Let act transitively on the set . A block is a nonempty subset such that for all either or .
 Prove that if is a block containing and we define then .
 Prove that if is a block then is a partition of .
 A (transitive) group is called primitive if the only blocks in are the trivial ones the singletons and itself. Prove that is primitive on . Prove that is not primitive as a permutation group on the four vertices of a square.
 Let act transitively on . Prove that is primitive on if and only if for all , is maximal in .
 Note that is not empty since . Now let . Note that , so that . Thus . By the subgroup criterion, .
Now suppose . We have , so that . Hence is nonempty, and we have since is a block. Thus . By a previous exercise, .
 First we show that . The direction is clear. : Let and . Since the action of is transitive, there exists such that . Then , so that .
Now suppose . Then there exist such that . Now , so that is not empty. Thus , and we have . So elements of are pairwise disjoint; hence is a partition of .

 Let be a proper nonempty subset. Then there exist elements and . Suppose is a block. Consider ; since , we have . Let and be the remaining elements of . If , then , a contradiction; similarly for . Thus . Now clearly itself is a block, and any proper block must be a singleton. Thus this action of is primitive.
 We saw previously that , where are labels on the vertices of a square (written clockwise). Consider the set . We see that
So that is a nontrivial block; hence this action is not primitive.
 We begin with some lemmas.
Lemma 1: Let act transitively on and let be a block under the action. Then if and only if . Proof: The direction is clear. Suppose is a proper subset and let , . Now and , so that . Since is a block, we have . But since and , a contradiction. So is not proper, and we have .
Now we move to the main result.
Suppose is primitive on ; then the only blocks of are and the singletons. Now let and let be a subgroup with . We consider the set .
We claim that is a block. Proof of claim: is not empty since . Now let . If , then . If , then . Suppose that ; say that for some . Then , so that . But then , a contradiction. Thus if , then . Hence is a block.
Next we claim that . Proof of claim: If , then . Suppose . Then for some , hence . Then , so that .
Since is primitive on and , we have or . If , we have , so that . If , we have . Thus is a maximal subgroup of .
Suppose that for all , is a maximal subgroup in . Let be a block with . Now by part 1 above. Since is maximal, there are two cases.
If and such that , then since acts transitively there exists such that . Now , a contradiction. Thus .
If , then by Lemma 1 we have .
Thus is primitive on .