Let be a finite cyclic semigroup. Show that has a unique maximal subgroup and compute it. Compute the kernel of . Under what circumstances is a group? Deduce that every finite semigroup contains an idempotent.
Let be a finite cyclic subgroup with generator . Now the map given by is a surjective semigroup homomorphism (where is a semigroup under ). Since cannot be injective, there exist pairs of distinct positive natural numbers such that . Let be such that is minimal among these pairs. (For such , there is a unique with , since otherwise is not minimal.)
We now prove a lemma.
Lemma 1: If with , then there exists a unique such that . Proof: (Existence) Note that . By the Division Algorithm in , there exist natural numbers (that is, nonnegative integers) and such that and . Letting and , we have , so that , and thus . (Uniqueness) Suppose and , we have , so that , and so .
This number will turn out to be very special.
Lemma 2: For all and , we have . Proof: We proceed by induction on . For the base case , if , then . If , then . Now for the inductive step, suppose the result holds for some . Then . So the result holds for all .
Corollary 2.1: If and mod , then . Proof: Without loss of generality, say . Now for some , and thus and for some .
Corollary 2.2: If , then there exists such that and . Proof: It suffices to show that there exists such that and mod . To that end, let be the least residue of mod . That is, , where and are nonnegative. Letting , we have Then and , so .
Corollary 2.3: consists of the elements , and these are pairwise distinct.
Lemma 3: If and , then mod . Proof: Since and , there exist nonnegative integers such that , , and . Since , we have . Whether and are zero (trivially) or not (Lemma 2), we have . Now , so that . By the minimalness of , we have . Thus mod , and so mod .
Lemma 4: If and such that , then . Proof: If and , we violate the minimalness of . If , then by Corollary 2.2, there is an integer with and , again violating the minimalness of .
Using Lemmas 3 and 4 and Corollary 2.1, we can now characterize the kernel of as follows. The equivalence class is if , and is if .
Lemma 5: If is idempotent, with , then . (Where is as defined in Lemma 1.) Proof: If , then we have , which has no solution in . If , then mod . So mod , and then for some . By Lemma 1, .
So has precisely one idempotent.
Lemma 6: Let be a semigroup. Let denote the set of idempotents in , and let denote the set of maximal subgroups of . Let be defined by . Then is bijective. Proof: (Surjective) If is idempotent, then the set is the -greatest subgroup of having identity by 1.4.11. In particular, is a maximal subgroup, and . If is a subgroup with identity , then . If is maximal, then . So is injective, and thus bijective.
Since our cyclic semigroup has only one idempotent, it has a unique maximal subgroup. Namely, where .
Lemma 7: if and only if . Proof: The direction was proved in Lemma 2. Now suppose . Now , so that (using our characterization of ).
Lemma 8: If , then there exists such that . Proof: Choose such that mod .
Lemmas 7 and 8 demonstrate that .
In particular, is a group if and only if .
Recall now that the kernel of a semigroup is the intersection of its two sided ideals.
Lemma 9: If is an ideal, then . Proof: Suppose . For all , there exists such that and mod ; then . So .
Finally, using Corollary 2.2 if necessary, we can see that is itself an ideal of . Thus the kernel of is .
Finally, suppose is a finite semigroup and let . Now is a cyclic subsemigroup of , and so must be finite. As we argued above, , and thus , contains an idempotent.