Algebra

Step 1: Understanding the Concept:
This question relates to the structure of finite rings. An integral domain is a commutative ring with a multiplicative identity (unity) and no zero-divisors (if , then or ). A key theorem in abstract algebra connects finite integral domains to fields.

Step 2: Key Formula or Approach:
The fundamental theorems we need are: 1. Every finite integral domain is a field. 2. The order (or cardinality) of any finite field must be a power of a prime number, i.e., for some prime and integer .
Therefore, the cardinality of a finite integral domain must be a prime power. We need to check which of the given numbers are not prime powers.

Step 3: Detailed Explanation:
Let's analyze the given cardinalities:
A. 5: 5 is a prime number, so it can be written as . This is a prime power. Thus, an integral domain of cardinality 5 is possible (it would be the field ).
B. 6: 6 can be factored as . This is not a power of a single prime. Therefore, no integral domain (or field) of cardinality 6 can exist.
C. 7: 7 is a prime number, so it can be written as . This is a prime power. An integral domain of cardinality 7 is possible (the field ).
D. 10: 10 can be factored as . This is not a power of a single prime. Therefore, no integral domain of cardinality 10 can exist. The question asks for which cardinalities are not possible. Based on our analysis, these are 6 and 10.

Step 4: Final Answer:
The cardinalities that are not possible for an integral domain are 6 (B) and 10 (D).

Step 1: Understanding the Concept:
A mathematical structure forms a group if it satisfies four axioms: 1. Closure: For all , . 2. Associativity: For all , . 3. Identity Element: There exists an element such that for all , . 4. Inverse Element: For each , there exists an element such that .

Step 2: Detailed Explanation:
Let's check each option:
(A) , where : - Associativity: . And . It is associative. - Identity: We need an element such that and . The first part, , is always true. The second part requires . This must hold for all , but cannot be equal to all . No unique identity element exists. Not a group.
(B) , where : - Associativity: . And . Since in general, it is not associative. Not a group.
(C) , where : - Closure: If , then . Closure holds. - Associativity: . And . It is associative. - Identity: We need , so , which gives . Let's check: . The identity element is , which is in . - Inverse: We need , so , which gives . Since , is also in . An inverse exists for every element. All axioms are satisfied. This is a group.
(D) , where : - Identity: We need and . The first part: . This identity element depends on , so there is no unique identity element for the whole set. Not a group.
Step 3: Final Answer:
The structure with is the only one that forms a group.