\\{b,c\\}\in\mathcal{B}$ or
$\\{a,c\\}\in\mathcal{B}$. We will consider each case separately.

Case 1: $\\{a,b\\}\in\mathcal{B}$.

Applying the projection map $P_{a,b,c}:V\otimes U\to\langle
A(u_{a},u_{b})\otimes u_{c}\rangle$ to (2.7.1), we obtain

$P_{a,b,c}\big{(}\sum\limits_{\\{i,j,k\\}\in\mathcal{W}(\mathcal{B})}\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\})\big{)}=0.$
(2.7.2)

Now we will show that $P_{a,b,c}(\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\}))=0$
whenever $\\{i,j,k\\}\neq\\{a,b,c\\}$. By Lemma 2.6 (ii),
$P_{a,b,c}(f(\\{i,j,k\\}))=0$ for $\\{i,j,k\\}\in\mathcal{W}(\mathcal{B})$
with $c\notin\\{i,j,k\\}$. Thus we need only to consider elements
$f(\\{i,j,c\\})$. If $\\{i,j,c\\}>\\{a,b,c\\}$, then
$\lambda_{\\{i,j,c\\}}=0$, and hence
$P_{a,b,c}(\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\}))=0$. If
$\\{i,j,c\\}<\\{a,b,c\\}$, then $\\{i,j\\}<\\{a,b\\}$, by Lemma 2.5. Applying
Lemma 2.6 (iii), we get that $P_{a,b,c}(f(\\{i,j,c\\}))=0$. Thus (2.7.2)
becomes
$P_{a,b,c}\big{(}\sum\limits_{\\{i,j,k\\}\in\mathcal{W}(\mathcal{B})}\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\})\big{)}=\lambda_{\\{x,y,z\\}}A(u_{a},u_{b})\otimes
u_{c}=0$, whence $\lambda_{\\{x,y,z\\}}=0$.

Case 2: $\\{b,c\\}\in\mathcal{B}$.

Applying the projection map $P_{b,c,a}:V\otimes U\to\langle
A(u_{b},u_{c})\otimes u_{a}\rangle$ to (2.7.1), we obtain

$P_{b,c,a}\big{(}\sum\limits_{\\{i,j,k\\}\in\mathcal{W}(\mathcal{B})}\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\})\big{)}=0.$
(2.7.3)

Now the proof follows mutatis mutandis the proof of Case 1.

Case 3: $\\{a,c\\}\in\mathcal{B}$.

Applying the projection map $P_{a,c,b}:V\otimes U\to\langle
A(u_{a},u_{c})\otimes u_{b}\rangle$ to (2.7.1), we obtain

$P_{a,c,b}\big{(}\sum\limits_{\\{i,j,k\\}\in\mathcal{W}(\mathcal{B})}\lambda_{\\{i,j,k\\}}f(\\{i,j,k\\})\big{)}=0.$
(2.7.4)

Now the proof follows mutatis mutandis the proof of Case 1.

  2. (ii)

To prove the map $f:\mathcal{W}(\mathcal{B})\to W(\mathcal{B})$ is bijective,
we need to check only injectivity. The linear independence of $f(\\{a,b,c\\})$
where $\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})$ shows that $f$ is injective.
This completes the proof.

∎

### 2.3 Estimating the size of $W(\mathcal{B})$

In this section, our aim is to estimate the lower bound for $\dim\text{Im
}{(\Psi)}.$ Towards this aim, we state our next proposition.

######  Proposition 2.8.

Let $U$ and $V$ be finite-dimensional $F$ vector spaces with an alternating
bilinear map $A:U\times U\to V$ such that $\text{Im }A$ spans V. Let $X$ be a
basis of $U$ with a total order and let $\mathcal{B}$ be the set constructed
in Section 2.1. Let $\dim U=n$, $\dim V=m$, and $r\in\mathbb{N}$ be such that
${r\choose 2}\leq{n\choose 2}-m\leq{r+1\choose 2}$. If $W(\mathcal{B})$ be the
set as in (2.4.2), then the number of linearly independent elements in
$|W(\mathcal{B})|$ is at least ${n\choose 3}-{r\choose
3}-\scalebox{0.55}{$\dbinom{{n\choose 2}-m-{r\choose 2}}{2}$}$. In particular,
$\dim\text{Im }{(\Psi)}\geq{n\choose 3}-{r\choose
3}-\scalebox{0.55}{$\dbinom{{n\choose 2}-m-{r\choose 2}}{2}$}$

###### Proof.

By Proposition 2.7 (i), we have that $W(\mathcal{B})$ is linearly independent.
We have that $f:\mathcal{W}(\mathcal{B})\to W(\mathcal{B})$ is a bijection.
Since $\\{a,b\\}\in\mathcal{B}$ implies that $a\neq b$, we obtain that
$\mathcal{W}(\mathcal{B})=\\{\\{a,b,c\\}\mid 1\leq a,b,c\leq n,\text{
}\\{a,b\\}\in\mathcal{B},c\notin\\{a,b\\}\\}$ is a subset of
$X=\\{\\{a,b,c\\}\mid 1\leq a,b,c\leq n,c\notin\\{a,b\\},a\neq b\\}$. Noting
that $X$ has size $\binom{n}{3}$, we get that
$|\mathcal{W}(\mathcal{B})^{c}|={n\choose 3}-|\mathcal{W}(\mathcal{B})|$. Thus
an upper bound on the size of $\mathcal{W}(\mathcal{B})^{c}$ gives a lower
bound on the size of $\mathcal{W}(\mathcal{B})$. Observe that

$\displaystyle\mathcal{W}(\mathcal{B})^{c}=\\{\\{i,j,k\\}\in
X\mid\\{i,j\\}\notin\mathcal{B},\\{i,k\\}\notin\mathcal{B},\\{j,k\\}\notin\mathcal{B}\\}$

To estimate the size of $\mathcal{W}(\mathcal{B})^{c}$, we consider a graph
with $n$ vertices numbered $1,\ldots,n$ and an edge connecting vertices $i$
and $j$ if $\\{i,j\\}\notin\mathcal{B}$. So the number of edges is ${n\choose
2}-m$. Observe that $|\mathcal{W}(\mathcal{B})^{c}|$ is the number of
triangles in this graph. Hence we estimate the number of triangles in this
graph. Suppose that ${r\choose 2}\leq{n\choose 2}-m\leq{r+1\choose 2}$. Then
by [11, Theorem 2.5], the number of triangles in the graph is at most
${r\choose 3}+\scalebox{0.55}{$\dbinom{{n\choose 2}-m-{r\choose 2}}{2}$}$.
Thus the size of $\mathcal{W}(\mathcal{B})$ has a lower bound given by
${n\choose 3}-{r\choose 3}-\scalebox{0.55}{$\dbinom{{n\choose 2}-m-{r\choose
2}}{2}$}$. Since $f$ is a bijection onto $W(\mathcal{B})$, we have that the
size of $W(\mathcal{B})$ has the same lower bound. ∎

######  Proposition 2.9.

Let $U$ and $V$ be finite-dimensional $F$ vector spaces with an alternating
bilinear map $A:U\times U\to V$ such that $\text{Im }A$ spans V. Let $X$ be a
basis of $U$ with a total order and let $\mathcal{B}$ be the set constructed
in Section 2.1. If the set $\mathcal{B}$ is not a tree of height one, then
$|\mathcal{W}(\mathcal{B})|\geq(\sum\limits_{i=2}^{m+1}(n-i)+(m-2),$ where
$n=\text{dim }U$ and $m=\text{dim }V.$ In particular, $\dim\text{Im
}{(\Psi)}\geq(\sum\limits_{i=2}^{m+1}(n-i)+(m-2)$.

###### Proof.

Recall the set $\mathcal{W}(\mathcal{B}):=\\{\\{a,b,c\\}\mid 1\leq a,b,c\leq
n,\\{a,b\\}\in\mathcal{B},c\notin\\{a,b\\}\\}.$ Let
$\\{i_{1},i_{2}\\}\in\mathcal{B}$ be any element in $\mathcal{B}$. We consider
two cases:

Case 1: For any $\\{k,l\\}\in\mathcal{B}$,
$\\{k,l\\}\cap\\{i_{1},i_{2}\\}\neq\varnothing$.

For $\\{k,l\\}\in\mathcal{B}$, $i_{1}\in\\{k,l\\}$ or $i_{2}\in\\{k,l\\}$.
Thus,
$\mathcal{B}=\\{\\{i_{1},i_{2}\\},\\{i_{1},a_{1}\\},\dots,\\{i_{1},a_{r}\\},\newline
\\{i_{2},b_{1}\\},\dots,\\{i_{2},b_{s}\\}\\},$ for some
$a_{i},b_{j}\in\\{1,\dots,n\\}.$ Since $|\mathcal{B}|=m$, we have $r+s+1=m$.
If $r=0$ or $s=0$, then $\mathcal{B}$ is a tree of height one. Since
$\mathcal{B}$ is not a tree of height one, we have $r\geq 1$ and $s\geq 1$.
Define $\mathcal{W}_{0}=\\{\\{i_{1},i_{2},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1},i_{2}\\}\\}$. For $1\leq i\leq r$, define
$\mathcal{W}_{i}^{1}=\\{\\{i_{1},a_{i},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1},i_{2},a_{1},\dots a_{i}\\}\\}$. Similarly
for $1\leq j\leq s$, define $\mathcal{W}_{j}^{2}=\\{\\{i_{2},b_{j},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1},i_{2},b_{1},\dots b_{j}\\}\\}$. The sets
$\mathcal{W}_{0},\mathcal{W}^{1}_{i},\mathcal{W}_{j}^{2}$ are contained in
$\mathcal{W}(\mathcal{B})$ and $|\mathcal{W}_{0}|=n-2$,
$|\mathcal{W}_{i}^{1}|=n-(i+2)$, $|\mathcal{W}_{j}^{2}|=n-(j+2).$ Moreover,
they are pairwise disjoint. Thus we have,

$\displaystyle|\mathcal{W}(\mathcal{B})|$
$\displaystyle\geq|\mathcal{W}_{0}|+\sum\limits_{i=1}^{r}|\mathcal{W}_{i}^{1}|+\sum\limits_{i=1}^{s}|\mathcal{W}_{i}^{2}|$
$\displaystyle=(n-2)+\sum\limits_{i=3}^{r+2}(n-i)+\sum\limits_{i=3}^{s+2}(n-i)$
$\displaystyle=(n-2)+\sum\limits_{i=3}^{r+2}(n-i)+(n-3)+\sum\limits_{i=1}^{s-1}(n-(3+i))$

Since $r\geq 1,$ we have $n-(3+i)\geq n-(r+2+i)$. Therefore

$\displaystyle|\mathcal{W}(\mathcal{B})|$
$\displaystyle\geq(n-2)+\sum\limits_{i=3}^{r+2}(n-i)+(n-3)+\sum\limits_{i=1}^{s-1}(n-(r+2+i))$
$\displaystyle=(n-2)+(n-3)+\sum\limits_{i=3}^{r+2+(s-1)}(n-i)$
$\displaystyle=(n-2)+(n-(m+1))+(m-2)+\sum\limits_{i=3}^{r+s+1}(n-i)$
$\displaystyle=\sum\limits_{i=2}^{r+s+1}(n-i)+(n-(m+1))+(m-2).$

Noting that $r+s+1=m$ completes the proof.

Case 2: There exists $\\{r,s\\}\in\mathcal{B}$, with
$\\{r,s\\}\cap\\{i_{1},i_{2}\\}=\varnothing$.

Let $\mathcal{B}=\\{\\{i_{1}^{j},i_{2}^{j}\\}\mid 1\leq j\leq m\\}$. Without
loss of generality, we may assume $i_{1}^{1}=i_{1},\ i_{2}^{1}=i_{2}$ and
$i_{1}^{2}=r,\ i_{2}^{2}=s$. For each $1\leq j\leq m$, our aim is to
inductively define $\mathcal{W}_{j}\subseteq\\{\\{i_{1}^{j},i_{2}^{j},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}\\}$ such that
$\mathcal{W}_{1},\dots,\mathcal{W}_{j-1},\mathcal{W}_{j}$ are contained in
$\mathcal{W}(\mathcal{B})$ and they are pairwise disjoint. Define
$\mathcal{W}_{1}=\\{\\{i_{1}^{1},i_{2}^{1},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{1},i_{2}^{1}\\}\\}$ and
$\mathcal{W}_{2}=\\{\\{i_{1}^{2},i_{2}^{2},c\\}\mid
c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{2},i_{2}^{2}\\}\\}$. We have
$|\mathcal{W}_{1}|=|\mathcal{W}_{2}|=n-2$. Note that $\mathcal{W}_{1}$ and
$\mathcal{W}_{2}$ are pairwise disjoint. Assume we have defined
$\mathcal{W}_{i}$ for $i<j$, we will now define $\mathcal{W}_{j}$ for $3\leq
j\leq m$. Observe that if $\\{i_{1}^{j},i_{2}^{j},c\\}\in\mathcal{W}_{i}$ for
some $i<j$, then $\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$. We define
$C_{j}=\\{c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}\mid\\{i_{1}^{j},c\\}\neq\\{i_{1}^{i},i_{2}^{i}\\}\text{
and }\\{i_{2}^{j},c\\}\neq\\{i_{1}^{i},i_{2}^{i}\\}\text{ for any }i<j\\}$ and
$\mathcal{W}_{j}=\\{\\{i_{1}^{j},i_{2}^{j},c\\}\mid c\in C_{j}\\}$. We now
prove $\mathcal{W}_{1},\dots,\mathcal{W}_{j}$ are pairwise disjoint. By
induction, $\mathcal{W}_{1},\dots,\mathcal{W}_{j-1}$ are pairwise disjoint. If
$\\{i_{1}^{j},i_{2}^{j},c\\}\in\mathcal{W}_{i}\cap\mathcal{W}_{j}$ for some
$i<j$, then $\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$, a contradiction as $c\in
C_{j}.$ Thus $\mathcal{W}_{i}\cap\mathcal{W}_{j}=\varnothing$ for all $i<j$.
Thus $\mathcal{W}_{1},\dots,\mathcal{W}_{j}$ are pairwise disjoint. Our aim is
to estimate the size of $\mathcal{W}_{j}$ for $j\geq 3$. First note that
$|\mathcal{W}_{j}|=|C_{j}|$, and hence it is enough to estimate the size of
$C_{j}$. In order to obtain a lower bound for the size of $C_{j}$, we will
give an upper bound on the size of the complement of $C_{j}$ in
$\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$. If $c\notin C_{j}$, then
$\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some $1\leq i<j$. Now we
define $D=\\{i_{1}^{1},i_{2}^{1}\\}\cup\\{i_{1}^{2},i_{2}^{2}\\}$, and
consider the following cases:

Case 2a: $\\{i_{1}^{j},i_{2}^{j}\\}\cap D=\varnothing$.

For any $c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$, we have
$\\{i_{1}^{j},c\\}\neq\\{i_{1}^{1},i_{2}^{1}\\}$,
$\\{i_{2}^{j},c\\}\neq\\{i_{1}^{1},i_{2}^{1}\\},\\{i_{1}^{j},c\\}\neq\\{i_{1}^{2},i_{2}^{2}\\}$
and $\\{i_{2}^{j},c\\}\neq\\{i_{1}^{2},i_{2}^{2}\\}$. Thus if $c\notin C_{j}$,
then $\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some $3\leq i<j$. We will
show that for any $i$ with $3\leq i<j$, there exist at most one
$c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ such that
$\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}.$ Towards that end, assume there
exits $c,c^{\prime}\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ with
$c\neq c^{\prime}$ such that $\\{i_{r}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ and
$\\{i_{s}^{j},c^{\prime}\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some
$r,s\in\\{1,2\\}$. This implies the set
$\\{i_{r}^{j},i_{s}^{j},c,c^{\prime}\\}$ whose cardinality is at least three
is a subset of $\\{i_{1}^{i},i_{2}^{i}\\}$, a contradiction. Thus the size of
the complement of $C_{j}$ in
$\\{1,\ldots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ is at most $j-3$, and
therefore $|\mathcal{W}_{j}|=|C_{j}|\geq(n-2)-(j-3)=n-j+1$.

Case 2b: $|\\{i_{1}^{j},i_{2}^{j}\\}\cap D|=1$

Assume for some $r\in\\{1,2\\}$, $i_{r}^{j}\in D$ and $i_{s}^{j}\notin D$,
where $s\in\\{1,2\\}\setminus\\{r\\}$. Since
$\\{i_{1}^{1},i_{2}^{1}\\}\cap\\{i_{1}^{2},i_{2}^{2}\\}=\varnothing$, we have
$\\{i_{1}^{j},i_{2}^{j}\\}\cap\\{i_{1}^{t},i_{2}^{t}\\}=\varnothing$ for some
$t\in\\{1,2\\}$. Thus if $c\notin C_{j}$, then
$\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some
$i\in\\{1,\dots,j-1\\}\setminus\\{t\\}$. Thus the size of the complement of
$C_{j}$ in $\\{1,\ldots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ is at most
$j-2$, and therefore $|\mathcal{W}_{j}|=|C_{j}|\geq(n-2)-(j-2)=n-j$.

Case 2c: $\\{i_{1}^{j},i_{2}^{j}\\}\subseteq D$.

We will first consider the case $j\geq 3$ being the smallest integer such that
$\\{i_{1}^{j},i_{2}^{j}\\}\subseteq D$. If $c\notin C_{j}$, then
$\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some
$i\in\\{1,\dots,j-1\\}$. Thus $|\mathcal{W}_{j}|\geq(n-2)-(j-1)=n-j-1$. Now
consider the case where $j$ is not the smallest integer such that
$\\{i_{1}^{j},i_{2}^{j}\\}\subseteq D$. Thus there exists $j^{\prime}\geq 3$
with $j^{\prime}<j$ such that
$\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}\subseteq D$. Note that
$\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}=\\{i_{1}^{j},i_{2}^{j}\\}$ is not
possible. If
$\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}\cap\\{i_{1}^{j},i_{2}^{j}\\}=\varnothing$,
then $\\{i_{1}^{j},c\\}\neq\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}$ and
$\\{i_{2}^{j},c\\}\neq\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}$ for any
$c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$. Thus if $c\notin
C_{j}$, then $\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some
$i\in\\{1,\dots,j-1\\}\setminus\\{j^{\prime}\\}$, and therefore
$|\mathcal{W}_{j}|\geq(n-2)-(j-2)=n-j$. Assume
$\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}\cap\\{i_{1}^{j},i_{2}^{j}\\}\neq\varnothing$,
and let $i_{r}^{j^{\prime}}$ be the element in the intersection. Let
$s\in\\{1,2\\}\setminus\\{r\\},$ observe that

$\\{i_{s}^{j^{\prime}},i_{1}^{j}\\}\in\\{\\{i_{1}^{1},i_{2}^{1}\\},\\{i_{1}^{2},i_{2}^{2}\\}\\}\text{
or
}\\{i_{s}^{j^{\prime}},i_{2}^{j}\\}\in\\{\\{i_{1}^{1},i_{2}^{1}\\},\\{i_{1}^{2},i_{2}^{2}\\}\\}.$
(2.9.1)

If $c\notin C_{j}$, then $\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ for some $1\leq i<j$. Moreover
as in Case 2a, for any $i$ with $1\leq i<j$, there exist at most one
$c\in\\{1,\dots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ such that
$\\{i_{1}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{i},i_{2}^{i}\\}.$ Note that
$c=i_{s}^{j^{\prime}}$ satisfies
$\\{i_{1}^{j},c\\}=\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{j^{\prime}},i_{2}^{j^{\prime}}\\}$. By (2.9.1),
$c=i_{s}^{j^{\prime}}$ also satisfies
$\\{i_{1}^{j},c\\}=\\{i_{1}^{t},i_{2}^{t}\\}$ or
$\\{i_{2}^{j},c\\}=\\{i_{1}^{t},i_{2}^{t}\\}$ for some $t\in\\{1,2\\}$. Thus
the size of the complement of $C_{j}$ in
$\\{1,\ldots,n\\}\setminus\\{i_{1}^{j},i_{2}^{j}\\}$ is at most $j-2$, and
therefore $|\mathcal{W}_{j}|=|C_{j}|\geq(n-2)-(j-2)=n-j.$

Thus $|\mathcal{W}_{j}|\geq n-j$ for $3\leq j\leq m$, except possibly for one
$j$. For such a $j$, $|\mathcal{W}_{j}|\geq n-j-1.$ Therefore

$\displaystyle|\mathcal{W}(\mathcal{B})|$
$\displaystyle\geq\sum\limits_{j=1}^{m}|\mathcal{W}_{j}|\geq(n-2)+(n-2)+(\sum_{j=3}^{m}(n-j))-1$
$\displaystyle=(\sum\limits_{i=2}^{m+1}n-i)+(m-2).$

∎

## 3 Bounds on the order of the Schur multiplier of class 2

Let $G$ be finite $p$-group of nilpotency class $c$ and let
$X=\\{g_{1},\cdots,g_{\delta}\\}\subseteq G$ be such that its image forms a
minimal generating set of $\overline{G}:=G/Z$. The union of $X$ and a
generating set for $Z$ will be a generating set for $G$. It is well known that
a generating set of $\gamma_{2}G/\gamma_{3}G$ consisting of simple commutators
can be obtained from a generating set for $G$. Thus the generating set of $G$
provides us with a generating set of $\gamma_{2}G/\gamma_{3}G$ consisting of
simple commutators $\overline{[a,b]}$ where $a$ and $b$ belongs to either $X$
or the generating set of $Z$. But as the generating set of $Z$ commutes with
every element in $G$, the non-trivial elements of the generating set for
$\gamma_{2}G/\gamma_{3}G$ are simple commutators $\overline{[a,b]}$ where
$a,b\in X$. By abuse of notation, we shall omit the use of the bar notation,
but it will be clear from the context what the image is. Consider the
homomorphism defined in [4, Proposition 1]:

$\displaystyle\Psi_{2}$
$\displaystyle:\overline{G}^{ab}\otimes\overline{G}^{ab}\otimes\overline{G}^{ab}\to\gamma_{2}G/\gamma_{3}G\otimes\overline{G}^{ab}$
$\displaystyle\overline{x}\otimes\overline{y}\otimes\overline{z}$
$\displaystyle\mapsto\overline{[x,y]}\otimes\overline{z}+\overline{[y,z]}\otimes\overline{x}+\overline{[z,x]}\otimes\overline{y}.$

A more general version of the above homomorphism was considered in [3] on page
$2516$. We have the natural projection maps
$P_{1}:\gamma_{2}G/\gamma_{3}G\to\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}$
and
$P_{2}:\overline{G}^{ab}\to\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$,
which induces the surjective homomorphism $P_{1}\otimes
P_{2}:\gamma_{2}G/\gamma_{3}G\otimes\overline{G}^{ab}\to\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes_{\mathbb{Z}}\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$.
The next lemma is standard and we omit its proof.

######  Lemma 3.1.

Let $R$ be a commutative ring with unit and $I$ be an ideal of $R$. Let
$M_{1}$ and $M_{2}$ be $R$-modules. If $IM_{1}=IM_{2}=0$, then
$M_{1}\otimes_{R}M_{2}\cong M_{1}\otimes_{R/I}M_{2}$.

By Lemma 3.1, we get that
$\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes_{\mathbb{Z}}\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\cong\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes_{\mathbb{F}_{p}}\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$.
We consider $U=\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$,
$V=\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}$ and
$A:\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\times\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\to\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}$
be the commutator map. With this setup from Section 2.2, we obtain the map
$\Psi:\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\to\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$.
The proof of the next result is obvious and we omit the proof.

######  Lemma 3.2.

Let $G$ be a finite $p$-group with nilpotency class $c\geq 2$. The following
diagram is commutative,

${\overline{G}^{ab}\otimes\overline{G}^{ab}\otimes\overline{G}^{ab}}$${\gamma_{2}G/\gamma_{3}G\otimes\overline{G}^{ab}}$${\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}}$${\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}}$$\scriptstyle{\Psi_{2}}$$\scriptstyle{P_{2}\otimes
P_{2}\otimes P_{2}}$$\scriptstyle{P_{1}\otimes P_{2}}$$\scriptstyle{\Psi}$

Assume that the basis of $\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$,
$X$, has a total order. From Section 2.1, one obtains $\mathcal{B}$ and a
basis, $B$, for
$\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}$. Let
$W(\mathcal{B})$ be as defined in (2.4.2) where
$\Psi:\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}\to\frac{\gamma_{2}G/\gamma_{3}G}{\Phi(\gamma_{2}G/\gamma_{3}G)}\otimes\frac{\overline{G}^{ab}}{\Phi(\overline{G}^{ab})}$.
By Lemma 3.2, we have that $W(\mathcal{B})$ is contained inside $(P_{1}\otimes
P_{2})(\text{Im }{(\Psi_{2})})$. Using (1.0.1), Rai derives the following in
the proof of [15, Theorem 1.1]:

######  Proposition 3.3 (Rai, [15]).

Let $G$ be a non-abelian $p$-group of nilpotency class $c$ with
$|\gamma_{2}G|=p^{k}$ and $d=d(G)$. The following inequality holds:

$\displaystyle|M(G)||\text{Im }\Psi_{2}|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)}.$ (3.3.1)

Now we prove the following result.

######  Theorem 3.4.

Let $G$ be a $p$-group of order $p^{n}$ and nilpotency class $c\geq 2$. Assume
that $d(G)=d$, $d(G/Z)=\delta$, $d(\gamma_{2}G/\gamma_{3}G)=k^{\prime}$ and
$|\gamma_{2}G|=p^{k}$. Let $r\in\mathbb{N}$ be such that ${r\choose
2}\leq{\delta\choose 2}-k^{\prime}<{r+1\choose 2}$. Then,

$|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)-\binom{\delta}{3}+\binom{r}{3}+\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}}.$

###### Proof.

By Proposition 3.3, we have $|M(G)||\text{Im }\Psi_{2}|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)}$. Now we will show that $|\text{Im
}\Psi_{2}|\geq
p^{\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}}$.
Let $X=\\{g_{1},\ldots,g_{\delta}\\}$ be elements of $G$ such that its images
form a minimal generating set of $G/Z(G)$. Let
$g_{i_{1}}<\dots<g_{i_{\delta}}$ be any total order on $X$ and let
$\mathcal{B}$ be the set constructed in Section 2.1. Corresponding to this
$\mathcal{B}$, we have $W(\mathcal{B})$ as defined in (2.4.2). By Proposition
2.8, we have that $W(\mathcal{B})$ has at least
$\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}$
linearly independent elements. Noting that
$W(\mathcal{B})\subseteq(P_{1}\otimes P_{2})(\text{Im }\Psi_{2})$, we obtain
that $(P_{1}\otimes P_{2})(\text{Im }\Psi_{2})$ has at least
$p^{\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}}$
elements. Thus $|\text{Im }\Psi_{2}|\geq
p^{\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}}$.
Plugging this estimate in (3.3.1) gives the required bound. ∎

######  Lemma 3.5.

The bound given in Theorem 3.4 is sharper than the bound in [15, Theorem 1.4]

###### Proof.

The bound given in [15, Theorem 1.4] is

$|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-\sum\limits_{i=2}^{\min(d,k^{\prime}+1)}(d-i)}.$
(1.2.1)

But the proof of [15, Theorem 1.4] shows that the author obtains (1.2.1) from
the following bound:

$|M(G)|\leq
p^{\frac{1}{2}(d-1)(n+k)-k(d-\delta)-\sum\limits_{i=2}^{\min(\delta,k^{\prime}+1)}(\delta-i)}.$
(1.2.2)

It is easy to see that (1.2.2) is sharper than (1.2.1). Now we will show that
the bound obtained in Theorem 3.4 is sharper than (1.2.2). It is enough to
show that
$\sum\limits_{i=2}^{\min(\delta,k^{\prime}+1)}(\delta-i)\leq\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}$.
We have the following binomial identity which will be used throughout the
proof,

${n\choose k}={n-1\choose k}+{n-1\choose k-1}.$ (3.5.1)

Case 1: $\delta<k^{\prime}+1$

Clearly
$\sum\limits_{i=2}^{\min(\delta,k^{\prime}+1)}(\delta-i)=\sum\limits_{i=2}^{\delta}(\delta-i)={\delta-1\choose
2}$. We will show that ${\delta-1\choose
2}<\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}$.
We will show that

${\delta\choose 2}-k^{\prime}-{r\choose 2}<r\leq\delta-2$ (3.5.2)

By Theorem 3.4, we have that ${r\choose 2}\leq{\delta\choose
2}-k^{\prime}<{r+1\choose 2}$. By (3.5.1), we have that ${\delta\choose
2}-k^{\prime}={\delta-1\choose 2}+\delta-1-k^{\prime}$. Since
$\delta<k^{\prime}+1$, we have that $\delta-1-k^{\prime}<0$, and hence
${\delta\choose 2}-k^{\prime}<{\delta-1\choose 2}$. From this, we obtain that
$r\leq\delta-2$. By definition of $r$, ${\delta\choose
2}-k^{\prime}<{r+1\choose 2}$ which yields ${\delta\choose
2}-k^{\prime}-{r\choose 2}<r$ by (3.5.1). This proves (3.5.2). By (3.5.2), we
have

$\displaystyle\binom{\delta}{3}-\binom{r}{3}-\binom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}>$
$\displaystyle\binom{\delta}{3}-\binom{\delta-2}{3}-\binom{\delta-2}{2}$
$\displaystyle=$
$\displaystyle\binom{\delta}{3}-\binom{\delta-1}{3}=\binom{\delta-1}{2},$

Case 2: $\delta\geq k^{\prime}+1$

Clearly,
$\sum\limits_{i=2}^{\min(\delta,k^{\prime}+1)}(\delta-i)=\sum\limits_{i=2}^{k^{\prime}+1}(\delta-i)=\binom{\delta-1}{2}-\scalebox{0.55}{$\dbinom{\delta-(1+k^{\prime})}{2}$}$.
We will show that
$\binom{\delta-1}{2}-\scalebox{0.55}{$\dbinom{\delta-(1+k^{\prime})}{2}$}=\binom{\delta}{3}-\binom{r}{3}-\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}$.
We begin by determining $r$ which depends on $\delta$ and $k^{\prime}$. By
Theorem 3.4, we have that ${r\choose 2}\leq{\delta\choose
2}-k^{\prime}<{r+1\choose 2}$. By (3.5.1), we have that ${\delta\choose
2}-k^{\prime}={\delta-1\choose 2}+\delta-1-k^{\prime}$. Since $\delta\geq
k^{\prime}+1$, we have that $0\leq\delta-1-k^{\prime}<\delta-1$. Therefore
${\delta-1\choose 2}\leq{\delta\choose 2}-k^{\prime}<{\delta\choose 2}$ and we
have that $r=\delta-1$. Thus

$\displaystyle\binom{\delta}{3}-\binom{r}{3}-\binom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}=$
$\displaystyle\binom{\delta}{3}-\binom{\delta-1}{3}-\binom{\binom{\delta}{2}-k^{\prime}-\binom{\delta-1}{2}}{2}$
$\displaystyle=$
$\displaystyle\binom{\delta-1}{2}-\binom{\delta-(1+k^{\prime})}{2}.$

Hence the proof. ∎

The next lemma follows from computations in Lemma 3.5, Case 2.

######  Lemma 3.6.

Let $r\in\mathbb{N}$ be such that ${r\choose 2}\leq{\delta\choose
2}-k^{\prime}<{r+1\choose 2}$. If $\delta>k^{\prime}+1$, then
$\binom{\delta}{3}+\binom{r}{3}+\scalebox{0.55}{$\dbinom{\binom{\delta}{2}-k^{\prime}-\binom{r}{2}}{2}$}=\sum\limits_{i=1}^{k^{\prime}+1}(\delta-i).$

In the next result, we aim to improve the bound for the case $\delta\geq
k^{\prime}+1$ for special $p$ groups mentioned in the introduction.

######  Theorem 3.7.

Let $G$ be a special $p$-group of order $p^{n}$, and that $d(G)=d$,
$d(\gamma_{2}G)=k>2$. If $d>k+1$, then

$|M(G)|\leq p^{\frac{1}{2}(d-1)(n+k)-\sum\limits_{i=2}^{k+1}(d-i)-(k-2)}$

###### Proof.

Note that for a special group $Z=\gamma_{2}G=\Phi(G)$, and hence
$d(G)=d(G/Z)$, $d(\gamma_{2}G)=k,|\gamma_{2}G|=p^{k},\Phi(G/Z)=1$ and
$\overline{G}^{ab}=G/Z$. By Proposition 2.3, there exists a basis
$\\{g_{1},\ldots,g_{d}\\}$ of $G/Z$ with the order $g_{1}<\cdots<g_{d}$ such
that the set $\mathcal{B}$ constructed using the algorithm given in Section
2.1 is not a tree of height one or satisfies the following properties:

$\displaystyle\mathcal{B}=\\{\\{1,2\\},\dots,\\{1,{k+1}\\}\\}.$ (3.7.1)
$\displaystyle[g_{1},g_{i}]=0\text{ for all }k+2\leq i.$ (3.7.2)
$\displaystyle[g_{i},g_{j}]=0\text{ for all }i,j\text{ with }k+1<i<j.$ (3.7.3)
$\displaystyle[g_{i},g_{j}]\in\text{span }\\{[g_{1},g_{i}]\\}\text{ for all
}i,j\text{ with }2\leq i\leq k+1<j\leq d.$ (3.7.4)

If $\mathcal{B}$ is not a tree of height one, then by Proposition 2.7 and
Proposition 2.9, $W(\mathcal{B})$ defined in (2.4.2) has size at least
${\left(\sum\limits_{i=2}^{k+1}d-i\right)+(k-2)}$ linearly independent
elements. Note that $W(\mathcal{B})\subseteq\text{Im }\Psi_{2},$ and hence the
proof follows from Proposition 3.3. Thus we can assume $\mathcal{B}$ satisfies
(3.7.1),(3.7.2),(3.7.3) and (3.7.4). For $2\leq s\leq k+1$, let
$[g_{s},g_{k+2}]=\alpha_{s}[g_{1},g_{s}]$ for some
$\alpha_{s}\in\mathbb{F}_{p}$. We will show that there exist $2\leq s,t\leq
k+1$ such that $\alpha_{s}\neq\alpha_{t}$. Suppose if
$\alpha_{s}=\alpha_{t}=\alpha$ for every $2\leq s,t\leq k+1$, then for all
$1\leq i\leq d$, we have $[g_{1}^{\alpha}g_{k+2},g_{i}]=0$ by (3.7.2) and
(3.7.3). Thus $g_{1}^{\alpha}g_{k+2}\in Z$, a contradiction. Thus for each
$2\leq i\leq k+1$, choose a $c(i)\in\\{s,t\\}$ such that
$\alpha_{i}\neq\alpha_{c(i)}$. Note that $\mathcal{W}(\mathcal{B})$ defined in
(2.4.1) can be written as
$\mathcal{W}(\mathcal{B})=\bigsqcup\limits_{i=2}^{k+1}\mathcal{W}_{i}$, where
$\mathcal{W}_{i}=\\{\\{1,i,c\\}\mid i<c\leq d\\}$. By Proposition 2.7, the set
$W(\mathcal{B})$ defined in (2.4.2) is linearly independent. Define
$\mathcal{W}=\\{\\{i,c(i),k+2\\}\mid 2\leq i\leq k+1,i\neq t\\}$ and observe
that if $2\leq i,j\leq k+1$ and $i\neq j$ then
$\\{i,c(i),k+2\\}\neq\\{j,c(j),k+2\\}$. We now show that the set
${W(\mathcal{B})}\cup\\{\Psi_{2}(g_{i}\otimes g_{c(i)}\otimes g_{k+2})\mid
2\leq i\leq k+1,i\neq t\\}$ is linearly independent. Let
$\lambda_{\\{a,b,c\\}}\in\mathbb{F}_{p}$ be such that

$\sum\limits_{i=2}^{k+1}\sum\limits_{\\{1,i,c\\}\in\mathcal{W}_{i}}\lambda_{\\{1,i,c\\}}\Psi_{2}(g_{1}\otimes
g_{i}\otimes
g_{c})+\sum\limits_{\\{i,c(i),k+2\\}\in\mathcal{W}}\lambda_{\\{i,c(i),k+2\\}}\Psi_{2}(g_{i}\otimes
g_{c(i)}\otimes g_{k+2})=0.$ (3.7.5)

Our aim is to prove that $\lambda_{\\{a,b,c\\}}=0$ for all
$\\{a,b,c\\}\in\mathcal{W}(\mathcal{B})\cup\mathcal{W}$. For $2\leq j\leq
k+1$, consider the projection maps
$P_{1,j,c(j)}:\gamma_{2}G/\gamma_{3}G\otimes G/Z\to\langle[g_{1},g_{j}]\otimes
g_{c(i)}\rangle$. Let $\\{1,a,b\\}\in\mathcal{W}(\mathcal{B}).$ If
$c(j)\notin\\{1,a,b\\}$, then by Lemma 2.6 (ii), we have
$P_{1,i,c(j)}(\Psi_{2}(g_{1}\otimes g_{a}\otimes g_{b})=0$. If
$j\notin\\{1,a,b\\}$, then by (3.7.1) and (3.7.2) the coefficient of
$[g_{1},g_{j}]$ is $0$, when $[g_{1},g_{a}]$ and $[g_{1},g_{b}]$ written as
linear combination of basis $B=\\{[g_{1},g_{s}]\mid 2\leq s\leq k+1\\}$.
Therefore $P_{1,j,c(j)}(\Psi_{2}(g_{1}\otimes g_{a}\otimes g_{b})=0$. Hence,
if $\\{1,a,b\\}\in\mathcal{W}(\mathcal{B})$ and
$P_{1,j,c(j)}(\Psi_{2}(g_{1}\otimes g_{a}\otimes g_{b})\neq 0$, then
$\\{1,a,b\\}=\\{1,j,c(j)\\}$. Thus
$P_{1,j,c(j)}\big{(}\sum\limits_{i=2}^{k+1}\sum\limits_{\\{1,i,c\\}\in\mathcal{W}_{i}}\lambda_{\\{1,i,c\\}}\Psi_{2}(g_{1}\otimes
g_{i}\otimes g_{c})\big{)}=\tau\lambda_{\\{1,j,c(j)\\}}[g_{1},g_{j}]\otimes
g_{c(j)}$, where $\tau=1$ if $j<c(j)$ and $\tau=-1$ if $j>c(j)$. Similarly,
using Lemma 2.6(ii), (3.7.4), (3.7.1) and (3.7.2), we obtain
$P_{1,j,c(j)}(\Psi_{2}(g_{i}\otimes g_{c(i)}\otimes g_{k+2})=0$, if $i\neq j$.
Thus applying $P_{1,j,c(j)}$ to equation (3.7.5) yields

$\tau\lambda_{\\{1,j,c(j)\\}}[g_{1},g_{j}]\otimes
g_{c(j)}-\alpha_{i}\lambda_{\\{i,c(j),k+1\\}}[g_{1},g_{j}]\otimes g_{c(j)}=0.$
(3.7.6)

Similarly applying projection map $P_{1,c(j),j}:\gamma_{2}G/\gamma_{3}G\otimes
G/Z\to\langle[g_{1},g_{c(j)}]\otimes g_{j}\rangle$ to (3.7.5), we obtain

$-\tau\lambda_{\\{1,j,c(j)\\}}[g_{1},g_{c(j)}]\otimes
g_{j}+\alpha_{c(j)}\lambda_{\\{j,c(j),k+2\\}}[g_{1},g_{c(j)}]\otimes g_{i}=0.$
(3.7.7)

The equation (3.7.6) and (3.7.7) yields $\lambda_{\\{j,c(j),k+2\\}}=0$ for all
$\\{j,c(j),k+2\\}\in\mathcal{W}$. By Proposition 2.7, we have
$\lambda_{\\{1,i.c\\}}=0$ for all $\\{1,i,c\\}\in\mathcal{W}(\mathcal{B})$.
Thus, $W(\mathcal{B})\cup\\{\Psi_{2}(g_{i}\otimes g_{c(i)}\otimes g_{k+2})\\}$
is linearly independent. By Proposition 2.8 and Lemma 3.6, the size of
${W}(\mathcal{B})$ is at least $\left(\sum\limits_{i=2}^{k+1}d-i\right)$.
Clearly, $|\\{\Psi_{2}(g_{i}\otimes g_{c(i)}\otimes g_{k+2})\mid 2\leq i\leq
k+2,i\neq t\\}|\geq k-2$. Note that
$W(\mathcal{B})\cup\\{\Psi_{2}(g_{i}\otimes g_{c(i)}\otimes g_{k+2})\mid 2\leq
i\leq k+2,i\neq t\\}\subseteq\text{Im }\Psi_{2},$ and hence $|\text{Im
}\Psi_{2}|\geq p^{\left(\sum\limits_{i=2}^{k+1}d-i\right)+(k-2)}.$ Now the
proof follows from Proposition 3.3.

∎

## Acknowledgements

Viji Z. Thomas acknowledges research support from SERB, DST, Government of
India grant MTR/2020/000483. Sathasivam K acknowledges the Ministry of
Education, Government of India, for the doctoral fellowship under the Prime
Minister’s Research Fellows (PMRF) scheme.

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