COMEDK · Maths · 17. Mathematical Reasoning

Identify the false statement.

A A non-empty subset \(H\) of group \(G\) is a subgroup of \(G\) if and only if for every a, \(b \in H \rightarrow a * b^{-1} \in H\)
B The intersection of two subgroups of a group \(G\) is again a subgroup.
C A group of order three is not abelian
D If in a group \(F,(a b)^{2}=a^{2} b^{2}, \forall a, b \in G\) then \(G\) is abelian

Verified Solution

Answer & Solution

Correct Answer

(C) A group of order three is not abelian

Step-by-step Solution

Detailed explanation

It is a false option because every group of order 3 is abelian because it is closed as well.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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