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Part BCSIR NET December 2025inverse-pairing-forces-h-g-odd-always-then-hunt-for-a-single-witness-not-a-universal-pattern

Inverse pairing forces h g odd always then hunt for a single witness not a universal pattern

For a finite group G, let H_G = {g ∈ G ∣ g15=eg^{15} = e}. Which of the following statements is necessarily true?

  1. A.There exists a finite group G such that |H_G| is even.
  2. B.There exists a finite group G such that |H_G| = 4n+1 for some n ≥ 3.✓
  3. C.For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+1.
  4. D.For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+3.

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.

See pricing

50 are analysed free — try those first.

Related counterexample: If d divides |G| then G has a subgroup of order d

More on this topic

The chapter behind this: Lagrange, orders and cyclic groups — free to read

From Groups › Subgroups, cosets, Lagrange, cyclic groups

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