How to prove $x \in H$ How to prove that
Let H be a normal subgroup of a finite group G. If $\gcd(|x|, |G/H|)$ = 1,
show that $x \in H$.
 A: Hints:
1) For any $g\in G$, $ord(g)| |G|$ (divides)
2) If $\varphi:G_1\to G_2$ is a group homomorphism, then for all $g\in G$, $ord_{G_2}(\varphi(g))|ord_{G_1}(g)$. In particular this is true for the canonical homomorphism, $p:G\to G/H$
3) |G/H|=|G|/|H|
A: You can try to prove:


*

*The order of the class of $x$ in $G/H$ divides the order of $x$
in $G$,    

*The order of the class of $x$ in $G/H$ divides the
order of $G/H$

*The order of the class of $x$ in $G/H$ is $1$,
so that $x\in H$.

A: Consider the sequence $ x H, x^2 H, \dots, H$.  The length of this sequence divides $|G/H|$ because its a cyclic subgroup.  On the other hand it must also divide $|x|$ since $x^{|x|} H = H$ and the sequence repeats.  Therefore the length of the sequence is $1$ and so $x \in H$.
A: Let $n$ be the order of $x$ in $G$, and let $k$ be the order of $xH$ in $G/H$.
By Lagrange, $k$ divides $[G:H]=|G|/|H|=|G/H|$.
Now in $G/H$, we have $(xH)^n=x^nH=H$. So $k$ divides $n$.
Conclusion: $k$ divides $[G:H]$ and $n$ which are relatively prime, hence $k=1$. This means $xH=H$, i.e. $x$ belongs to $H$.
Note: the only nontrivial fact we used, besides Lagrange, is the following. In a group $\Gamma$, for every element $x$, the set $\{n\in\mathbb{Z}\;;\;x^n=e\}$ is an ideal in $\mathbb{Z}$, hence a principal ideal. We say that $x$ has finite order if this ideal is nontrivial. In this case, the order of $x$ is defined to be the positive generator of this ideal. That is the least positive integer $k$ such that $x^k=e$. Now what we used is obvious with this ideal structure 
$$k\mathbb{Z}= \{n\in\mathbb{Z}\;;\;x^n=e\}$$ 
in mind: if $x^n=e$ then $k$ divides $n$.
