Part CCSIR NET June 2025normality-is-not-transitive-and-prime-index-only-forces-it-for-the-smallest-prime

Normality is not transitive and prime index only forces it for the smallest prime

Let $G_{1}$ and $G_{2}$ be subgroups of a group G. Which of the following statements are true?

A.If $G_{1}$ is normal in G, then $(G_{2} G_{1}) / G_{1} ≅ G_{2} / (G_{2} \cap G_{1})$ .✓
B.If $H_{1}$ and $H_{2}$ are normal subgroups of $G_{1}$ and $G_{2}$ , respectively, then $(G_{1} \times G_{2}) / (H_{1} \times H_{2}) ≅ (G_{1} / H_{1}) \times (G_{2} / H_{2})$ .✓
C.If $G_{1}$ is normal in $G_{2}$ and $G_{2}$ is normal in G, then $G_{1}$ is normal in G.
D.Every subgroup of prime index in G is normal.

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 property not inherited.

See pricing

50 are analysed free — try those first.

The trap it tests

Property not inherited

A property assumed to pass to subobjects, quotients, or through a chain. It does not.

Drill statements like this

Related counterexample: Converse of Lagrange: d | |G| ⇒ subgroup of order d

Normality is not transitive and prime index only forces it for the smallest prime

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