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Part CCSIR NET December 2024the-congruence-q-equals-one-mod-p-is-what-breaks-it

The congruence q equals one mod p is what breaks it

For a finite group G, let S(G) denote the set of all its Sylow subgroups. A finite group G is said to have property (J) if G is isomorphic to the direct product ∏(H∈S(G))H\prod(H\in{}S(G)) H. Which of the following statements are true?

  1. A.Any group with p(p + 2) elements, where p and p + 2 are both prime numbers, has property (J).✓ Correct
  2. B.Any finite abelian group has property (J).✓ Correct
  3. C.The symmetric group S3S_{3} has property (J).
  4. D.Any group with 77 elements has property (J).✓ Correct

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 standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: A group of order pq (p < q) is always cyclic

More on this topic

The chapter behind this: Sylow theorems and classifying small groups — free to read

From Groups › Sylow theorems and groups of small order

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