$H$ is a normal subgroup of $G$ which $[G:H]=n$. Show that for all $g \in G$ which $(o(g), n)= 1$ and $o(g)<\infty$ we can conclude $g \in H$. $H$ is a normal subgroup of $G$ which $[G:H]=n$. Show that for all $g \in G$ which $(o(g), n)= 1$ and $o(g)<\infty$ we can conclude $g \in H$.
Now Suppose $H$ is not normal, Can we conclude the same ($g \in H$)?
I'm stuck in using $(o(g), n)= 1$, How should I use it in the proof?
 A: For not normal subgroups the statement does not hold. Take $S_3$ as the group $G$ and a subgroup of order $2$ as $H$.
For normal subgroups the statement is true. We have $o(g)u+nv=1$ for some integers $u,v$. Then $g=g^{o(g)u} \cdot g^{nv}=(g^n)^v$. Since $G/H=n$ we have $g^n\in H$. Hence $g\in H$.
A: Outline:
Note that $o(gH)$ divides $o(g)$ (?)  and $\gcd(o(g), n)=1$, so we must have $\color{red}{\gcd(o(gH), n)=1}$. Also $gH$ is an element of a group $G/H$, so $o(gH)$ divides $o(G/H)=n$
So $o(gH)$ must be...? and hence.......
A: 
Proposition Let $H$ be a non-trivial subgroup of the finite group $G$, with $n = [G:H]$. Assume that $\gcd(|H|,n)=1$. Then the following are equivalent.
(a) For all $g \in G$: $g^n \in H$.
(b) $H \unlhd G$.

Proof (b)$\Rightarrow$(a) is trivial by Lagrange's Theorem. So let us prove (a)$\Rightarrow$(b) (Sketch) We are going to use induction on $|G|$. To start the induction, we argue that $\operatorname{core}_G(H) \neq \{1\}$. For suppose $\operatorname{core}_G(H) =\{1\}$ and pick $g\in G$ and $h\in H$. By the assumption (a) $(g^{-1}hg)^n=g^{-1}h^ng  \in H$, so $h^n \in H^{g^{-1}}$. We conclude that $h^n \in \operatorname{core}_G(H)$, hence $h^n=1$ and the order of $h$ must divide $n$. But the order also divides $|H|$ and since $\gcd(|H|,n)=1$, we conclude $h=1$. But $h$ was arbitrary, so $H$ must be trivial, which contradicts the assumption. If $H$ is normal there is nothing to prove, so we can safely assume that $\operatorname{core}_G(H)$ is a proper subgroup of $H$. Now write $\bar{G}$ for $G/\operatorname{core}_G(H)$ and $\bar {H}$ for $H/\operatorname{core}_G(H)$, then $\bar {G}$  and $\bar {H}$ satisfy all the conditions of the proposition. By induction we get $\bar {H} \unlhd \bar {G}$, and this implies $H \unlhd G$.

This provides you a slew of counterexamples: just take a group $G$ with a Sylow $p$-subgroup $H$ that is *not* normal.
A: Consider $[g]\in G/H$, the image of $g$ in the quotient $G/H$.
Its order $o=o([g])$ must divide $o(g)$ and must also divide $n=|G/H|$. So it also divides $(n,o(g))=1$.
Thus, $[g]$ has order 1 and is equal to the identity in $G/H$. It follows that $g\in H$.
