If a prime $p$ divides the order of a group $G$ and $p^2 > |G|$, then there is a normal subgroup of order $p$ in $G$

I was wondering if there were a way to prove this without invoking the full force of the Sylow theorems. Here is my attempt:

Suppose $$G$$ is a group, a $$p$$ prime divides its order, and $$p^2 > |G|$$. By Cauchy's theorem, there exists a subgroup $$H$$ such that $$|H| = p$$. Suppose there exists another subgroup $$K$$, $$|K| = p$$, $$K \neq H$$. Since $$H$$ and $$K$$ are of prime order, any nontrivial element in each generates the entire subgroup, so we must have $$H \cap K = \{e\}$$. By counting argument, we have $$|HK| = \frac{|H| |K|}{|H \cap K|} = p^2 > |G|,$$ contradicting $$HK \subseteq G$$. Therefore, such a $$K$$ does not exist, and we conclude $$H$$ is the unique subgroup of $$G$$ of order $$p$$.

This uniqueness implies $$H \triangleleft G$$: for all $$g \in G$$, $$gHg^{-1}$$ is a subgroup of order $$p$$, so we must have $$gHg^{-1} = H, \forall g \in G$$.

Is there anything inconsistent in this proof? If not, are there even more elementary proofs?

Another way to prove it: Let $$|G|=pm$$ and $$H$$ be an order p subgroup. Then the action of $$G$$ on the cosets of $$H$$ gives a homomorphism from $$G$$ to $$S_m$$. The action is transitive on the $$m$$ cosets of $$H$$. Also, $$H$$ is in the kernel since p does not divide $$|S_m|$$. Since $$H$$ has index m, it must be the entire kernel, and thus normal