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Let $H$ and $K$ denote two subgroups of a group $G$. Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$.

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    $\begingroup$ Hint: Since $gcd(s,O(n))=1$ we can find integer $a$ and $b$ such that $as+bO(n)=1$. $\endgroup$
    – Siming Tu
    Oct 22, 2014 at 21:24
  • $\begingroup$ This is essentially the same question as math.stackexchange.com/questions/334405 $\endgroup$
    – Derek Holt
    Oct 22, 2014 at 23:42

1 Answer 1

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I shall use multiplicative notation for the group law.

If $H$ or $K$ is the trivial subgroup $\{1\}$ consisting only of the multiplicative identity, the result is trivial. So, take $H, K$ to be subgroups of $G$ such that $H,K \neq \{1\}$.

$\Leftarrow$ If WLOG $H \subset K$, then $H \cup K=K$. Because $K$ is a subgroup of $G$, $H \cup K$ is a subgroup of $G$.

$\Rightarrow$ Suppose for sake of contradiction that we have neither $H \subset K$ nor $K \subset H$ but that $H \cup K$ is a subgroup of $G$. This implies that there exists $h \in H$ such that $h \notin K$; likewise, $\exists k \in K$ such that $k \notin H$. Because $H \cup K$ is a subgroup of $G$, $hk \in H \cup K$, which implies $hk \in H \vee hk \in K$. Say, WLOG, $hk \in H$. Well, since $h \in H$, $h^{-1} \in H$. Thus, $h^{-1}hk=k \in H$, contradiction.

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