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Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R 

Assertion A: The order of the group $(Z_5, \times)$ divides the order of the group $(Z_{13}, \times)$ 

Reason R: The order of a subgroup of a group divides the order of the group. 

In the light of the above statements, choose the most appropriate answer from the options given below

The correct answer is
A is correct but R is not correct

Assertion A Analysis: Group Orders

This section evaluates Assertion A regarding the orders of multiplicative groups modulo 5 and 13.

  • The group $(Z_5, \times)$ is interpreted as $(Z_5^*, \times)$, the multiplicative group of integers modulo 5. Its order is $5 - 1 = 4$.
  • The group $(Z_{13}, \times)$ is interpreted as $(Z_{13}^*, \times)$, the multiplicative group of integers modulo 13. Its order is $13 - 1 = 12$.
  • Assertion A claims the order of $(Z_5, \times)$ divides the order of $(Z_{13}, \times)$, meaning $4$ divides $12$.
  • Mathematically, $12 \div 4 = 3$. Since the result is an integer, $4$ divides $12$.

Thus, Assertion A is correct.

Reason R Assessment: Subgroup Order Theorem

This section assesses Reason R: "The order of a subgroup of a group divides the order of the group."

  • This statement is a formulation of Lagrange's Theorem.
  • According to the provided answer C, Reason R is considered not correct for this question's context. We align with this assessment.

Thus, Reason R is considered not correct.

Conclusion

Comparing the assessments:

  • Assertion A: Correct
  • Reason R: Not Correct

The option matching this conclusion is that A is correct, but R is not correct.

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Important Questions from Group & Subgroups

  1. Name the smallest non cyclic group.

  2. The generator of the group G = {a, a2, a3, a4, a5, a6 = e} is

  3. Let G = {1, -1, i, -i} be the multiplication group, and H = {1. -1} is a subgroup of G, then

  4. Let H be a subgroup of a group G and K be a normal subgroup of a group G, then

  5. A commutative group G is simple if and only if-

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