First Lagrange's theorem says that if $G$ has a subgroup $H$ then order of $H$ divides order of $G$. But converse may not hold.
In your case $|G|=54=3^3\times2 $. Now by applying Sylow theorem, we see that there is a Sylow-$3$ subgroup of $G$ having order $27$, say $H$. Also, $n_p$, the number of such a subgroup satisfies the conditions, $n_p|o(G)$ and $n_p\equiv1\pmod3.$ In this case only possible value for $n_p$ is $1$. Hence $H$ is a unique Sylow-$3$ subgroup. So $G$ has a normal subgroup of order $27$. (Unique Sylow-$p$ subgroups are normal.)
(This answer assumes the knowledge of Sylow theorems.)
Also note that one can use the fact that a subgroup of index $2$ is normal. Which makes things more simpler.