Baer's Theorem: Let $x$ be a $p$-element of a finite group $G$. Suppose that $\langle x,x^g\rangle$ is a $p$-subgroup for every $g\in G$. Then $x\in O_p(G)$.
Here, $O_p(G)$ denotes the largest normal $p$-subgroup of $G$.
My attempt: Since $\langle x,x^g\rangle$ is a $p$-subgroup, it is also nilpotent for all $g\in G$. By definition, all subgroups of a nilpotent group are subnormal. In particular, $\langle x\rangle$ is subnormal in $\langle x,x^g\rangle$ for all $g\in G$. Therefore, by Wielandt's theorem, $\langle x\rangle$ is subnormal in $G$. Also, as $x$ is a $p$-element, $\langle x\rangle$ is a $p$-subgroup.
I don't know how to proceed from here. To show that $x\in O_p(G)$, I'm left to show that $\langle x\rangle$ is normal in $G$. However, I could only show subnormality.
So, let $H$ be subnormal in $G$. If $H$ is normal, then we're done. Otherwise, there exists a proper normal subgroup of $K$ of $G$ such that $H$ is subnormal in $K$. Thus, $|K|<|G|$ and so by induction, $H\subseteq O_p(K)$. Now, $O_p(K)$ is characteristic in $K$ which is normal in $G$, so $O_p(K)$ is a $p$-subgroup which is normal in $G$. Thus, $O_p(K)\subseteq O_p(G)$. It follows that $H\subseteq O_p(G)$.