ition in that set. However, while N is a monoid under addition, it is not a group. Why not? The problem is (G4). We know that every natural number has an additive inverse; after all, 2 + (−2) = 0. Nevertheless, (G4) is not satisfied because −2 6∈ N! It’s not enough to have an inverse in some set; (G4) requires that the opposite is in the same game! For this reason, N is not a group.
We can now classify all groups of order two. We will do this in a “brute force” manner: building the Cayley table for a “generic” group of order two. As a consequence, we show that regardless of the set and its operation, every group of order 2 behaves the same way: there is only one structure possible for its Cayley tab.
We did not say that e represents the only identity. For all we know, a might also be an identity; is that possible? It is not possible; why? Remember that a group is a monoid. We showed in Proposition 2.9 that the identity of a monoid is unique; thus, the character of a group is unique; therefore, there can be only one identity, e.
By the transitive property of equality, c = d. This shows that if a appears in one column of the row headed by b, then that column is unique; a does not appear in a different column. We still have to show that looks in at least one row of the addition table headed by b. This follows from the fact that each row of the Cayley table contains |G| elements. What applies to an above refers to the other items, so each part of G can appear at most once. Thus, if we do not use a, then only n − 1 pair are defined, which contradicts either the definition of operation (b x must be set for all x ∈ G) or closure (that b x ∈ G for all x ∈ G). Hence a must appear at least once visit this website.