Quiz: Groups, Rings, and Fields — 23 questions

1. Which condition makes a binary operation an internal composition law on a set E?

A. It maps pairs from any set into a larger containing set

B. It maps every element of E to a pair in E × E

C. It assigns each element of E a distinct element of E

D. It maps every pair in E × E to an element of E

Answer: (D)

Explanation

An internal composition law is a function from E × E into E, so combining any two elements of E produces an element that remains in E. A general binary operation need not preserve the chosen set E.

2. Why is composition an internal composition law for functions from E to E?

A. The composition is defined only when the two functions have different domains

B. The composition of two functions from E to E always produces a constant function

C. The composition of two functions from E to E is another function from E to E

D. The composition reverses the domain and codomain of both original functions

Answer: (C)

Explanation

Applying one function from E to E after another still maps E into E, so the result belongs to the same set of functions. Producing a constant function is not required, and composition does not reverse domains and codomains.

3. Which equation expresses that an internal law * is commutative?

A. $x*y=y*x=e$

B. $a*b=b*a$

C. $e*x=x*e=x$

D. $(a*b)*c=a*(b*c)$

Answer: (B)

Explanation

Commutativity means that exchanging the two operands does not change the result, expressed by $a*b=b*a$. The equation involving parentheses defines associativity, while the other equations describe identity and inverse relationships.

4. Which property allows the parentheses to be changed in a product without changing its value?

A. Inverses, expressed as $x*y=y*x=e$

B. Identity, expressed as $e*x=x*e=x$

C. Associativity, expressed as $(a*b)*c=a*(b*c)$

D. Commutativity, expressed as $a*b=b*a$

Answer: (C)

Explanation

Associativity concerns the placement of parentheses, so it permits the regrouping shown in $(a*b)*c=a*(b*c)$. Commutativity instead concerns changing the order of operands.

5. What must an element e satisfy to be an identity element for an internal law *?

A. It must satisfy $e*x=x*e$ for at least one x in E

B. It must satisfy $e*x=x^{-1}*e$ for every x in E

C. It must satisfy $e*x=x*e=x$ for every x in E

D. It must satisfy $e*x=x*e=e$ for every x in E

Answer: (C)

Explanation

An identity leaves every element unchanged when placed on either side, so $e*x=x*e=x$ for every x in E. Returning e or merely commuting with selected elements does not establish the identity property.

6. If e is the identity for *, what condition makes y an inverse of x?

A. It satisfies $x*y=y*x=e$

B. It satisfies $x*y=e*x=x$

C. It satisfies $x*y=y*x=x$

D. It satisfies $x*y=y*x=y$

Answer: (A)

Explanation

An inverse combines with the original element on either side to produce the identity, giving $x*y=y*x=e$. Producing x or y instead describes neither the inverse condition nor the identity requirement.

7. Which collection of properties defines a group?

A. An internal law that is associative, lacks an identity, and gives some elements inverses

B. An internal law that is distributive, has an identity, and contains finitely many elements

C. An internal law that is associative, has an identity, and gives every element an inverse

D. An internal law that is commutative, has an identity, and contains a zero element

Answer: (C)

Explanation

A group requires closure under an internal law, associativity, an identity element, and an inverse for every element. Commutativity and finiteness are not required, and inverses cannot be restricted to only some elements.

8. When is a group called an abelian group?

A. When its operation is also commutative

B. When every element has two distinct inverses

C. When its operation is associative but has no identity

D. When its underlying set contains only real numbers

Answer: (A)

Explanation

An abelian group is a group whose operation additionally satisfies commutativity. Associativity and an identity are already group requirements, while the type of elements does not determine whether a group is abelian.

9. Which statement correctly classifies the following structures under their indicated operations?

A. $(ℕ,+)$ is not a group, while $(ℤ,+)$ and $(ℚ^*,\times)$ are groups

B. $(ℕ,+)$ is a group, while $(ℤ,+)$ and $(ℚ^*,\times)$ are not groups

C. $(ℕ,+)$ is not a group, while $(ℤ,+)$ is not a group and $(ℚ^*,\times)$ is a group

D. $(ℕ,+)$ and $(ℤ,+)$ are groups, while $(ℚ^*,\times)$ is not a group

Answer: (A)

Explanation

Natural numbers under addition lack additive inverses for nonzero elements, whereas integers under addition have additive inverses and nonzero rationals under multiplication have multiplicative inverses. Therefore the stated classification follows from the inverse requirement for groups.

10. Which statement correctly describes identity and inverse elements in a group?

A. The group has one identity, and each element has its own unique inverse.

B. Each element has one identity, and the group has several possible inverses.

C. The group has one inverse, and each element may have several identities.

D. Every element shares one inverse, while the identity depends on the element.

Answer: (A)

Explanation

A group has a unique identity element, and every particular element has a unique inverse. The inverse is associated with the chosen element, so it is not one common element for the entire group.

11. For elements x and y in a group, which expression gives the inverse of their product?

A. $x^{-1}*y^{-1}$

B. $y^{-1}*x^{-1}$

C. $y*x^{-1}$

D. $x*y^{-1}$

Answer: (B)

Explanation

Taking the inverse of a product reverses the order, giving $(x*y)^{-1}=y^{-1}*x^{-1}$. The expression $x^{-1}*y^{-1}$ generally has the factors in the wrong order.

12. In a group, what is the unique solution of the equation $a*x=b$?

A. $a^{-1}*b$

B. $a*b^{-1}$

C. $b^{-1}*a$

D. $b*a^{-1}$

Answer: (A)

Explanation

Left-multiplying the equation by $a^{-1}$ gives $x=a^{-1}*b$. The other expressions do not follow from isolating x on the left in a potentially noncommutative group.

13. Which condition makes a subset H of a group G a subgroup?

A. H contains at least one element and is closed under the group operation but not necessarily inverses.

B. H contains the identity and inverses, but products of elements may lie outside H.

C. H contains products of elements, while the identity and inverses may belong to G outside H.

D. H contains the identity, is closed under inverses, and is closed under the group operation.

Answer: (D)

Explanation

A subgroup must contain the identity of G, contain the inverse of each of its elements, and remain closed under the group operation. Closure under products alone does not ensure that H has the required subgroup structure.

14. What property distinguishes a group morphism f from an arbitrary function between groups?

A. It maps every source element to the identity element of the target group.

B. It preserves the group operation: $f(x*y)=f(x)\bullet f(y)$ for all x and y.

C. It preserves inverses but need not preserve products of source elements.

D. It assigns a different target element to every source element, regardless of the group operation.

Answer: (B)

Explanation

A group morphism preserves multiplication through the relation $f(x*y)=f(x)\bullet f(y)$. An arbitrary function can fail to respect products, even if it is defined between the same two groups.

15. Which description correctly distinguishes an endomorphism, an automorphism, and an isomorphism?

A. An endomorphism is always bijective; an automorphism maps two different groups; an isomorphism maps a group to itself.

B. An endomorphism maps two different groups; an automorphism need not be bijective; an isomorphism is any operation-preserving function.

C. An endomorphism is an inverse map; an automorphism is any morphism; an isomorphism is a morphism from a group to itself.

D. An endomorphism maps a group to itself; an automorphism is a bijective endomorphism; an isomorphism is a bijective morphism between groups.

Answer: (D)

Explanation

An endomorphism has the same group as its source and target, an automorphism is a bijective endomorphism, and an isomorphism is a bijective morphism between groups. Thus an isomorphism can connect distinct groups, unlike an endomorphism.

16. If f is a group morphism and x is an element of its source group, which identity must hold?

A. $f(x^{-1})=f(x^{-1})^{-1}$

B. $f(x^{-1})=f(x)$

C. $f(x^{-1})=f(x)^{-1}$

D. $f(x^{-1})=f(x)*f(x)$

Answer: (C)

Explanation

A morphism sends inverses to inverses, so $f(x^{-1})=f(x)^{-1}$. This follows because applying f to the relation $x*x^{-1}=e$ produces the identity in the target group.

17. For a morphism $f:G_1\to G_2$, where does the kernel lie and how is it defined?

A. It is a subset of $G_2$ consisting of elements attained by f.

B. It is a subset of $G_1$ consisting of elements mapped to arbitrary elements of $G_2$.

C. It is a subset of $G_1$ consisting of elements mapped to $e_2$.

D. It is a subset of $G_2$ consisting of elements mapped to the identity of $G_1$.

Answer: (C)

Explanation

The kernel is $\operatorname{Ker}f=\{x\in G_1\mid f(x)=e_2\}$, so it contains source elements whose images are the target identity. The image, in contrast, is made of attained elements in the target group.

18. For a morphism $f:G_1\to G_2$, which set is its image?

A. The target elements that have no preimage in $G_1$.

B. The source elements sent to the identity element of $G_2$.

C. The target elements attained as $f(x)$ for some $x\in G_1$.

D. The source elements that remain unchanged under the group operation.

Answer: (C)

Explanation

The image is $\operatorname{Im}f=\{y\in G_2\mid \exists x\in G_1,\,y=f(x)\}$, so it consists of target elements actually reached by f. Source elements mapped to the target identity form the kernel instead.

19. When is a group morphism f injective?

A. When its kernel equals the whole source group $G_1$.

B. When its kernel is the trivial subgroup $\{e_1\}$.

C. When its image contains the target identity but not other target elements.

D. When its image is the whole target group $G_2$.

Answer: (B)

Explanation

A group morphism is injective exactly when the only source element mapped to the target identity is the source identity, so $\operatorname{Ker}f=\{e_1\}$. Having image equal to $G_2$ characterizes surjectivity rather than injectivity.

20. Which combination of properties defines a ring with identity?

A. Multiplication forms an abelian group, addition is associative and distributive, and addition has an identity

B. Both operations form abelian groups, and each operation is distributive over the other

C. Addition forms an abelian group, multiplication is associative and distributive, and multiplication has an identity

D. Addition forms a group, multiplication is commutative, and every element has a multiplicative inverse

Answer: (C)

Explanation

A ring requires an abelian additive group, associative multiplication, distributive laws, and a multiplicative identity. Multiplication need not form a group, so requiring multiplicative inverses would impose a stronger structure.

21. Which equation correctly expresses the distributive law for multiplication in a ring?

A. $(x+y)\times z=x\times(y\times z)$

B. $x+(y\times z)=(x+y)\times(x+z)$

C. $x\times(y\times z)=(x\times y)\times z$

D. $x\times(y+z)=x\times y+x\times z$

Answer: (D)

Explanation

Distributivity connects multiplication with addition by expanding a product over a sum. The second equation describes associativity of multiplication, not distributivity.

22. For an arbitrary element $a$ in a ring, what is the value of $a\times0$?

A. $0$

B. $-a$

C. $1$

D. $a$

Answer: (A)

Explanation

Multiplication by the additive identity produces the additive identity, so $a\times0=0$ in every ring. Multiplication by the multiplicative identity gives $a$, which is a different rule.

23. If $a$ and $b$ are elements of a ring, which identity relates the product of their additive inverses to their original product?

A. $(-a)\times(-b)=a+b$

B. $(-a)\times(-b)=-(a\times b)$

C. $(-a)\times(-b)=-(a+b)$

D. $(-a)\times(-b)=a\times b$

Answer: (D)

Explanation

Changing the sign of both factors introduces two additive inverses, whose effects cancel in the product. A single sign change would produce the additive opposite of the original product, but two sign changes return $a\times b$.