Solution If G contains element g of order n , then G contains cyclic subgroup <g> of order n .
Then ∣G∖<g>∣=∣G∣−∣<g>∣=n−n=0(G∖<g> is complement <g> in G ), that is G=<g> is a cyclic group. So if G is a non-cyclic group of order n , then G has no element of order n .
Z2⊕Z2 is a non-cyclic group with cyclic proper subgroups
{(0,0),(0,0),(0,1),(0,0),(1,0),(0,0),(1,1) }
Z2⊕Z2 does not have other proper subgroups, because
⟨(1,0),(0,1)⟩=⟨(1,0),(1,1)⟩=⟨(1,1),(0,1)⟩=Z2⊕Z2
So every proper subgroup of Z2⊕Z2 is a cyclic group.
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