FALSE: every primitive group has a unique minimal normal subgroup

The diagonal subgroup is maximal in $T\times T$: if $\mathrm{\Delta }(T)<L$, an element $(a,b)\in L\setminus \mathrm{\Delta }(T)$ yields the nontrivial element $(1,b{a}^{-1})\in L$ after multiplication by $(a^{-1},a^{-1})$; its diagonal conjugates generate $1\times T$ by simplicity, and then $L=T\times T$. Hence the coset action is primitive. Its kernel is the core of $\mathrm{\Delta }(T)$. If $(t,t)$ lies in that core, conjugation by every $(x,1)$ gives $(xt{x}^{-1},t)\in \mathrm{\Delta }(T)$, so $t\in Z(T)=1$ because $T$ is nonabelian simple. Thus the action is faithful. Its two factors $T\times 1$ and $1\times T$ are distinct minimal normal subgroups.

By [L1], those two minimal normal subgroups are regular. So this primitive action has two distinct minimal normal subgroups, contradicting uniqueness. The corollary [L2] shows that this exceptional size is the largest possible.