Can A ⊎ B, where |A| ≠ |B|, be a complete graph? There is a lemma about Hamiltonian cycles which states:
If G = (A ⊎ B| E) is a bipartite graph, |A| ≠ |B|, then there is no Hamiltonian cycle in G.
A union of disjoint graphs (as far as I know that is what ⊎ implies), cannot contain a Hamiltonian cycle, because the graph wouldn't be connected. If so, how is the graph being bipartite and |A| ≠ |B| even relevant?