= 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 $\Delta$, set
$$
N=\{b:(1,b)\in H\}.
$$
Conjugation by diagonal elements makes $N$ normal in $T$. Multiplying $(a,b)\in H$ by $(a,a)^{-1}$ shows $(1,a^{-1}b)\in H$, so
$$
H=\{(a,b):a^{-1}b\in N\}.
$$
Conversely 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. Hence
$$
\boxed{\text{The action is faithful and transitive, and primitive iff }T
\text{ is nonabelian simple}.}
$$
This 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
$$
\boxed{n=60=p+1,\qquad G=A_5\times A_5,\qquad |G|=3600,\quad59\nmid3600.}
$$
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