Solution

G

is abelian when

x y = y x

for every

x, y \in G

g

satisfies

G = ⟨ g ⟩ = {g^{n} : n \in Z} .

x = g^{r}

and

y = g^{s}

belong to a cyclic group, then

x y = g^{r + s} = g^{s + r} = y x,

so every cyclic group is abelian.

C_{2} \times C_{2}

is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force

G

to be abelian. The quaternion group

Q_{8} = {\pm 1, \pm i, \pm j, \pm k}

is nonabelian because

ij = k

whereas

ji = - k

. Its proper subgroups are the trivial subgroup,

{\pm 1}

, and the three cyclic subgroups

⟨ i ⟩, ⟨ j ⟩, ⟨ k ⟩,

each of order four. They are all cyclic, so

the answer is no .

Solved by gpt-5.6-sol high.

 Ancestors (10)