Proof:

To prove that a finite group G is the union of proper subgroups if and only if G is non-cyclic, we need to show both directions of the statement.

Direction 1: If G is the union of proper subgroups, then G is non-cyclic.

Assume that G is the union of proper subgroups. We will prove by contradiction that G is non-cyclic.

Suppose G is cyclic. Then there exists an element g in G such that every element in G can be expressed as a power of g. Let's denote this element g as the generator of G.

Since G is finite, the order of the generator g, denoted as |g|, must be a positive integer. Let |g| = n.

Consider the subgroup H generated by g^k, where k is a positive integer such that 1 < k="" />< n.="" since="" k="" />< n,="" g^k="" is="" a="" proper="" power="" of="" />

If G is cyclic, then every element in G can be expressed as a power of g. In particular, every element in H can be expressed as a power of g^k. This implies that H is a subgroup of G.

However, H is a proper subgroup of G because g^k is a proper power of g. This contradicts the assumption that G is the union of proper subgroups, as H is a proper subgroup not included in the union.

Therefore, our assumption that G is cyclic must be false, and G is non-cyclic.

Direction 2: If G is non-cyclic, then G is the union of proper subgroups.

To prove this direction, we need to show that for a non-cyclic finite group G, it can be expressed as the union of proper subgroups.

Since G is non-cyclic, there does not exist a generator g in G such that every element in G can be expressed as a power of g.

Consider the set of proper subgroups of G, denoted as S = {H1, H2, ..., Hn}. Each Hi in S is a proper subgroup of G, which means it does not contain all elements of G.

Now, let's define a new subgroup K as the union of all proper subgroups in S:

K = H1 ∪ H2 ∪ ... ∪ Hn.

Since each Hi is a proper subgroup of G, K is also a proper subgroup of G because it does not contain all elements of G.

We claim that K = G, i.e., every element in G is also in K.

If K = G, then G is the union of proper subgroups, which completes this direction of the proof.

To prove the claim, consider an arbitrary element x in G. Since G is non-cyclic, x cannot be expressed as a power of any generator in G.

Therefore, x is not in any proper subgroup Hi, and consequently, x is in K.

Hence, our claim is true, and G is the union of proper subgroups.

Conclusion:

From both directions of the proof, we can conclude that a finite group G is the union of proper subgroups if and only if G is non-cyclic.