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?

Consider the action of $G$ on the set of cosets $G/H$ by left multiplication. This induces a homomorphism
$$\phi:G\to \mathrm{Sym}(G/H)\cong S_4$$
As $|S_4|=24$, $S_4$ has no subgroups of order $20=|G|$. This means that $\ker\phi$ is nontrivial. Note that $\ker\phi\leq H$ since any $g\in \ker\phi$ must satisfy $gH=H$ (i.e. $g\in H$). Since $|H|=5$ and $\ker\phi\neq\{1\}$, we must have $\ker\phi=H$. This proves that $H$ is normal.
