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

[Mr Davis 97]

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.

 

[fresh_42]

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.