First we show that if for each , is a maximal subgroup, then is primitive. In particular, we prove the contrapositive that if is not a maximal subgroup for some then is not primitive. Let be a chain of proper subgroups. We claim that is a block. It's easy to see that if then . On the other hand, let , and suppose towards a contradiction that . Then there are such that and hence . But then , giving us , a contradiction. So is indeed a block. Furthermore, is nontrivial: for if then and are distinct elements of . But , because if it were then we would have for some and hence , which would contradict the fact that . Thus the first half of the proof is complete.
For the converse, let be a nontrivial block, and let be distinct. Then . We claim that both subgroup relations are proper: For let . Then there is with by the transitivity of , which means . So is a proper subgroup of . Also by the transitivity of there is with , which means . Since is a block with then it must be that , and hence . So is a proper subgroup of . We conclude that is not maximal, and this completes the second half of the proof.