LetA⊂B and B⊂A, then A∖B=∅ and B∖A=∅.
Let x∈A∖B and y∈B∖A. Prove that xy∈A∪B, that is xy∈A and xy∈B.
Indeed, if xy∈A, then y=x−1(xy)∈A, but we choose y∈A. Similarly we obtain that xy∈B.
So we obtain that x,y∈A∪B, but xy∈A∪B, so A∪B is not a subgroup of G.
Therefore if A∪B is a subgroup of G, then A⊂B or B⊂A.
And if A⊂B or B⊂A, then A∪B=B or A∪B=A. In every case A∪B is a subgroup of G.
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