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Part CCSIR NET December 2025complex-conjugation-supplies-the-transposition-for-free-cauchys-theorem-then-supplies-the-p-cycle-together-they-force-g-equals-s-p

Complex conjugation supplies the transposition for free cauchys theorem then supplies the p cycle together they force g equals s p

Let p≥3 be a prime number and f(x)Q[x]f(x)\in\mathbb{Q}[x] an irreducible polynomial of degree p. Suppose that a1,,apCa_{1},\dots,a_{p}\in\mathbb{C} are the roots of f and that a1R,a2Ra_{1}\notin\mathbb{R}, a_{2}\notin\mathbb{R}, and aiRa_{i}\in\mathbb{R} for all 3≤i≤p. Let K=Q(a1,,ap)K=\mathbb{Q}(a_{1},\dots,a_{p}) be the subfield of C\mathbb{C} generated by the roots of f. Consider the Galois group G of K over Q\mathbb{Q} as a subgroup of SpS_{p}, the group of permutations of {a1,,apa_{1},\dots,a_{p}}. Which of the following statements are true?

  1. A.The transposition (1 2) belongs to G.
  2. B.|G| is divisible by p.
  3. C.A p-cycle belongs to G.
  4. D.G=SpG = S_{p}

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.

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Related counterexample: Every extension of degree n has a Galois group of order n

More on this topic

The chapter behind this: Galois correspondence and the standard groups — free to read

From Rings and FieldsGalois theory essentials

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