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← Abstract Algebra: Structures levels

Abstract Algebra: Structures / LEVEL 1 · DIFFICULTY 4/5

Abstract Algebra: Structures

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE UABS 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Groups, Subgroups & Cosets

Useful preparation: Logic, Sets & Relations

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Groups describe operations that combine and undo. Subgroups collect self-contained operations; cosets divide the group into equally sized translated copies.

Group and abelian group
A group is a nonempty set with a closed associative binary operation, an identity e, and a two-sided inverse for every element. It is abelian when the operation also commutes.
Subgroup and generated subgroup
A subgroup H is a subset that is itself a group under the inherited operation. The subgroup generated by a, written ⟨a⟩, contains all integer powers of a, or integer multiples in additive notation.
Coset and index
The left coset gH consists of gh for h in H. The index [G:H]\displaystyle \left[G : H\right] counts distinct left cosets. In additive groups, write g+H\displaystyle g + H instead.

Conditions before calculation

  • A subgroup candidate must be nonempty. It suffices to show that ab−1\displaystyle ab^{-1} belongs to it whenever a and b do; in additive notation this becomes closure under differences.
  • Lagrange’s numerical statement ∣G∣=[G:H]\displaystyle \left|G\right| = \left[G : H\right]∣H∣\displaystyle \left|H\right| concerns finite G. Divisibility of group sizes is a necessary condition for a subgroup, not a general existence theorem.
  • Cosets form a quotient group under their natural multiplication only if H is normal: gHg⁻¹=H for every g. Every subgroup of an abelian group satisfies this condition.

Why the method works

1. Separate the operation from the elements

The same underlying set may form a group under one operation and fail under another. In modular addition the identity is the zero class and the inverse of a is −a.

2. Derive the order formula

In Z/nZ\mathbb{Z}/n\mathbb{Z} under addition, the order of a is the smallest positive k for which n divides ka. Dividing n and a by their greatest common divisor reveals that smallest k.

ord⁡([a])=ngcd⁡(n,a)\operatorname{ord}([a])=\frac n{\gcd(n,a)}

3. Count equal blocks

Multiplication by g maps H bijectively onto gH; multiplication by g−1\displaystyle g^{-1} reverses it. Cosets are either disjoint or equal, so counting their elements gives Lagrange’s formula.

∣G∣=[G:H]⋅∣H∣|G|=[G:H]\cdot|H|

Worked examples

EXAMPLE 1

In the additive group Z/18Z\mathbb{Z}/18\mathbb{Z}, find the subgroup generated by 6 and its index.

  1. Repeated addition produces the residue classes 0,6,12, then returns to 0.

  2. These three classes form the generated subgroup; the order formula agrees because gcd⁡ (18,6)=6\displaystyle \operatorname{gcd}\, \left(18, 6\right) = 6.

  3. Each coset contains three elements. Dividing 18 by 3 gives six cosets, represented by 0 through 5.

Conclusion: The subgroup is {0,6,12}, the element order is 3, and the index is 6.

EXAMPLE 2

In Z/20Z\mathbb{Z}/20\mathbb{Z}, let H={0,5,10,15}\displaystyle H = \left\{0, 5, 10, 15\right\}. Decide whether 3+H\displaystyle 3 + H and 8+H\displaystyle 8 + H are different cosets.

  1. Subtracting any two listed multiples of 5 gives another such class modulo 20, so H is a subgroup.

  2. The difference 8−3=5\displaystyle 8 - 3 = 5 lies in H, which means translation by 8 reaches the same block as translation by 3.

  3. Listing confirms the conclusion.

    3+H={3,8,13,18}=8+H3+H=\{3,8,13,18\}=8+H

Conclusion: The two expressions name the same coset, not two different blocks.

Pitfalls & counterexamples

Tempting claim: Every coset is itself a subgroup.

Correction: Only the coset containing the identity equals H and is a subgroup.

Test the claim: In Z/20Z\mathbb{Z}/20\mathbb{Z}, the coset 3+H\displaystyle 3 + H above does not contain 0, so it cannot be an additive subgroup.

Check your understanding

What is the additive order of 8 in Z/20Z\mathbb{Z}/20\mathbb{Z}?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

5.

The smallest positive k with 20 dividing 8k is 5; equivalently, 20gcd⁡ (20,8)=5\displaystyle \frac{20}{\operatorname{gcd}}\, \left(20, 8\right) = 5.

Connect this idea

Cosets are equivalence classes. Their equal sizes prepare kernel-image counting; normality makes quotient multiplication well-defined.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    In the additive group Z12 Z\displaystyle \frac{Z}{12}\, Z, find the order of the element 3.

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  2. Problem 2

    A finite group has 24 elements and a subgroup with 6 elements. How many left cosets are there?

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  3. Problem 3

    In Z/nZ has modulus n=5\displaystyle n = 5. What is the additive inverse of 2, represented between 0 and n−1\displaystyle n - 1?

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  4. Problem 4

    In the additive group Z8 Z\displaystyle \frac{Z}{8}\, Z, let H be the subgroup of even residue classes. How many elements does the quotient group have?

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  5. Problem 5New context

    A cyclic-state model uses Z18 Z\displaystyle \frac{Z}{18}\, Z, find the order of the element 3.

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  6. Problem 6New context

    A symmetry group has 36 elements and a subgroup with 9 elements. How many left cosets are there?

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  7. Problem 7New context

    An additive cyclic group has modulus n=6\displaystyle n = 6. What is the additive inverse of 2, represented between 0 and n−1\displaystyle n - 1?

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  8. Problem 8New context

    A cyclic group is Z12 Z\displaystyle \frac{Z}{12}\, Z, let H be the subgroup of even residue classes. How many elements does the quotient group have?

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Open stage UABS 1.1 in the student workspace →

STAGE UABS 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Homomorphisms, Kernels & Images

Useful preparation: Groups, Subgroups & Cosets

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Homomorphisms preserve operations while possibly merging inputs. The kernel records what disappears; every nonempty fiber is one translated copy of that kernel.

Group homomorphism
For groups G and H, a map ϕ:G→H\displaystyle \phi : G \to H is a homomorphism when ϕ (ab)=ϕ (a) ϕ (b)\displaystyle \phi\, \left(ab\right) = \phi\, \left(a\right)\, \phi\, \left(b\right) for all a,b in G. In additive groups, replace products by sums.
Kernel and image
The kernel is the set of inputs sent to the identity of H. The image is the set of all attained outputs; it need not equal the entire codomain.
Fiber and isomorphism
The fiber over y is the set of inputs mapped to y. A bijective homomorphism is an isomorphism and identifies the two group structures.

Conditions before calculation

  • Check that a proposed rule is well-defined on the declared elements. For residue classes, different integer representatives must produce the same output class.
  • The finite counting identity requires a finite domain group and a genuine homomorphism. An arbitrary function need not have fibers of equal size.
  • The quotient G/ker ϕ\displaystyle \phi is a group because the kernel is normal. The first isomorphism theorem identifies this quotient with the image, not automatically with the entire codomain.

Why the method works

1. Find what collisions mean

Two elements have the same output exactly when their relative difference belongs to the kernel. This follows from preservation of inverses and multiplication.

φ(a)=φ(b)  ⟺  φ(b−1a)=eH  ⟺  b−1a∈ker⁡φ\varphi(a)=\varphi(b)\iff\varphi(b^{-1}a)=e_H\iff b^{-1}a\in\ker\varphi

2. Describe a whole fiber

If ϕ (a)=y\displaystyle \phi\, \left(a\right) = y, then multiplying a by any kernel element leaves its output unchanged. Conversely every input giving y differs from a by a kernel element; the fiber is a ker ϕ\displaystyle \phi.

3. Count and interpret

Kernel cosets partition G and pair with image elements. For finite G, their equal sizes give the counting formula. A trivial kernel means no two inputs collide.

∣G∣=∣ker⁡φ∣ ∣im⁡φ∣,G/ker⁡φ≅im⁡φ|G|=|\ker\varphi|\,|\operatorname{im}\varphi|,\qquad G/\ker\varphi\cong\operatorname{im}\varphi

Worked examples

EXAMPLE 1

Let ϕ\displaystyle \phi:Z/18Z\mathbb{Z}/18\mathbb{Z}→Z/18Z\mathbb{Z}/18\mathbb{Z} send [x] to [6x]. Find its kernel and image.

  1. Changing x by 18 changes 6x by 108, a multiple of 18. Thus the rule is well-defined; distributivity proves it preserves addition.

  2. The kernel condition is 18 dividing 6x, equivalent to 3 dividing x.

    ker⁡φ={0,3,6,9,12,15}\ker\varphi=\{0,3,6,9,12,15\}

  3. The attained outputs are 0,6,12. Each has six preimages, consistent with the total 18.

    im⁡φ={0,6,12},18=6⋅3\operatorname{im}\varphi=\{0,6,12\},\qquad18=6\cdot3

Conclusion: The map is neither injective nor surjective onto its declared codomain.

EXAMPLE 2

Let ψ:(Z,+)→(Z,+) be ψ(n)=4 n\displaystyle \left(n\right) = 4\, n. Is it an isomorphism?

  1. For any integers m,n, multiplication distributes over addition.

    ψ(m+n)=4m+4n=ψ(m)+ψ(n)\psi(m+n)=4m+4n=\psi(m)+\psi(n)

  2. The only integer sent to 0 is 0, so the kernel is trivial and the map is injective.

  3. The integer 1 has no integer preimage; the image is 4Z. Restricting the codomain to 4Z makes the map bijective.

Conclusion: It is not an isomorphism onto Z, but it is an isomorphism from Z to 4Z.

Pitfalls & counterexamples

Tempting claim: A trivial kernel guarantees surjectivity.

Correction: It guarantees injectivity. Surjectivity must be tested against the specified codomain.

Test the claim: The map n↦4n on Z has trivial kernel but misses 1, 2, and 3.

Check your understanding

A homomorphism has a domain group of order 42 and a kernel of order 7. How many elements are in its image?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

6.

The image indexes the six equal kernel cosets, since 427=6\displaystyle \frac{42}{7} = 6.

Connect this idea

Kernels group indistinguishable inputs. Linear maps have analogous kernels and images, with dimensions replacing finite group sizes.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    A homomorphism has domain order 12 and kernel order 2. Find its image order.

    Practise this exact problem →
  2. Problem 2

    The reduction map ϕ:Z6 Z→Z3 Z\displaystyle \phi : \frac{Z}{6}\, Z \to \frac{Z}{3}\, Z sends each class to its remainder modulo 3. How many elements are in its kernel?

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  3. Problem 3

    An additive homomorphism ϕ:Z→Z\displaystyle \phi : Z \to Z satisfies ϕ (1)=2\displaystyle \phi\, \left(1\right) = 2. Find ϕ (4)\displaystyle \phi\, \left(4\right).

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  4. Problem 4

    A group homomorphism from a group of order 5 has kernel size 1. Is it injective? Enter 1 for yes, 0 for no.

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  5. Problem 5New context

    A finite-group map has domain order 18 and kernel order 3. Find its image order.

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  6. Problem 6New context

    A residue map ϕ:Z9 Z→Z3 Z\displaystyle \phi : \frac{Z}{9}\, Z \to \frac{Z}{3}\, Z sends each class to its remainder modulo 3. How many elements are in its kernel?

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  7. Problem 7New context

    A map preserving addition ϕ:Z→Z\displaystyle \phi : Z \to Z satisfies ϕ (1)=3\displaystyle \phi\, \left(1\right) = 3. Find ϕ (4)\displaystyle \phi\, \left(4\right).

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  8. Problem 8New context

    An operation-preserving map from a group of order 6 has kernel size 1. Is it injective? Enter 1 for yes, 0 for no.

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Open stage UABS 1.2 in the student workspace →

STAGE UABS 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Rings, Ideals & Fields

Useful preparation: Homomorphisms, Kernels & Images

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Rings support addition and multiplication; fields also support division by nonzero elements. Ideals make quotient arithmetic consistent.

Ring convention
Here rings have an abelian additive group, associative multiplication, identity 1≠0\displaystyle 1 \ne 0, and both distributive laws. Multiplication need not commute; ring homomorphisms preserve 1.
Unit, integral domain, and field
A unit has a two-sided multiplicative inverse. An integral domain is a commutative ring with no product of two nonzero elements equal to zero. A field is a commutative ring whose every nonzero element is a unit.
Two-sided ideal
An ideal I is an additive subgroup such that ri and ir belong to I whenever i belongs to I and r belongs to the ring. These absorption conditions allow multiplication of residue classes.

Conditions before calculation

  • Under our convention 1≠0\displaystyle 1 \ne 0, use a proper ideal I for RI\displaystyle \frac{R}{I}. Proper ideals contain neither 1 nor any other unit.
  • For integer n≥2\displaystyle n \ge 2, the ring Z/nZ\mathbb{Z}/n\mathbb{Z} is a field exactly when n is prime. A class [a] is a unit exactly when gcd⁡ (a,n)=1\displaystyle \operatorname{gcd}\, \left(a, n\right) = 1.
  • Every finite integral domain is a field. Finiteness is essential: Z has no nonzero zero divisors but its element 2 has no integer multiplicative inverse.

Why the method works

1. Distinguish the two inverses

The additive inverse of a always exists and sums with a to zero. A multiplicative inverse must instead produce one; the additive group axioms do not supply it.

a+(−a)=0,aa−1=a−1a=1a+(-a)=0,\qquad aa^{-1}=a^{-1}a=1

2. Explain ideal absorption

If representatives change by ideal elements, the product changes by sums containing those ideal elements. Absorption keeps that difference in I, making the quotient product independent of representatives.

(a+i)(b+j)−ab=aj+ib+ij∈I(a+i)(b+j)-ab=aj+ib+ij\in I

3. Explain the finite-domain theorem

For nonzero a in a domain, ax=ay\displaystyle ax = ay implies x=y\displaystyle x = y because a (x−y)=0\displaystyle a\, \left(x - y\right) = 0. On a finite set an injective self-map is surjective, so multiplication by a reaches 1 and supplies an inverse.

Worked examples

EXAMPLE 1

Find the additive and multiplicative inverses of [7] in Z/11Z\mathbb{Z}/11\mathbb{Z}.

  1. The additive inverse must make a sum divisible by 11; 7+4=11\displaystyle 7 + 4 = 11.

  2. For multiplication, the integer identity 8⋅7−5⋅11=1\displaystyle 8 \cdot 7 - 5 \cdot 11 = 1 exhibits the required inverse modulo 11.

    [7]+[4]=[0],[7][8]=[1][7]+[4]=[0],\qquad[7][8]=[1]

  3. Since 11 is prime, every nonzero class has a multiplicative inverse; here [4] and [8] play different roles.

Conclusion: The additive inverse is [4]; the multiplicative inverse is [8].

EXAMPLE 2

Explain why Z/10Z\mathbb{Z}/10\mathbb{Z} is a quotient ring but not a field.

  1. The multiples 10Z form an additive subgroup of Z. Multiplying a multiple of 10 by any integer gives another multiple, so it is an ideal.

  2. It is proper because 1 is not a multiple of 10. Its ten cosets inherit well-defined addition and multiplication.

  3. The nonzero classes [2] and [5] multiply to [0]. If [2] had an inverse, multiplying this equation by it would force [5]=[0]\displaystyle \left[5\right] = \left[0\right], a contradiction.

Conclusion: The quotient is a commutative ring with zero divisors, so it is not a field.

Pitfalls & counterexamples

Tempting claim: Every nonzero ring element can be canceled from a product.

Correction: Cancellation needs an additional condition, such as being a unit or working in an integral domain.

Test the claim: Modulo 10, [2] [1]=[2] [6]\displaystyle \left[2\right]\, \left[1\right] = \left[2\right]\, \left[6\right], but [1] and [6] are different classes.

Check your understanding

Can a proper ideal in a ring contain 1?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

If 1 belongs to I, then every r=r⋅1\displaystyle r = r \cdot 1 belongs to I by absorption, so I equals the whole ring.

Connect this idea

Ideals enable quotient rings. Fields support linear algebra; zero divisors explain failed cancellation.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    In the ring Z4 Z\displaystyle \frac{Z}{4}\, Z, multiply the nonzero classes 2 and 2. What residue between 0 and 3 results?

    Practise this exact problem →
  2. Problem 2

    In Z/nZ n=4\displaystyle n = 4. What is the multiplicative inverse of n−1\displaystyle n - 1, as a residue between 0 and n−1\displaystyle n - 1?

    Practise this exact problem →
  3. Problem 3

    The ideal 3Z. How many residue classes are in Z3 Z\displaystyle \frac{Z}{3\, Z}?

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  4. Problem 4

    In the ring Z5 Z\displaystyle \frac{Z}{5}\, Z, is 2 a unit? Enter 1 for yes, 0 for no.

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  5. Problem 5New context

    A modular arithmetic ring is Z6 Z\displaystyle \frac{Z}{6}\, Z, multiply the nonzero classes 2 and 3. What residue between 0 and 5 results?

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  6. Problem 6New context

    A modular ring has n=5\displaystyle n = 5. What is the multiplicative inverse of n−1\displaystyle n - 1, as a residue between 0 and n−1\displaystyle n - 1?

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  7. Problem 7New context

    An ideal in Z is 4Z. How many residue classes are in Z4 Z\displaystyle \frac{Z}{4\, Z}?

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  8. Problem 8New context

    A modular inverse test uses the ring Z7 Z\displaystyle \frac{Z}{7}\, Z, is 2 a unit? Enter 1 for yes, 0 for no.

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