Let $K \lhd G$. Prove are equivalent:
a) $K\le H \le G \Rightarrow H=K $ or $H = G$
b) $|G/K| = p$ with $p$ prime
This is what I know so far about the problem
$K \lhd G$ (stands for '$K$ is normal subgroup of $G$') means that $gKg^{-1}= K, \forall g \in G$
A (left) lateral class of $H \le G$ has the form $xH$ with $x\in G$.
Actually the lateral classes of $H$ forms a partition in $G$.
$|G/K|$ is the order of $G/K$, which is the same as the cardinal or size of $G/K$.
$G/K$ is the set of all lateral classes of $K \le G$. When $K \lhd G$ then $G/K$ is a group.
This is from a homework.
I guess I can assume $G$ is finite. If that is the case, I know that $|G/K| = |G|/|K|$
but also, I guess I have to use the next theorem : let $G$ be a group (not neccesarily finite), and $H,K \le G$ then $|HK| = [H : H\cap K] |K|$. If $G$ is finite then $|HK||H \cap K| = |H||K|$
