 Solution

Use the displayed formula as a right group action: applying $(a,b)$ and then $(c,d)$ gives $(ac,bd)$. It is transitive since $(1,t)$ sends the identity to $t$. An element fixing every $t$ must have $a=b$ by evaluating at the identity, and then $a^{-1}ta=t$ for every $t$. Its kernel is therefore $\{(z,z):z\in Z(T)\}$, which is trivial by the center of a group hypothesis.

The identity stabilizer subgroup is $\Delta =\{(a,a):a\in T\}$. For every subgroup $H$ containing $\Delta$, setConjugation by diagonal elements makes $N$ normal in $T$. Multiplying $(a,b)\in H$ by $(a,a{)}^{-1}$ shows $(1,a^{-1}b)\in H$, soConversely each normal subgroup $N$ gives such an $H$. Thus the subgroups strictly between $\Delta$ and $T\times T$ correspond exactly to proper nontrivial normal subgroups of $T$.

For a transitive action, blocks containing a chosen point correspond to subgroups containing its stabilizer subgroup: a block is the orbit of that point under its setwise stabilizer subgroup, and conversely the orbit under an intermediate subgroup gives a block. Therefore the diagonal action on a centerless group is primitive exactly when $T$ is simple. A nontrivial centerless simple group cannot be abelian, because an abelian group's center of a group is the whole group. HenceThis uses the usual nontrivial degree convention. If one allows a one-point action to count as primitive, $T=1$ is the additional degenerate exception.

For the requested order example use $T=A_5$. Its simplicity follows from its conjugacy-class sizes $1,15,20,12,12$: a normal subgroup is a union of classes containing the identity, and no proper nontrivial such sum divides 60. Set $p=59$, a prime. The primitive diagonal action has