In which condition will the union of two subgroups be a subgroup?
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I shall use multiplicative notation for the group law.
If HH or KK is the trivial subgroup {1}{1} consisting only of the multiplicative identity, the result is trivial. So, take H,KH,K to be subgroups of GG such that H,K≠{1}H,K≠{1}.
⇐⇐ If WLOG H⊂KH⊂K, then H∪K=KH∪K=K. Because KK is a subgroup of GG, H∪KH∪K is a subgroup of GG.
⇒⇒ Suppose for sake of contradiction that we have neither H⊂KH⊂K nor K⊂HK⊂H but that H∪KH∪K is a subgroup of GG. This implies that there exists h∈Hh∈H such that h∉Kh∉K; likewise, ∃k∈K∃k∈K such that k∉Hk∉H. Because H∪KH∪K is a subgroup of GG, hk∈H∪Khk∈H∪K, which implies hk∈H∨hk∈Khk∈H∨hk∈K. Say, WLOG, hk∈Hhk∈H. Well, since h∈Hh∈H, h−1∈Hh−1∈H. Thus, h−1hk=k∈Hh−1hk=k∈H, contradiction.
If HH or KK is the trivial subgroup {1}{1} consisting only of the multiplicative identity, the result is trivial. So, take H,KH,K to be subgroups of GG such that H,K≠{1}H,K≠{1}.
⇐⇐ If WLOG H⊂KH⊂K, then H∪K=KH∪K=K. Because KK is a subgroup of GG, H∪KH∪K is a subgroup of GG.
⇒⇒ Suppose for sake of contradiction that we have neither H⊂KH⊂K nor K⊂HK⊂H but that H∪KH∪K is a subgroup of GG. This implies that there exists h∈Hh∈H such that h∉Kh∉K; likewise, ∃k∈K∃k∈K such that k∉Hk∉H. Because H∪KH∪K is a subgroup of GG, hk∈H∪Khk∈H∪K, which implies hk∈H∨hk∈Khk∈H∨hk∈K. Say, WLOG, hk∈Hhk∈H. Well, since h∈Hh∈H, h−1∈Hh−1∈H. Thus, h−1hk=k∈Hh−1hk=k∈H, contradiction.
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