The question is
Let $G$ be a group such that $|G| = pk$, where $p$ is a prime, $k < p$. Then $G$ contains a normal subgroup of order $p$.
It is easy to use Cauchy's Theorem to see that there exists a subgroup $H$ of $G$ with order $p$, but how to continue from here? Thanks.
Let $n_p$ be the number of subgroups of order $p$. Then by Sylow's theorem $$n_p | k$$ and$$n_p\equiv1 \ mod \ p$$ As $k<p$, therefore $n_p =1$. Hence, the group is normal.
Edit You mentioned you don't know Sylow's so I'm adding a different proof. From future please mention what you know about the subject in the question itself.
Let $H$ be a subgroup of order $p$, $g \in G$ and $K=gHg^{-1}$. Then $|K|=p$. Now $$|HK|=\frac{|H||K|}{|H \cap K|}$$ If $|H \cap K|= 1$, then $|HK|=p^2$. But $|G| < p^2$, therefore $|H \cap K|\neq 1$. This forces $|H \cap K|= p$ which shows $H =K$. As $g \in G$ was arbitrary, therefore $H$ is normal.
Don't know where you are in learning group theory, but I would consider the Sylow Theorems.
suppose $H=<h>$ and $S=<s>$ are distinct subgroups of order $p$. consider the complex $HS$. since $$ |HS| \le pk \lt p^2 $$ there must be indices $a\ne b,c \ne d$ in $[0,p-1] \cap \mathbb{N}$ such that: $$ h^as^c = h^bs^d $$ but this gives $$ h^{a-b} = s^{c-d} $$ from which it follows (since $a-b$ is invertible in $\mathbb{Z}_p$) that $H$ and $S$ are the same subgroup, counter to our assumption.
since the subgroup of order $p$ guaranteed by Cauchy's theorem is unique, it is normal. can you show that it is in fact a characteristic subgroup?
