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Groups: Binary Operations & Group Properties Syllabus

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GROUPS
Compiled by: Nyasha P. Tarakino (Trockers)
+263772978155/+263717267175
ntarakino@gmail.com
30 JANUARY 2020
Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175
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SYLLABUS (6042) REQUIREMENTS
 define a binary operation
 define closure, commutation, association,
distribution, identity and inverse element
 define a group
 use the basic properties to show that a
given structure is, or is not, a group
 solve problems involving binary operations
and properties of a group
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BINARY OPERATION
Definition of a Binary Operation
Binary operations on the set are calculations that combine two elements of the set (called
operands) to produce another element of the same set.
Algebraic Definition
If G is a nonempty set, a binary operation on
is a function
NOTES
 A binary operation is represented by
 Binary operations
are:
.
 The properties of a binary operation are: Closure, Commutative, Associative, Identity
Inverse and Distributive.
 A binary operation
can be represented on a table called Cayley table/ Latin square
Modular Arithmetic
Definition
For any integers,
by
and , we write
to denote the remainder when
is divided
.
So in this case, the answer is the remainder when it is divided by the modulo number.
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Hence it is a set of remainders/residues when dividing two Integers together.
Example
remainder
Example
Find the value of:
(i)
(ii)
(iii)
(iv)
Solution
(i)
(ii)
(iii)
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(iv)
Properties of a Binary Operation
A binary operation
has the following properties: Closure, Associativity, Distribution,
Commutative, Identity and Inverse.
Property 1: Closure
A set is said to be closed under a binary operation
the result of the binary operation
if for any two members from the set,
returns a member of the same set.
Algebraically we define closure property as follows: If
is a set with elements
the
Example
Show that addition is closed under the set
of real numbers.
Solution
Let
and
.
This is true for all real numbers,
addition is closed under the set
of real numbers.
Example
If
and
. Is
closed under ?
Solution
Let
Now
and one would say it is closed.
NOTE: When attempting these particular questions on a given binary definition we have to
be very careful before we give a conclusion. To circumvent the problem of giving a false
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conclusion, we should construct the Cayley table. If there are no new values in the table then
the set is said to be closed under a binary ( ) operation, otherwise it is not closed.
How to find the elements?
Remember that
Let
.
Now
Members in blue are not contained in
is not closed under
Example
If
and
. Is closed under ?
Solution
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How to find the elements?
Remember that
Let
.
.
Now
All values in the table are contained in
is closed under
Property 2: Commutative
A binary operation
defined on set
of real numbers is commutative if
for all
Note
Common binary operations which are commutative are addition
and multiplication
.
Example
(Addition is commutative)
(Multiplication is commutative)
(Subtraction is not commutative)
(Division is not commutative)
Example
Is
commutative for all
?
Solution
Since addition
and multiplication
Since
are commutative then:
is commutative.
Note
If a binary operation is commutative and when given a table, the elements must reflect each
other along the leading diagonal.
Example
Is the following binary operation commutative?
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Solution
The elements are reflecting along the leading diagonal therefore it is commutative.
Example
Is the following binary operation commutative?
Solution
The elements are not reflecting along the leading diagonal therefore it is not commutative.
Example
Complete the table so that it is commutative?
1
Solution
Members should reflect along the leading diagonal. Therefore the table is given by:
1
Property 3: Associative
A closed binary operation
defined on set
of real numbers is associative if
for all
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Note
Common binary operations which are associative are addition
and multiplication
.
Example
and
(Addition is associative)
and
(Multiplication is associative)
and
(Subtraction is not associative)
and
(Division is not associative)
Example
If
, is associative.
Solution
and
.
Since
is not associative.
Example
If
, is associative.
Solution
.
Since
is associative.
Property 4: Distributive
A closed binary operation
and
are defined on set
of real numbers if
: Left Distributive (LD) and
: Right Distributive (RD).
Then the operator is said to be distributive over the operator .
Note
Common binary operation multiplication
is distributive over addition
.
Example
and
(LD).
and
(RD).
Also
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Since both LD and RD hold the operation of multiplication is distributive over addition.
Example
Given the binary operation
and
where
and
. Is the operation
distributive over ?
Solution
We have to show that
and if they agree then it is distributive.
and
Since
it is LD
Also
and
Since
it is RD
Since both LD and RD hold
the operation distributive over .
Property 5: Identity Element
For a binary operation , if there exist just one unique element
then
such that:
is called an identity element for the operation
Common Identity Elements
For a binary operation, addition
multiplication
, the identity element,
, the identity element,
and for a binary operation,
.
Example
For addition,
For multiplication,
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Identifying/finding the identity element in a Cayley Table
Look for where the elements match the outer elements i.e. the row that matches the top upper
row in green and the column that matches the outside column in green. The intersection of
the two lines (red) gives the identity element.
The identity element,
.
Finding the identity when given a binary definition
Example
If
, where
, find the identity element .
Solution
We know that
Now
and replacing
we have:
Also
The identity element is
.
Property 6: Inverse Element
An element
is called an inverse of
under the binary operation if:
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where
is the identity element under the operation
Common Inverse Elements
For a binary operation, addition
Thus the inverse of
is
, where the identity element,
under addition.
If we take any real number, say , the inverse is the negative of that particular number
e.g. The inverse of
is –
under addition.
ALSO
For a binary operation, multiplication
Thus the inverse of
, where the identity element,
is under multiplication.
If we take any real number, say , the inverse is the reciprocal of that particular number
e.g. The inverse of
NB:
is . under multiplication
does not have an inverse under multiplication.
Identifying/finding the identity element in a Cayley Table
We have to find the identity element first (as illustrated under identity element section above)
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Therefore, in this case the identity element
.
To find the inverse of each element you identify the position where the identity element ( in
red) is located in that particular row/column, and that is the inverse.
Since we know that
thus in our case:
, where
.
THEREFORE:
(i)
(ii)
(iii)
Finding the inverse when given a binary definition
Example
If
, where
, find
, the inverse of .
Solution
NB: We have to find the identity element first.
We know that
Now
and replacing
we have:
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The identity element is .
Now let the inverse be
and we know that
and in our case
GROUP
Definition of a Group
A group
is a non-empty set
with a binary operation
which satisfies the following
properties: Closure, Associativity, Identity and Inverse.
Properties of a Group
A group has 4 properties which are: Closure, Associativity, Identity and Inverse. The other
special property is called Commutative.
Property 1: Closure
For every elements
and
where
in , the result
must also lie in
i.e
.
Example
Given that
then suppose
,
. Show that
is closed under
multiplication modulo .
Solution
Since
therefore that
is closed under multiplication modulo .
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Property 2: Associative
For every
Example
Show that
, under multiplication modulo
is associative.
Solution
The Cayley table for
is:
For Associativity,
Let
LHS
;
RHS
Since LHS
RHS
the associative axiom/property holds.
Property 3: Identity element
An element
such that
, for all
, is called an identity element for
.
Example
Find the identity element for
, under multiplication modulo
Solution
The Cayley table for
is:
If you are asked to find the identity element in an exam you must look for the column which
is identical to the 1st column (in blue). The number or element on the 1st row of that column is
the identity element. OR Look for where the elements match the outer elements i.e. the row
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that matches the top upper row and the column that matches the outside column. The
intersection of that column and the row gives the identity element.
The identity element is .
Property 4: Inverse
Every element has its own inverse.
For every
in
there exists
such that
.
Example
Find the inverse element for
, under multiplication modulo
Solution
The Cayley table for
is:
If you are asked to find the inverse element in an exam you must look for the identity element
(in blue) for each element. The number or element on the 1st row of that column or 1st
column of that row is the inverse of that element.
The inverse of:
is (self inverse)
is (self inverse)
is (self inverse)
is (self inverse).
HOW TO SHOW THAT A GIVEN SET IS A GROUP?
If you are asked to show that something is a group in an exam you must tick off each of the
above criteria one-by-one i.e. it has to satisfy all the above properties. If one of the properties
is not met then it is not a group.
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Special Property: Commutative
If a group is commutative, then it’s called an Abelian group.
Definition
A group
is abelian if
for all elements
.
NOTE: When using a Latin Square/ Cayley table, a group is Abelian when it is symmetrical
along the leading diagonal or when the elements are reflected in the leading diagonal.
How to observe symmetry along the leading diagonal?
The elements on the leading diagonal are the same
OR
The elements on the leading diagonal are the symmetric
either
or
How to identify commutation using the leading diagonal?
The elements are reflecting along the leading diagonal
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The order of a group is the number of elements the group contains. If a group contains an
infinite number of elements it is said to be of infinite order.
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SUBGROUPS
Definition of a Subgroup
If is a nonempty subset of a group , then
same operation as .
is a subgroup of
if
is a group under the
Properties of a Subgroup
If the following three points are met then we say
1.
is a nonempty subset of a group of
2.
is a group itself
3.
and
is subgroup
use the same binary operation.
Types of Subgroups
1. Trivial subgroup - A subgroup includes the subset containing just the identity set
2. Improper subgroup – the group
itself since
is a subset
itself.
3. A proper subgroup is any subgroup with order not one or the same as the original
group. The order of an element, say
is the smallest positive integer ( ) in which we
raise the element ( ) to get the identity element i.e.
If there is no such an element then
and it is denoted:
.
has an infinite order.
Example
Find all the subgroups for the set
, under multiplication modulo .
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Solution
The Cayley table for
is:
NB: A subgroup has to satisfy the following properties:
1.
is a nonempty subset of a group of
2.
is a group itself
3.
and
use the same binary operation.
For the set
(i) the identity
(ii) the group
, the subgroups are:
- trivial subgroup
itself
- improper subgroup
(iii) The proper subgroups are:
a)
b)
c)
NB: For each Subgroup we have to check if the closure property holds since it is
also a Group.
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ASSIGNMENT-ZIMSEC QUESTIONS
ZIMSEC SPECIMEN QUESTION NOVEMBER 2018 PAPER 2
Let
be the set of matrices of the form
where
.
Show that
(i)
does not form a group under matrix addition.
[4]
(ii)
forms an Abelian group under matrix multiplication.
[6]
[Assume associativity.]
ZIMSEC NOVEMBER 2018 PAPER 2
a)(i) Show that the set of integers
under multiplication
forms a group.
(ii) Write down any three subgroups.
b) The set
[7]
[3]
under the operation of composition of functions forms a group H,
where
Show the operation table for
.
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[6]
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ZIMSEC JUNE 2019 PAPER 2
Show that the set
forms a group under multiplication.
[6]
ZIMSEC NOVEMBER 2019 PAPER 2
(i) If
is a set of all
matrices of the form
where
a group under addition of matrices.
(ii) Verify if the set
forms a group under addition modulo .
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, show that
forms
[7]
[9]
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SOLVED QUESTIONS & ANSWERS
Question 1
The set of matrices
under matrix multiplication. Construct its Cayley table/Latin square.
Suggested Solution
The Cayley table for this group is:
NB:
Question 2
The set of complex numbers
forms a group under complex number
multiplication. Construct its Cayley table/Latin square. Hence show that
is a group and
write down all subgroups of .
Suggested Solution
1. The Cayley table for this group is:
NB:
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2. a) Checking for closure: Checking in the Cayley table we see that there are no new values
i.e all values in set
are there. Hence it is closed.
b) For Associativity,
Let
LHS
;
RHS
Since LHS RHS the associative axiom holds.
c) The identity element is .
d)
The inverse of:
is
is
(self inverse)
is
(self inverse)
is
Since all properties hold
is a group.
3. For the set
(i) the identity
, the subgroups are:
- trivial subgroup
(ii) the group
itself
- improper subgroup
(iii) The proper subgroups are:
a)
NB This is not a proper subgroup because it does not satisfy closure
b)
NB This is the only proper subgroup
c)
NB This is not a proper subgroup because it does not satisfy closure
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Question 3
(i)Show that the set of numbers
, under multiplication
, does not form a
group.
(ii) The set of numbers
, under multiplication
, forms a group. Write
down the value of .
Suggested Solution
(i) The Cayley table for this set is:
NB:
then write down the remainder.
(Check the cell containing a letter with different colour)
The leading diagonal contains 1’s which are new elements not found in the original set.
Hence the closure property does not hold and therefore it does not form a group.
(ii) The Cayley table for this set is:
If
, then closure property holds and the set forms a group.
Question 4
The set S consists of all non-singular
real numbers such that
, where
.
(i) Prove that each matrix A must be of the form
.
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(ii) State clearly the restrictions on the value of
such that
is in
Suggested Solution
(i) Let
.
and
Now
Comparing corresponding elements
(a)
(b)
(c)
(d)
(Matrix
(ii) Since
is non-singular,
conforms to the required form).
.
Question 5
The set S consists of the numbers
, where
. (
denotes the set of integers
). Prove that the element of S, under multiplication forms a commutative
group G. (You may assume that addition of integers is associative and commutative.)
Suggested Solution
(i)
hence it is closed under multiplication.
(ii) For Associativity,
LHS
RHS
Since LHS = RHS
it is associative.
(iii) The identity element is
(iv) The inverse element is
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(v) For commutative property to hold,
Let
.
LHS
RHS
Since LHS
RHS
element of S, under multiplication forms a commutative group G.
or simply
it is commutative.
NB: A group which is commutative is called an Abelian group.
Question 6
A group G of order 6 has the combination table shown below.
(i) State, with a reason, whether or not
is commutative.
(ii) Write down the inverse of each element
Suggested Solution
(i) It is not commutative because:
(a) the leading diagonal elements (in yellow) are not symmetric, i.e they are
instead of
. or
(b) For commutative property to hold,
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Let
:
LHS
(in grey)
RHS
(in green)
Since LHS
RHS
is not commutative
(ii) The identity element is
NB: To find the identity element you look for a column which is identical to the column
outside the table
The inverse of:
is
is
is
is (self inverse)
is (self inverse)
is (self inverse).
NB: To find the inverse you look for a cell in every row which contains the identity
element and in this case the identity is .
Question 7
For the group of matrices of the form
under matrix addition, where
, state the
identity element and the inverse of
Suggested Solution
(i) The identity element is
(ii) The inverse of
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Question 8
consists of the set
with the operation of multiplication
Write down the operation table and, assuming Associativity, show that
.
is a group.
Suggested Solution
(i) The Cayley table for
is:
a) Checking for closure: Checking in the Cayley table we see that there are no new values i.e
all values in set
are there. Hence it is closed.
b) The associative property holds (from the assumption) OR
For Associativity,
Let
LHS
;
RHS
Since LHS RHS the associative axiom holds.
c) The identity element is .
d)
The inverse of:
is (self inverse)
is (self inverse)
is (self inverse)
is (self inverse).
Since all properties hold
is a group.
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Question 9
The group G consists of the set
combined under multiplication
. Find the inverse of each element.
Suggested Solution
The Cayley table for set G is given by:
a) The identity element is .
b) The inverse of:
is (self inverse)
is
is
is (self inverse)
is (self inverse)
is
is
is
(self inverse).
Question 10
Elements of the set
are combined according to the operation table shown below.
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(i) Verify that
.
(ii) Assuming that the associative property holds for all elements, prove that the set
, with the operation table shown, forms a group .
Suggested Solution
(i) LHS
.
NB:
in purple and
in green
RHS
.
NB:
in yellow and
Since LHS = RHS
in grey
it is associative.
(ii) Checking for basic axioms of a group.
a) Checking for closure: Checking in the Cayley table we see that there are no new values i.e
all values in the given set are there. Hence it is closed.
b) The associative property holds (check part (i))
c) The identity element is .
d) The inverse of:
is
is
is (self inverse)
is (self inverse).
is (self inverse).
Since all properties hold
is a group.
Question 10
If
is the set of all matrices of the form
(i) Prove that
:{
and
},
forms a group under matrix multiplication.
[Assume matrix multiplication is associative]
(ii) Determine whether
is
1. Abelian,
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2. closed under matrix addition.
Suggested Solution
If
is the set of all matrices of the form
(i) Proving that
:{
and
},
forms a group under matrix multiplication.
[Assuming matrix multiplication is associative]
Testing for Closure
Let :{
{
and
and
}
}
is closed under matrix multiplication
Testing for Association
is associative (from the assumption)
Testing for Identity
Let the identity element be
where {
and
}
has an identity
Testing for Inverse
Let the inverse element be
where {
and
}
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has an identity
Alternatively, since the identity is
we can find the inverse using the
normal procedure of finding an inverse of a
matrix.
Since all four properties/ Axioms hold
forms a group under matrix
multiplication.
(ii) Determining whether
is
1. Abelian,
Let :{
and
{
}
and
}
{
and
}
Since
and
is Abelian.
Also
2.
closed under matrix addition.
Testing for Closure
Let :{
and
}
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{
and
} and also
conforms to
is closed under matrix addition
Question 11
consists of the set of matrices of the form
where
and
are real and
, combined under operation of matrix multiplication.
(i) Prove that
is a group. You may assume that matrix multiplication is associative.
(ii) Determine whether
(i)
is commutative.
consists of the set of matrices of the form
Checking for closure
Let
and
Now:
Also:
Hence
is closed under matrix multiplication
Checking for association
is associative under matrix multiplication (Assumption)
Checking for identity
Let
be the identity element of .
Since
,
is invertible
Checking for inverse
Since
is invertible therefore its inverse is given by:
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Since all the four properties hold
is a group.
(ii) Checking for commutation
Let
and
Now:
Also:
Since
is commutative under matrix multiplication
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PRACTISE QUESTIONS
Question 1
The function
is defined by
is defined by
for
. The function g
.
(i) Show that
and that
It is given that
.
are elements of a group
under the operation of composition of
functions.
The element
is the identity, where e :
(iii) The inverse of the element
for
.
is denoted by h. Find
(iv) Construct the operation table for the elements
.
of the group .
Question 2
Given that
under addition
.
a) Show that
i.
forms a group,
ii.
is an Abelian group.
b) Find all subgroups of
Question 3
(i) The operation
is defined by
, where
and
are real numbers and
is a real constant.
(a) Prove that the set of real numbers, together with the operation , forms a group.
(b) State with a reason, whether the group is commutative.
(ii) The operation by
, where
and
are positive real numbers. By giving
a numerical example in each case, show that two of the basic groups properties are not
necessarily satisfied.
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Question 4
(i)
on
(ii) If
,a set of rational numbers. Is commutative?
and
(iii) Find the values of
, find
such that
.
if
.
Question 5
The set M consists of the six matrices
forms a group
, where
. It is given that
under matrix multiplication, with numerical addition and
multiplication both being carried out
(i) Determine whether
.
is a commutative group, justifying your answer.
(ii) Write down the identity element of the group and find the inverse of
Question 6
Find all Latin squares of side
in standard form with respect to the sequence
. For
each square found determine whether or not it is the multiplication table of a group.
Question 7
The set of polynomials
, where
and
, is denoted by
. Assuming the
Associativity property holds, prove that , under addition, is a group.
Question 8
Show that the Latin Square
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is not the multiplication table of a group.
Question 9
Groups A, B, C and D are defined as follows:
A: The set
under multiplication modulo
B: The set
under multiplication modulo
C: The set
.
.
under multiplication modulo
D: The set
.
under multiplication.
(i) Write the identity element for each of groups A, B, C and D
(ii) Prove the closure property for D
(iii) Elements of the set
are combined under
addition. State which of the four basics group properties are not satisfied.
(Justification is not required.)
Question 10
consist of the set of matrices of the form
, where
and
are real and
, combined under the operation of matrix multiplication.
(i) Prove that
is a group. You may assume that the matrix is associative.
(ii) Determine whether
is commutative.
Question 11
Given that
.
(a) Show that
(i)
is a group under multiplication modulo .
(ii)
is an abelian group.
(b) Find all subgroups of
.
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Question 12
(i) If
, where
, addition
. Determine whether
forms a
group or not.
(ii) Find the value of
(a)
.
(b)
(iii) If
, evaluate
.
Question 13
(a) Is the binary operation
closed on
(b) If
, find the identity element .
where
(c) Show that
.
does not form a group under multiplication
.
Question 14
Consider the set
of all matrices of the form
, where
is a non-zero rational
number. Show that , under operation of matrix multiplication, forms a group,
(You may
assume that matrix multiplication is associative).
Question 15
A group
has distinct elements
, where
is the identity element and
is the
binary operation.
(i) Prove that if
and
, the set of elements
forms a subgroup
of .
(ii) Prove that if
,
and
, the set elements
does not
form a subgroup of .
Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175
Page 38
Question 16
The group
consists of the set
multiplication modulo
combined under
. Find the inverse of each element.
Question 17
If
, where
, multiplication
. Show that
is not a group.
Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175
Page 39
ASANTE SANA
Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175
Page 40
*******DONT BE SATISFIED BY MEDIOCRE WHILST EXCELLENCE
IS THERE*******
CONSTRUCTIVE COMMENTS ON THE FORM
OF THE PRESENTATION, INCLUDING ANY
OMISSIONS OR ERRORS, ARE WELCOME.
***ENJOY***
Nyasha P. Tarakino (Trockers)
+263772978155/+263717267175
ntarakino@gmail.com
Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175
Page 41
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