$G$ is a group and $H \leq G$ with $|G:H|=3$. Show that $x$ is an element of $H$ if $x \in G$ with $|x|=7$. Hint: let $\langle x \rangle$ act on $G/H$ by left multiplication and look at the orbits.
Since the index of $H$ in $G$ is $3$ we can name the elements of $G/H$:
$1H=H$, $g_1H$ and $g_2H$
Except for the neutral element all the elements of $\langle x \rangle$ have order $7$ since $7$ is a prime.
$x^n \cdot H$ is a subset of one of $H$, $g_1H$ or $g_2H$
$x^n \cdot g_1H$ is a subset of one of $H$, $g_1H$ or $g_2H$
$x^n \cdot g_2H$ is a subset of one of $H$, $g_1H$ or $g_2H$
So far for my thoughts. I don't know how I can conclude anything about the orbits let alone conclude that $x$ is an element of $H$. How to proceed?
The action mentioned in the hint gives a homomorphism $\phi: \langle x \rangle \to S_3$. We must have $\phi(x)=1$, because the order of $\phi(x)$ must divide both $7$ and $3!=6$, which are coprime. In other words, $x$ acts on the cosets of $H$ like the identity, and so $xH=H$. This can only happen if $x \in H$.