Conditions on H and K if H ∪ K is a subgroup

  • #1
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Homework Statement


Let H and K be subgroups of G. Prove that if ##H \cup K## is a subgroup of ##G## then ##H \subseteq K## or ##K \subseteq H##

Homework Equations




The Attempt at a Solution


Suppose that ##H \cup K \le G##. For contradiction, suppose that neither H nor K is a subset of the other. This implies that there exists an ##h \in H## that is not in ##K## and a ##k \in K## that is not in ##H##. Since ##H## and ##K## are individually subgroups, ##h^{-1} \in H## and ##k^{-1} \in K##. Now since ##h \in H \cup K## and ##k \in H \cup K##, we know that ##hk \in H \cup K##, since we supposed it's a subgroup. We have three cases to check: If ##hk \in (H \cap K)##, then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - H## then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - K## then ##h^{-1}(hk) = k \in H##, a contradiction.
 

Answers and Replies

  • #2
14,413
11,726

Homework Statement


Let H and K be subgroups of G. Prove that if ##H \cup K## is a subgroup of ##G## then ##H \subseteq K## or ##K \subseteq H##

Homework Equations




The Attempt at a Solution


Suppose that ##H \cup K \le G##. For contradiction, suppose that neither H nor K is a subset of the other. This implies that there exists an ##h \in H## that is not in ##K## and a ##k \in K## that is not in ##H##. Since ##H## and ##K## are individually subgroups, ##h^{-1} \in H## and ##k^{-1} \in K##. Now since ##h \in H \cup K## and ##k \in H \cup K##, we know that ##hk \in H \cup K##, since we supposed it's a subgroup.
I wouldn't say suppose here as it sounds as if it could be wrong. It is our given condition, i.e. we accept it as given truth. "... since ##H \cup K## is a subgroup." would be better.
We have three cases to check: If ##hk \in (H \cap K)##, then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - H## then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - K## then ##h^{-1}(hk) = k \in H##, a contradiction.
I think you don't need the cases. Just look at ##hk##. It's either in ##H## in which case ##k## is as well, or vice versa for the other possibility. You can say w.l.o.g. ##hk \in H## (the problem is symmetric in ##H## and ##K##) and have only one case to consider.
 
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