Union of proper subgroups

• A.WrongStandard counterexample

  False.

  Any cyclic group refutes it: a generator lies in no proper subgroup

  so the proper subgroups cannot cover G.

  For $\mathbb{Z}/6$, SymPy gives the divisors 1, 2, 3, 6 — so the proper subgroups have orders 1, 2, 3, and their union misses the totient(6) = 2 generators, namely 1 and 5.

• B.WrongStandard counterexample

  False, and $\mathbb{Z}/6$ is the counterexample the option seems designed to survive.

  |$\mathbb{Z}/6$| = 6 has two distinct prime divisors

  yet $\mathbb{Z}/6$ is cyclic and so is not a union of proper subgroups.

  Having several prime factors says nothing about being cyclic.

• C.WrongStandard counterexample

  False.

  $\mathbb{Z}/p$ is abelian and cyclic

  so it is not a union of proper subgroups — in fact it has none besides the trivial one.

  Abelian is much weaker than non-cyclic.

• D.Correct

  True, and it is an iff so both directions matter.

  If G is cyclic, a generator belongs to no proper subgroup

  so the union cannot be all of G.

  If G is not cyclic, then every element g generates a proper subgroup ⟨g⟩, and G is the union of these.

  The trigger: 'union of proper subgroups' is precisely 'no single element generates'

  which is precisely 'not cyclic'.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.