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?
 A: There are already some good answers in the comments. Here is a less conceptual approach based on counting. If there are two distinct subgroups $H_1, H_2$ of order $p$, then $H_1\cap H_2=\{1\}$, since any nonidentity element of $H_i$ has order $p$ hence must be a generator of $H_i$. Now it's easy to check that $H_1\times H_2\rightarrow H_1H_2$ defined by $(h_1, h_2)\mapsto h_1h_2$ is an injection (or to cite the counting formula $|H_1H_2|=\frac{|H_1||H_2|}{|H_1\cap H_2|}$). Therefore $G$ has at least $p^2$ elements.
A: Let $x\in G$ be an element of order $p$. Let $y\neq 1$ be an arbitrary element in $G$ of order $p$. If $y$ is an element of $\langle x\rangle$ then we are good; if not, then $\langle x\rangle\cap\langle y\rangle = \{1\}$. Recall that $|G| = kp$ with $k < p$, so if we consider the cosets
$$\langle x\rangle, y\langle x\rangle, y^2\langle x\rangle, \cdots, y^{p-1}\langle x\rangle$$
there exists two identical cosets,  namely $y^a\langle x\rangle = y^b\langle x\rangle$ (here $0\leq a < b\leq p-1$). This gives $\langle x\rangle = y^{b-a}\langle x\rangle$, contradicts with the condition $\langle x\rangle \cap\langle y\rangle = \{1\}$.
So every nontrivial element of order $p$ is contained in $\langle x\rangle$, which is what you want to obtain.
A: 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
