Let $H$ be a normal subgroup of a finite group $G$ and $x \in G$. Show that if the greatest common divisor $(| x |, | H |) = 1$ , then $x \in H$.
Since $H$ is the normal subgroup of $G$ then we know that $xH = Hx$ for every $x \in G$. Thus, $xh \in Hx$ for all $h \in H$. Then $xh = h_{1}x$ for some $h_{1} \in H$. I want to show that $x \in H$. Is it enough to show that $H = xH = Hx$?
In general I have difficulty in applying the greatest common divisor. I know for $x^{|x|} = 1_{G}$ for some $x \in <x>$ and also $h^{|H|} = 1_{H} = 1_{G}$ for some $h \in H$. How to use the fact that the greatest common divisor is $1$?
As given, the statement is false. Let $G$ be the integers mod $6$, $x$ be $2$, and $H$ be the subgroup $\{0, 3\}$.