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Part CCSIR NET December 2025the-subgroup-at-index-equal-to-the-smallest-prime-divisor-is-automatically-normal-not-just-at-index-2

The subgroup at index equal to the smallest prime divisor is automatically normal not just at index 2

Let G be a finite non-abelian group. Which of the following statements are necessarily true?

  1. A.If d is a positive integer that divides |G|, then G has a subgroup of order d.
  2. B.The map f : G×G → G given by f(a,b)=ab is not a group homomorphism.✓ Correct
  3. C.Suppose that for every positive integer d that divides |G|, there exists a subgroup of G of order d. Then G has at least three normal subgroups.✓ Correct
  4. D.|G| ≠ 16

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests property not inherited.

See pricing

50 are analysed free — try those first.

The trap it tests

Property not inherited

A property assumed to pass to subobjects, quotients, or through a chain. It does not.

Drill statements like this

Related counterexample: Converse of Lagrange: d | |G| ⇒ subgroup of order d

More on this topic

The chapter behind this: Normal subgroups, quotients and the isomorphism theorems — free to read

From Groups › Normal subgroups, quotients, isomorphism theorems

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