How can I Prove that H=G or K=G

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Let subgroups $H,K \subseteq G$ such that $G=H \cup K$

Prove $H=G$ or $K=G$

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In the special case when $G$ is a finite group, it is also possible to use a counting argument. (This is less general and a bit more complicated than the other answers, but I find the method interesting.)

Suppose $H$ and $K$ are both proper subgroups of $G$. Then $|H|$ and $|K|$ are proper divisors of $|G|$. Therefore, $$\begin{align}|G|=|H\cup K| &= |(H\setminus\{1\})\cup(K\setminus\{1\})\cup\{1\}|\\&\leq|H\setminus\{1\}|+|K\setminus\{1\}|+|\{1\}|\\&\leq\left(\frac{|G|}{2}-1\right)+\left(\frac{|G|}{2}-1\right)+1\\&=|G|-1,\end{align}$$ a contradiction.

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Proof that for any 2 subgroups $G_1,G_2$ of $G$: $G_1\cup G_2$ is a subgroup iff $G_1\subset G_2$ or $G_2\subset G_1$.

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Take an element of $H$ that's not in $K$. Take an element of $K$ that's not in $H$. Multiply them. Where is that product?