Let $G$ be a group and let prime $p\mid \vert G\vert$. Suppose further that $\vert G\vert < p^2 $. Show that $G$ has a normal subgroup of order $p$. Suppose $G$ is a finite group and $p$ is a prime that divides $\vert G\vert$. Suppose further that $\vert G\vert < p^2 $. Show that $G$ has a normal subgroup of order $p$.
Attempt: We know there is an element of order $p$ by Cauchy's theorem. Say $x\in G$ has order $p$. Let $H$ be the cyclic subgroup generated by $x$.
My idea was to show that the only elements of order $p$ in $G$ are in $H$ and then, since conjugation preserves order of an element, $gHg^{-1} = H$.
However I am finding this difficult to prove. I would appreciate some help as to whether this is a sensible idea or if I should be doing something else?

Let $G$ act on the set of cosets $G/H$ by left multiplication.   You get a homomorphism $\varphi $ from $G$ into $S_{|G|/p}$.
The kernel is non-trivial since $|G|\not\mid (|G|/p)! $.  That's because $p\mid |G|$ and $|G|/p\lt p$ by hypothesis.
Furthermore, if $x\not\in H$, then $xH\not=H$, so $x\not\in\rm{ker}\varphi $.
Thus $\rm{ker}\varphi \le H$.
Thus, since $|H|=p$, $\rm{ker}\varphi =H$.
Thanks to @Arturo Magidin
