Part CCSIR NET June 2025an-abelian-group-can-have-a-thoroughly-non-abelian-automorphism-group

An abelian group can have a thoroughly non abelian automorphism group

For a group G, let Aut(G) denote the group (under composition) of all bijective group homomorphisms from G onto itself. Which of the following statements are true?

A.If $G_{1}, G_{2}$ are two groups such that Aut $(G_{1})$ is isomorphic to Aut $(G_{2})$ , then $G_{1}$ is isomorphic to $G_{2}$ .
B.If |G| = 2, then Aut(G × G) is abelian.
C.If G is the group of complex numbers under addition, then Aut(G) is abelian.
D.If G is finite, then Aut(G) is finite.✓

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

An abelian group can have a thoroughly non abelian automorphism group

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