Part CCSIR NET June 2025in-an-abelian-quotient-every-subgroup-is-its-own-core

In an abelian quotient every subgroup is its own core

Let G be a group, H a subgroup of G, and T = {gH | g ∈ G}, the set of all left cosets of H in G. Let S_T be the set of all permutations of T and $\pi : G \to S_T$ be the map defined by $\pi(g)(g_{1}H) =$ gg $_{1}H$ . For a prime number p, let 𝔽_p denote the field with p elements. In which of the following cases is $\ker \pi$ trivial?

A.G = GL $_{2}($ 𝔽_p) and H is a subgroup of order p.✓ Correct
B.G = SL $_{2}($ 𝔽_p) and H is a subgroup of order p.✓ Correct
C.p ≡ 3 (mod 4), G = GL $_{2}($ 𝔽_p)/SL $_{2}($ 𝔽_p) and H is a subgroup of order 2.
D.p ≡ 1 (mod 4), G = GL $_{2}($ 𝔽_p)/SL $_{2}($ 𝔽_p) and H is a subgroup of order 2.

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.

See pricing

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: Every group of order p² is cyclic

In an abelian quotient every subgroup is its own core

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