Sylow Theorems not allowed alternative method to show $\lvert G\rvert = 20$ single subgroup $\lvert H\rvert = 5$ I know that I can use the Sylow Theorems but I have not been taught that in class. For instance I know using the Sylow Theorems that if $\lvert G\rvert = 20$ and $\lvert H\rvert = 5$ then there is a single subgroup of order $5$ and thus it is normal. But how is this shown without the Sylow Theorems?

Even simpler: assume that $H$ is not normal, then $H$ has a conjugate, say $K$, with $H \neq K$. Observe that $H \cap K=1$. But then for the set $|HK|=\frac{|H||K|}{|H \cap K|}=25$, contradicting $|G|=20$.
