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Assertion (A): A group can be isomorphic to its proper subgroup
Reason (R): The additive group Z of integer is isomorphic to (H, +) where
H = {mx : x ∈ Z and 0 ≠ m ∈ Z}
  • a)
    Both A and R true and R is the correct explanation of A
  • b)
    Both A and R are true but R is not the correct explanation of A
  • c)
     A is true, but R is false
  • d)
    A is false but R is true
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Assertion (A): A group can be isomorphic to its proper subgroupReason ...
This assertion is false.

Reason (R) is incorrect because the set H = {mx : x < 0,="" m="" is="" a="" positive="" integer}="" is="" not="" a="" proper="" subgroup="" of="" the="" additive="" group="" z="" of="" integers.="" />

Also, even if H were a proper subgroup, the assertion would still be false. A group cannot be isomorphic to its proper subgroup because an isomorphism preserves the group structure, and a proper subgroup has a different group structure than the original group.
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Community Answer
Assertion (A): A group can be isomorphic to its proper subgroupReason ...
But isn't H the proper subgroup of Z under addition operation?
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Assertion (A): A group can be isomorphic to its proper subgroupReason (R): The additive group Z of integer is isomorphic to (H, +) whereH = {mx : x ∈ Z and 0 ≠ m ∈ Z}a)Both A and R true and R is the correct explanation of Ab)Both A and R are true but R is not the correctexplanation of Ac)A is true, but R is falsed)A is false but R is trueCorrect answer is option 'B'. Can you explain this answer?
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