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Part BCSIR NET December 2025the-two-orderings-of-the-double-count-are-equal-not-one-bigger

The two orderings of the double count are equal not one bigger

Let X = {1, …, 17} and S17S_{17} be the group of permutations of X. For a subgroup G of S17S_{17} and x ∈ X, let S_G(x) = {σ∈G\sigma \in G ∣ σ(x)=x\sigma(x) = x}. Which of the following statements is true for every subgroup G of S17S_{17}?

  1. A.The number of pairs (σ,x)∈G×X(\sigma, x) \in G \times X such that σ(x)=x\sigma(x) = x is strictly greater than ∑x\sum_{x} |S_G(x)|.
  2. B.The number (1/|G|)⋅∑x)\cdot\sum_{x} |S_G(x)| is always an integer.✓
  3. C.For all x ∈ X, S_G(x) is a normal subgroup of G.
  4. D.For all x, y ∈ X, S_G(x) is isomorphic to S_G(y).

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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50 are analysed free — try those first.

Related counterexample: Every group of order p² is cyclic

More on this topic

The chapter behind this: Group actions, the class equation and p-groups — free to read

From Groups › Group actions, class equation, p-groups

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