I have a question i am trying to prove that if $H<G$ and $\dfrac{|G|}{|H|}$ is a prime number then H is a maximal subgroup.
I prove this by contradiction, thus i assume that $\exists K : H<K<G$ and $K\neq H \neq G$.
I use Langrage's theorem to show that:
$\exists a \in \mathbb{N}$ : $|K|=a|H|$
$\exists b \in \mathbb{N}$ : $|G|=b|K|$
Thus $|G|=ab|H|\Leftrightarrow ab=\dfrac{|G|}{|H|}$ so $ab$ has to be prime.
Now I say that $a=1$ and $b=2$ but then $|K|=|H|$ and we knew that $H<K$ thus $K=H$ and this gives a contradiction is this correct?
Your solution is more or less right. You can streamline it by remarking that $$|G:H|=|G:K| \cdot |H:K|$$ So if $|G:H|$ is prime, then either one of the factors equals $1$, that is $G=K$ or $K=H$.