Algebraic Structures

Overview

In our study of engineering mathematics, we frequently encounter sets upon which certain operations are defined. The systematic investigation of such sets, together with their operations and governing axioms, constitutes the field of abstract algebra. This chapter introduces the foundational concepts of algebraic structures, which provide a powerful framework for abstracting and analyzing a wide range of computational and mathematical problems. We shall focus on two of the most fundamental structures: monoids and groups. These are not merely abstract curiosities; they form the mathematical bedrock for numerous domains within computer science.

A mastery of these concepts is indispensable for the serious student of computer science. The principles of group theory, for instance, are central to modern cryptography and coding theory, ensuring the security and integrity of digital information. Similarly, the structure of a monoid provides the formal language for understanding automata theory and the behavior of sequential machines. For the purposes of the GATE examination, questions frequently test one's ability to verify the properties of a given structure, identify its type, and apply its principles to solve discrete problems. A thorough understanding of the axioms—closure, associativity, identity, and inverse—is therefore of paramount importance.

In this chapter, we will proceed formally, beginning with the definition of a monoid as a set with an associative binary operation and an identity element. We will then build upon this foundation to define a group, which adds the requirement of an inverse for every element. Throughout our discussion, we will examine concrete examples, such as integers under addition $(\mathbb{Z}, +)$ and non-zero rational numbers under multiplication $(\mathbb{Q}^*, \times)$, to solidify the abstract definitions and prepare for the types of problems encountered in the GATE examination.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Monoids | Defining identity and associative binary operations. |
| 2 | Groups | Structures with identity, inverses, and associativity. |

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Learning Objectives

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We now turn our attention to Monoids...

Part 1: Monoids

Introduction

In our study of discrete mathematics, we frequently encounter sets endowed with one or more binary operations. The exploration of these systems, known as algebraic structures, forms the bedrock of abstract algebra. These structures are not mere mathematical curiosities; they provide the formal language to describe computation, symmetry, and a vast array of concepts in computer science, from formal languages and automata theory to cryptography.

A monoid is one of the most fundamental and elegant of these algebraic structures. It represents a system with a single associative binary operation and an identity element. This structure is simple enough to be ubiquitous yet rich enough to possess interesting properties. Understanding the monoid is a crucial step towards mastering more complex structures such as groups, rings, and fields, which are frequently tested in the GATE examination. We shall develop a precise understanding of the properties that define a monoid and learn to systematically identify and analyze such structures.

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Key Concepts

To fully comprehend the definition of a monoid, we must first build a firm foundation by examining the properties of binary operations and the hierarchy of algebraic structures.

1. Algebraic Structures and Binary Operations

An algebraic structure is fundamentally a set combined with one or more operations on that set, along with the axioms that the operations must satisfy. The most basic type of operation is a binary operation.

2. Properties of Binary Operations

The character of an algebraic structure is determined by the properties its binary operation satisfies. For the GATE examination, a systematic verification of these properties is often required.

Associativity

Associativity ensures that the order of evaluation does not matter when combining three or more elements.

For an operation $*$ on a set $S$, associativity holds if:

Consider the set of integers $\mathbb{Z}$ with the operation of subtraction. We find that $(5 - 3) - 2 = 2 - 2 = 0$, whereas $5 - (3 - 2) = 5 - 1 = 4$. Since $0 \neq 4$, subtraction on $\mathbb{Z}$ is not associative. In contrast, addition and multiplication are associative.

Identity Element

The identity element is a special element that leaves any other element unchanged when combined with it.

An element $e \in S$ is the identity element for an operation $*$ if:

For the structure $(\mathbb{Z}, +)$, the identity element is $0$, since $0 + a = a + 0 = a$ for any integer $a$. For $(\mathbb{Z}, \times)$, the identity element is $1$. It is important to note that an identity element, if it exists, must be unique for a given structure.

Commutativity

Commutativity implies that the order of operands does not affect the outcome.

An operation $*$ on a set $S$ is commutative if:

Addition on integers is commutative ($a+b = b+a$), but matrix multiplication is famously not. A monoid with a commutative operation is called a commutative monoid or an Abelian monoid.

Distributivity

When an algebraic system has two binary operations, say $*$ and $\circ$, we can examine how they interact. The property of distributivity describes this interaction.

The operation $*$ is said to be left-distributive over $\circ$ if:

The operation $*$ is said to be right-distributive over $\circ$ if:

If an operation is both left- and right-distributive, we simply say it is distributive.

Worked Example:

Problem: Let the set be the set of real numbers $\mathbb{R}$. Consider two operations $\oplus$ and $\otimes$ defined as $a \oplus b = a + b + 1$ and $a \otimes b = ab$. Determine if $\otimes$ is distributive over $\oplus$.

Solution:

We must check for left and right distributivity.

Step 1: Check for left distributivity of $\otimes$ over $\oplus$. We need to verify if $a \otimes (b \oplus c) = (a \otimes b) \oplus (a \otimes c)$.

The left-hand side (LHS) is:

The right-hand side (RHS) is:

Step 2: Compare LHS and RHS.

The equality does not hold in general (e.g., for $a=2$). Therefore, $\otimes$ is not left-distributive over $\oplus$.

Answer: The operation $\otimes$ is not distributive over $\oplus$. \boxed{\text{Not distributive}}

3. The Hierarchy: From Semigroups to Monoids

Algebraic structures are often classified in a hierarchy based on the axioms they satisfy. This classification is essential for answering questions that ask to identify a given structure.

Semigroup

Monoid

Group

+ Identity
+ Inverse

Closure, Associativity
Closure, Associativity, Identity
Closure, Assoc., Identity, Inverse

Abelian Group

+ Commutativity

Semigroup: A non-empty set $S$ with an associative binary operation $\ast$∗. That is, $(S,$) must satisfy closure and associativity. Example: $(\mathbb{Z}^+, +)$, the set of positive integers under addition. It is associative, but there is no identity element since $0 \notin \mathbb{Z}^+$.

Monoid: A semigroup that has an identity element. Example: $(\mathbb{N}, +)$, the set of natural numbers (including 0) under addition. The identity is $0$. Another example is the set of all strings over an alphabet $\Sigma$, with the operation of string concatenation. The identity element is the empty string $\epsilon$.

Group: A monoid in which every element has an inverse. For each $a \in S$, there must exist an element $a^{-1} \in S$ such that $a$ a^{-1} = a^{-1} a = e. Example: $(\mathbb{Z}, +)$. The inverse of any integer $a$ is $-a$. Note that $(\mathbb{N}, +)$ is a monoid but not a group, because positive integers do not have additive inverses in $\mathbb{N}$.

Abelian Group: A group whose binary operation is commutative. Example: $(\mathbb{Z}, +)$ is an Abelian group. The group of invertible $n \times n$ matrices under matrix multiplication is a non-Abelian group.

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Problem-Solving Strategies

When a GATE question presents a set and a binary operation and asks you to classify the resulting algebraic structure, a systematic, step-by-step verification of the properties is the most reliable approach.

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Common Mistakes

Students often make predictable errors when analyzing algebraic structures under time pressure. Being aware of these can significantly improve accuracy.

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Practice Questions

:::question type="MCQ" question="Let $S$ be the set of all $2 \times 2$ matrices of the form $\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ where $a \in \mathbb{R}$. Consider the algebraic structure $(S, \times)$, where $\times$ is standard matrix multiplication. Which of the following is true?" options=["$(S, \times)$ is a monoid but not a group.","$(S, \times)$ is a group.","$(S, \times)$ is a semigroup but not a monoid.","The operation $\times$ is not associative on $S$."] answer="$(S, \times)$ is a group." hint="Test the properties sequentially: closure, associativity, identity, and inverse. Matrix multiplication is known to be associative." solution="
Step 1: Check Closure
Let $M_a = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ and $M_b = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$ be two matrices in $S$.

Since $a, b \in \mathbb{R}$, their sum $a+b$ is also in $\mathbb{R}$. The resulting matrix is of the required form, so $S$ is closed under matrix multiplication.

Step 2: Check Associativity
Matrix multiplication is inherently associative. So, the associative property holds. $(S, \times)$ is a semigroup.

Step 3: Find Identity Element
We seek a matrix $M_e = \begin{pmatrix} 1 & e \\ 0 & 1 \end{pmatrix}$ such that $M_a \times M_e = M_a$.

For this to be equal to $M_a = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$, we must have $a+e=a$, which implies $e=0$.
The identity matrix is $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, which is in $S$ (for $a=0$). Thus, an identity element exists, and $(S, \times)$ is a monoid.

Step 4: Check for Inverses
For any matrix $M_a = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$, we seek an inverse $M_b = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$ such that $M_a \times M_b = I$.

This requires $a+b=0$, which means $b = -a$. Since $a \in \mathbb{R}$, $-a$ is also in $\mathbb{R}$. So, for every matrix $M_a$, its inverse $M_{-a}$ exists in $S$.

Result:
Since all group axioms (Closure, Associativity, Identity, Inverse) are satisfied, $(S, \times)$ is a group.
Answer: \boxed{$(S, \times)$ is a group.}
"
:::

:::question type="NAT" question="Consider the set of non-negative integers $\mathbb{N} = \{0, 1, 2, ...\}$. A binary operation $\star$ is defined as $a \star b = a + b + 3$. What is the identity element of the monoid $(\mathbb{N}, \star)$?" answer="-3" hint="The identity element $e$ must satisfy $a \star e = a$. Solve for $e$. Then check if this element belongs to the set $\mathbb{N}$. If it does not, no identity element exists within the set." solution="
Step 1: Define the condition for the identity element.
Let $e$ be the identity element. By definition, for any $a \in \mathbb{N}$, it must satisfy:

Step 2: Apply the definition of the operation $\star$.

Step 3: Solve for $e$.

Step 4: Verify if the identity element belongs to the set $\mathbb{N}$.
The set is $\mathbb{N} = \{0, 1, 2, ...\}$. The calculated identity element is $e = -3$.
Since $-3 \notin \mathbb{N}$, there is no identity element for the operation $\star$ within the set $\mathbb{N}$. Therefore, $(\mathbb{N}, \star)$ is not a monoid. The question implies it is a monoid and asks for the identity. If no identity exists in the set, a numerical answer cannot be provided that satisfies the monoid property on the given set. However, if the question implicitly asks for the identity in the larger structure of Integers where it is a monoid, the answer is -3.

Result:
The identity element is $-3$.
Answer: \boxed{-3}
"
:::

:::question type="MSQ" question="Let $\Sigma = \{0, 1\}$ and let $L = \Sigma^*$ be the set of all binary strings, including the empty string $\epsilon$. Let $\circ$ be the concatenation operation. Which of the following statements about the algebraic structure $(L, \circ)$ is/are correct?" options=["$(L, \circ)$ is a monoid.","The operation $\circ$ is commutative.","Every string in $L$ has an inverse.","$(L, \circ)$ is a semigroup."] answer="A,D" hint="Analyze the properties of string concatenation. What is the identity element? Is it possible to 'undo' concatenation to get back to the identity?" solution="
Let's check the properties for $(L, \circ)$.

Closure: If $s_1$ and $s_2$ are binary strings, their concatenation $s_1 \circ s_2$ is also a binary string. So, $L$ is closed under $\circ$.

Associativity: String concatenation is associative. For any strings $s_1, s_2, s_3$, we have $(s_1 \circ s_2) \circ s_3 = s_1 \circ (s_2 \circ s_3)$. For example, ("01" $\circ$ "11") $\circ$ "0" = "0111" $\circ$ "0" = "01110", and "01" $\circ$ ("11" $\circ$ "0") = "01" $\circ$ "110" = "01110". This holds in general. Thus, $(L, \circ)$ is a semigroup. (Option D is correct).

Identity Element: The empty string $\epsilon$ acts as the identity element. For any string $s \in L$, $s \circ \epsilon = s$ and $\epsilon \circ s = s$. Since an identity element exists, $(L, \circ)$ is a monoid. (Option A is correct).

Commutativity: String concatenation is not commutative. For example, let $s_1 = "01"$ and $s_2 = "10"$. Then $s_1 \circ s_2 = "0110"$, but $s_2 \circ s_1 = "1001"$. Since $s_1 \circ s_2 \neq s_2 \circ s_1$, the operation is not commutative. (Option B is incorrect).

Inverse Element: For a non-empty string like $s = "01"$, there is no string $s^{-1}$ such that $s \circ s^{-1} = \epsilon$. Concatenation only makes strings longer (or keeps them the same length if one is $\epsilon$). Therefore, elements other than $\epsilon$ do not have inverses. The structure is not a group. (Option C is incorrect).

Result:
The structure is a semigroup and a monoid.
Answer: \boxed{A,D}
"
:::

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Summary

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What's Next?

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Part 2: Groups

Introduction

In the study of discrete mathematics and abstract algebra, we frequently encounter sets endowed with one or more binary operations. The properties of these operations dictate the structure of the set. An algebraic structure known as a group is one of the most fundamental and pervasive of these. A group is a set combined with an operation that satisfies a few elementary axioms, yet this simple foundation gives rise to a rich and profound theory.

For the GATE examination, a firm understanding of group theory is indispensable. It provides the language for analyzing symmetry, a concept that appears in various domains of computer science, including cryptography, coding theory, and algorithm design. Our study will focus on the core definitions, essential theorems such as Lagrange's theorem, and the properties of specific types of groups like cyclic and Abelian groups, which are frequently tested. We shall develop the necessary tools to rigorously analyze a given algebraic structure and determine its properties.

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Key Concepts

1. Fundamental Properties and Types of Groups

Once a structure is established as a group, it inherits several important properties. We begin by considering the property of commutativity, which distinguishes a significant class of groups.

The size of the set $G$ is also a critical characteristic.

Let us now apply these definitions to a concrete example, demonstrating the process of verifying the group axioms.

Worked Example:

Problem: Consider the set of all subsets of $X = \{1, 2\}$, which is the power set $2^X$. Let the binary operation be symmetric difference, $\Delta$, defined as $A \Delta B = (A \setminus B) \cup (B \setminus A)$. Determine if $(2^X, \Delta)$ forms a group.

Solution:

The set is $2^X = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$. We must verify the four group axioms.

Step 1: Verify Closure

Let $A, B \in 2^X$. By definition, $A$ and $B$ are subsets of $X$. The set difference $A \setminus B$ and $B \setminus A$ are also subsets of $X$. The union of two subsets of $X$ is also a subset of $X$.
Therefore, for any $A, B \in 2^X$, $A \Delta B \in 2^X$. The closure property holds.

Step 2: Verify Associativity

The symmetric difference operation is associative. For any sets $A, B, C$, it holds that $(A \Delta B) \Delta C = A \Delta (B \Delta C)$. This is a standard property of set operations.

Step 3: Find the Identity Element

We seek an element $E \in 2^X$ such that for any $A \in 2^X$, $A \Delta E = A$. Let us test the empty set, $E = \emptyset$.

Thus, the empty set $\emptyset$ is the identity element.

Step 4: Find the Inverse Element

For each $A \in 2^X$, we seek an element $A^{-1} \in 2^X$ such that $A \Delta A^{-1} = \emptyset$. Let us test $A^{-1} = A$.

Every element is its own inverse. Since an inverse exists for every element, the inverse property holds.

Conclusion:
Since all four axioms are satisfied, $(2^X, \Delta)$ is a group. Furthermore, since every element is its own inverse, this group is also Abelian (as we shall prove next).

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2. Conditions for a Group to be Abelian

While the definition of an Abelian group requires checking $a$ b = b a for all pairs, certain group-wide properties can guarantee commutativity. These are frequently tested in GATE.

Property 1: Self-Inverse Elements
If every element in a group $G$ is its own inverse (excluding the identity, which is always its own inverse), then the group is Abelian. That is, if $x^2 = e$ for all $x \in G$, then $G$ is Abelian. (Here, $x^2$ means $x * x$ and $e$ is the identity element).

Proof:
Let $a, b \in G$. We are given that $x^2 = e$ for any $x \in G$. This implies $x = x^{-1}$ for all $x$.
Consider the element $ab \in G$. By the given property, $(ab)^2 = e$, which means $ab$ is its own inverse.

We also know the general "socks-shoes" property for inverses: $(ab)^{-1} = b^{-1}a^{-1}$.

Since every element is its own inverse, we have $b^{-1} = b$ and $a^{-1} = a$. Substituting these into the equation gives:

This holds for all $a, b \in G$, so the group is Abelian.

Property 2: The "Square of a Product" Rule
If for all $x, y \in G$, the relation $(xy)^2 = x^2y^2$ holds, then the group $G$ is Abelian.

Proof:
We are given $(xy)^2 = x^2y^2$. Let us expand both sides.

Now, we can apply the left cancellation law by multiplying by $x^{-1}$ on the left.

So, we have $(yx)y = (xy)y$. Now, apply right cancellation with $y^{-1}$.

This holds for all $x, y \in G$, so the group is Abelian.

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3. Subgroups and Lagrange's Theorem

Within a group, it is often possible to find smaller sets that also form a group under the same operation.

A powerful theorem connects the order of a finite group to the order of its subgroups.

A key consequence of Lagrange's theorem relates to groups of prime order. If $|G| = p$, where $p$ is a prime number, then the only possible orders for its subgroups are the divisors of $p$, which are $1$ and $p$. The subgroup of order $1$ is the trivial subgroup $\{e\}$, and the subgroup of order $p$ is $G$ itself.

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4. Cyclic Groups

Some groups can be entirely described by a single element and the group operation.

Key Properties of Cyclic Groups:

Every cyclic group is Abelian.

A subgroup of a cyclic group is also cyclic.

A group of prime order is always cyclic.

Proof of (3): Let $G$ be a group of prime order $p$. By Lagrange's theorem, the order of any element $a \in G$ (which is the order of the cyclic subgroup generated by $a$) must divide $p$. Since $p$ is prime, the order can only be $1$ or $p$. The only element of order $1$ is the identity $e$. For any other element $a \neq e$, its order must be $p$. The cyclic subgroup $\langle a \rangle$ thus has $p$ elements, which is the entire group $G$. Therefore, $G$ is cyclic.

Worked Example:

Problem: Let $G$ be a group of order $6$, and let $H$ be a subgroup of $G$ such that $1 < |H| < 6$. What can we conclude about $H$?

Solution:

Step 1: Apply Lagrange's Theorem

According to Lagrange's theorem, the order of the subgroup $H$ must divide the order of the group $G$.
Here, $|G| = 6$. The divisors of $6$ are $1, 2, 3, 6$.

Step 2: Apply the given constraints

We are given that $1 < |H| < 6$. This means $|H|$ cannot be $1$ or $6$.
The only possibilities for the order of $H$ are $|H| = 2$ or $|H| = 3$.

Step 3: Use properties of groups of prime order

Both $2$ and $3$ are prime numbers. We know that any group of prime order is cyclic.
Therefore, if $|H| = 2$ or $|H| = 3$, the subgroup $H$ must be cyclic.

Answer: The subgroup $H$ is always cyclic.
(Note: The group $G$ itself may not be cyclic. For example, the symmetric group $S_3$ has order $6$ but is non-Abelian, hence non-cyclic. The group of integers modulo $6$, $Z_6$, also has order $6$ and is cyclic.)

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#
## 5. Direct Product of Groups

We can construct a new group from existing groups using the Cartesian product.

The order of the direct product group is $|G_1 \times G_2| = |G_1| \times |G_2|$.

A common group used in this context is $(Z_n, +_{n})$, the group of integers $\{0, 1, \dots, n-1\}$ under addition modulo $n$. The identity is $0$ and the inverse of $a$ is $n-a$ (for $a \neq 0$).

Worked Example:

Problem: In the group $G = Z_4 \times Z_6$, find the inverse of the element $(3, 4)$.

Solution:

Step 1: Identify the operation and identity

The group is a direct product, so the operation is component-wise addition modulo the respective bases. The identity element is $(0, 0)$.

Step 2: Find the inverse in the first component, $Z_4$

We need to find an element $a \in Z_4$ such that $3 + a \equiv 0 \pmod 4$.
The inverse of $3$ in $Z_4$ is $4-3 = 1$.
Check: $(3+1) \pmod 4 = 4 \pmod 4 = 0$.

Step 3: Find the inverse in the second component, $Z_6$

We need to find an element $b \in Z_6$ such that $4 + b \equiv 0 \pmod 6$.
The inverse of $4$ in $Z_6$ is $6-4 = 2$.
Check: $(4+2) \pmod 6 = 6 \pmod 6 = 0$.

Step 4: Combine the results

The inverse of $(3, 4)$ in $Z_4 \times Z_6$ is $(1, 2)$.

Answer: $\boxed{(1, 2)}$

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Problem-Solving Strategies

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Common Mistakes

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:::question type="MCQ" question="Let $G$ be a group of order 15. Which of the following statements is necessarily true?" options=["$G$ is non-Abelian.","$G$ has a subgroup of order 5.","$G$ is cyclic.","$G$ has no proper non-trivial subgroups."] answer="$G$ is cyclic." hint="Consider the order of the group. It is a product of two distinct primes. A theorem by Sylow states that if $|G| = pq$ where $p, q$ are primes and $p < q$ and $p$ does not divide $q-1$, then $G$ is cyclic. Here $p=3, q=5$. $3$ does not divide $5-1=4$. Thus, $G$ is cyclic." solution="The order of the group is $|G| = 15 = 3 \times 5$. A known result in group theory states that any group of order $pq$, where $p$ and $q$ are distinct primes with $p < q$ and $p$ not dividing $q-1$, must be cyclic. Here, $p=3$ and $q=5$. We check if $3$ divides $5-1=4$. It does not. Therefore, the group $G$ must be cyclic. Since it is cyclic, it is also Abelian, so option A is false. Since it is cyclic of order 15, it has subgroups of order 3 and 5 (by Lagrange's theorem and properties of cyclic groups), so option B is true but option C is a stronger, necessary truth. Option D is false because subgroups of order 3 and 5 exist. Thus, the most definitive statement that is necessarily true is that $G$ is cyclic. Answer: \boxed{G \text{ is cyclic.}}"
:::

:::question type="NAT" question="Consider the group $G = Z_5 \times Z_8$ under component-wise addition. The number of elements in $G$ that are their own inverses is ________." answer="2" hint="An element $(a,b)$ is its own inverse if $(a+a \pmod 5, b+b \pmod 8) = (0,0)$. Solve $2a \equiv 0 \pmod 5$ and $2b \equiv 0 \pmod 8$ separately." solution="An element $(a, b) \in Z_5 \times Z_8$ is its own inverse if $(a, b) + (a, b) = (0, 0)$, which is the identity element. This gives us two separate congruences:

$$2a \equiv 0 \pmod 5$$

$$2b \equiv 0 \pmod 8$$

For the first congruence:
Since 5 is a prime and does not divide 2, we can multiply by the inverse of 2 modulo 5. The inverse is 3, since $2 \times 3 = 6 \equiv 1 \pmod 5$.

So, $a=0$ is the only solution in $Z_5$. (Number of solutions = 1)

For the second congruence:

This means $2b$ is a multiple of 8.
Possible values for $b$ in $Z_8 = \{0, 1, 2, 3, 4, 5, 6, 7\}$ are:

• If $b=0$, $2 \times 0 = 0$, which is a multiple of 8.


• If $b=4$, $2 \times 4 = 8$, which is a multiple of 8.

Other values do not work. For example, $2 \times 2 = 4$, $2 \times 6 = 12$.
So, $b=0$ and $b=4$ are the solutions in $Z_8$. (Number of solutions = 2)

The total number of elements is the product of the number of solutions in each component.
Total elements = (Solutions for $a$) $\times$ (Solutions for $b$) = $1 \times 2 = 2$.
The elements are $(0,0)$ and $(0,4)$.
Answer: \boxed{2}"
:::

:::question type="MSQ" question="Let $(G, *)$ be a group. Which of the following statements is/are always TRUE?" options=["If $|G| = 4$, then $G$ is Abelian.","Every subgroup of a non-Abelian group is non-Abelian.","If $a^{-1} = a$ for all $a \in G$, then $G$ is finite.","The identity element is unique."] answer="If $|G| = 4$, then $G$ is Abelian.,The identity element is unique." hint="Consider the properties of groups of small order and fundamental group axioms. For the second option, think about a cyclic subgroup within a non-Abelian group." solution="

• Option A: A group of order $p^2$ where $p$ is prime is always Abelian. Since $4=2^2$, any group of order 4 is Abelian. There are two such groups, $Z_4$ and the Klein-4 group $Z_2 \times Z_2$, both of which are Abelian. So, this statement is TRUE.


• Option B: This is false. Consider the non-Abelian group $S_3$ of order 6. The set $H = \{e, (12)\}$ forms a subgroup of order 2. Any group of order 2 is cyclic and hence Abelian. Thus, a non-Abelian group can have Abelian subgroups. So, this statement is FALSE.


• Option C: This is false. Consider the group $(2^X, \Delta)$ where $X$ is an infinite set (like the set of natural numbers $\mathbb{N}$). In this group, every element is its own inverse, but the group is infinite. So, this statement is FALSE.


• Option D: This is a fundamental property of groups. Assume there are two identity elements, $e_1$ and $e_2$. Then by definition of identity, $e_1$ e_2 = e_1e1​∗e2​=e1​ (since $e_2$ is an identity) and $e_1$ e_2 = e_2 (since $e_1$ is an identity). Therefore, $e_1 = e_2$. The identity element is always unique. So, this statement is TRUE.

Answer: \boxed{\text{If } |G| = 4 \text{, then } G \text{ is Abelian.,The identity element is unique.}}"
:::

:::question type="MCQ" question="Let $S$ be the set of all $2 \times 2$ matrices with real entries, and let the binary operation be standard matrix multiplication. Which is the primary axiom that fails for $(S, \times)$ to be a group?" options=["Closure", "Associativity", "Identity Element", "Inverse Element"] answer="Inverse Element" hint="Check each axiom systematically. For the inverse, you will find one element that does not have an inverse in the set $S$." solution="We check the four group axioms for the set $S$ of all $2 \times 2$ real matrices under matrix multiplication.

Closure: The product of two $2 \times 2$ matrices is another $2 \times 2$ matrix. This property holds.

Associativity: Matrix multiplication is associative. This property holds.

Identity Element: The identity element is the $2 \times 2$ identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, which is in the set $S$. This property holds.

Inverse Element: For an element to have an inverse, it must be invertible. A matrix is invertible if and only if its determinant is non-zero. The set $S$ includes singular matrices (matrices with a determinant of 0), such as the zero matrix $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. This matrix does not have a multiplicative inverse. Since not every element in $S$ has an inverse, the inverse axiom fails.

Answer: \boxed{\text{Inverse Element}}"
:::

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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Let $(G, \ast)$ be a group with identity $e$. Consider the following statements:
I. If $a^2 = e$ for all $a \in G$, then $G$ is Abelian.
II. The set of integers $\mathbb{Z}$ under the operation of subtraction ($-$) forms a group.
III. If $G$ is a finite group of order $n$, then for any $a \in G$, $a^n=e$.

Which of the above statements is/are correct?" options=["I and III only", "I only", "II and III only", "I, II, and III"] answer="A" hint="For statement I, consider the expression $(ab)^2$. For statement II, test the group axioms, especially associativity and identity. For statement III, recall Lagrange's Theorem and its corollary regarding the order of an element." solution="Let us analyze each statement individually.

Statement I: If $a^2 = e$ for all $a \in G$, then $G$ is Abelian.
This implies that every element is its own inverse, i.e., $a = a^{-1}$ for all $a \in G$.
Consider any two elements $a, b \in G$. We want to check if $ab = ba$.
Let's examine the expression $(ab)^2$. By the given property, $(ab)^2 = e$.
This means $(ab)(ab) = e$.
Multiplying on the left by $a$, we get $a(ab)(ab) = ae = a$.
Using associativity, $(a a)b(ab) = a$.
Since $a^2 = e$, this becomes $e b(ab) = a$, which simplifies to $b(ab) = a$.
Now, multiplying on the left by $b$, we get $b(b(ab)) = ba$.
Using associativity again, $(b b)(ab) = ba$.
Since $b^2 = e$, this becomes $e(ab) = ba$, which simplifies to $ab = ba$.
Since this holds for all $a, b \in G$, the group is Abelian. Thus, Statement I is correct.

Statement II: The set of integers $\mathbb{Z}$ under subtraction ($-$) is not a group.
Let us check the axioms.

• Closure: $a - b \in \mathbb{Z}$ for all $a, b \in \mathbb{Z}$. (Holds)


• Associativity: We must have $(a - b) - c = a - (b - c)$.

Let $a=3, b=2, c=1$.
LHS: $(3 - 2) - 1 = 1 - 1 = 0$.
RHS: $3 - (2 - 1) = 3 - 1 = 2$.
Since $0 \neq 2$, the associative property fails.
Therefore, $(\mathbb{Z}, -)$ is not a group. Thus, Statement II is incorrect.

Statement III: If $G$ is a finite group of order $n$, then for any $a \in G$, $a^n=e$.
Let the order of the element $a$ be $k$. From the corollary of Lagrange's Theorem, we know that the order of any element must divide the order of the group.
Therefore, $k$ divides $n$. This means $n = mk$ for some integer $m \ge 1$.
We know by the definition of the order of an element that $a^k = e$.
Now, consider $a^n$:

Thus, Statement III is correct.

Since statements I and III are correct, the correct option is A.
Answer: \boxed{A}"
:::

:::question type="NAT" question="Consider the group $(\mathbb{Z}_{30}, +_{30})$, which is the group of integers modulo 30 under addition. What is the order of the element 24 in this group?" answer="5" hint="The order of an element $k$ in the group $(\mathbb{Z}_n, +_n)$ is given by the formula $\frac{n}{\operatorname{gcd}(k, n)}$." solution="We are asked to find the order of the element $24$ in the group $(\mathbb{Z}_{30}, +_{30})$.

The order of an element $a$ in a group is the smallest positive integer $k$ such that $a^k = e$, where $e$ is the identity element. In the group $(\mathbb{Z}_{30}, +_{30})$, the operation is addition modulo 30, and the identity element is $0$. Therefore, we are looking for the smallest positive integer $k$ such that:

The general formula for the order of an element $a$ in $(\mathbb{Z}_n, +_n)$ is given by:

Here, $n=30$ and $a=24$.
First, we need to calculate the greatest common divisor (gcd) of 24 and 30.
The prime factorization of 24 is $2^3 \times 3$.
The prime factorization of 30 is $2 \times 3 \times 5$.
The common prime factors are 2 and 3. The lowest powers are $2^1$ and $3^1$.

Now, we can use the formula to find the order:

Therefore, the order of the element 24 is 5.

We can verify this by direct calculation:

• $1 \times 24 \equiv 24 \pmod{30}$


• $2 \times 24 = 48 \equiv 18 \pmod{30}$


• $3 \times 24 = 72 \equiv 12 \pmod{30}$


• $4 \times 24 = 96 \equiv 6 \pmod{30}$


• $5 \times 24 = 120 \equiv 0 \pmod{30}$

The smallest positive integer is indeed 5.
Answer: \boxed{5}"
:::

:::question type="MSQ" question="Let $S$ be the set of all $2 \times 2$ matrices of the form $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ where $a, b, c$ are real numbers. Let $\ast$ be the standard matrix multiplication. Which of the following statements about the algebraic structure $(S, \ast)$ is/are true?" options=["$(S, \ast)$ is a monoid.", "$(S, \ast)$ is a group.", "The identity element is $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.", "The operation $\ast$ is commutative on $S$."] answer="A, C" hint="Check all the group axioms. For the inverse, what condition must $a$ and $c$ satisfy for the matrix to be invertible? Does this condition hold for all elements in $S$?" solution="Let us evaluate the properties of the algebraic structure $(S, \ast)$.

1. Closure:
Let $M_1 = \begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix}$ and $M_2 = \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix}$ be two matrices in $S$.
Their product is:

The resulting matrix is also of the form $\begin{pmatrix} a' & b' \\ 0 & c' \end{pmatrix}$ where $a', b', c'$ are real numbers. So, closure holds.

2. Associativity:
Matrix multiplication is known to be associative. This property is inherited from the general set of $2 \times 2$ matrices. So, associativity holds.

3. Identity Element:
The standard identity matrix for multiplication is $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. This matrix is of the required form (with $a=1, b=0, c=1$), so it belongs to $S$. We can verify that for any $M \in S$, $M \ast I = I \ast M = M$. Thus, an identity element exists in $S$. Statement C is correct.

Since $(S, \ast)$ is a semigroup (closure + associativity) with an identity element, $(S, \ast)$ is a monoid. Statement A is correct.

4. Inverse Element:
For $(S, \ast)$ to be a group, every element must have an inverse. A matrix has an inverse if and only if its determinant is non-zero. The determinant of a matrix $M = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ is $\det(M) = ac - b(0) = ac$.
The inverse exists only if $ac \neq 0$.
However, the set $S$ includes matrices where $a=0$ or $c=0$. For example, the matrix $\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}$ is in $S$, but its determinant is 0, so it does not have an inverse.
Since not every element in $S$ has an inverse, $(S, \ast)$ is not a group. Statement B is incorrect.

5. Commutativity:
Let's check if $M_1 \ast M_2 = M_2 \ast M_1$.
We already calculated $M_1 \ast M_2 = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{pmatrix}$.
Now, let's calculate $M_2 \ast M_1$:

For commutativity, we need $a_1b_2 + b_1c_2 = a_2b_1 + b_2c_1$ for all choices of $a_1, b_1, c_1, a_2, b_2, c_2$.
Let $M_1 = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$ and $M_2 = \begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix}$.
$M_1 \ast M_2 = \begin{pmatrix} 2 & 2+3 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix}$.
$M_2 \ast M_1 = \begin{pmatrix} 1 & 1+1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$.
Since $M_1 \ast M_2 \neq M_2 \ast M_1$, the operation is not commutative. Statement D is incorrect.

Therefore, the correct statements are A and C.
Answer: \boxed{A, C}"
:::

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