If G conains element g of order n, then G conains 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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If A and B are two sets such that AUB=Fie, then A intersection B=fie