Let $G$ be a finite group with a subgroup $H$ of index $5$. Also $5$ is the smallest prime divisor of $|H|$. Let $X=\{gH:g\in G\}$ be the set of left cosets of $H$ in $G$, $H$ acts on $X$ by left multiplication.
Every orbit of $X$ has length $1$ (See here). Using this fact that every orbit has length 1, show that $hg\in gH$ for $h\in H$, and $g\in G$.
I'm not sure how I would show this, I tried by assuming it was normal and therefore $gH=Hg$ but I think this is wrong because I don't think I can assume that orbits of length 1 means it's normal?
Let $gH$ be one coset in $X$. Since it's orbit has only one element (i.e., one coset), you just need to figure out which coset it is. You can show fairly easily that $gH$ must be in its own orbit, so the orbit of $gH$ is just $\{gH\}$. Then you know that for any $h\in H$, $h(gH)$ must be in the orbit of $gH$, which means that $h(gH)=gH$. But $h(gH)=\{hgk|k\in H\}$, so you can use that to deduce that $hg\in gH$.