Abstract Algebra | 23. Groups - Definition and Examples

In this video we introduce the notion of a group and give lots of examples of groups.

[angelmendez-rivera]

An alternative definition of a group is a structure (G, e, ~, *) on G, where e is a 0-ary function, e : {0} —> G, ~ is a 1-ary function, G —> G, and * is a 2-ary function, GxG —> G, such that, with all universal quantifiers implied, and no existential quantifiers, x*e = e*x = x, x*~x = ~x*x = e, x*(y*z) = (x*y)*z. A motivation for this alternative definition, aside from having a completely equational signature, is the fact that group homorphisms can now be defined more concisely in terms of compositions.

In my opinion, this should have been taught earlier in the series, prior to rings. This is because, once you have been exposed to group theory, defining a ring axiomatically is extremely simple. A ring (R, 0, 1, –, +, ·) on R is a structure such that (R, 0, –, +) is an Abelian group, (R, 1, ·) is a monoid, and x·(y + z) = (x·y) + (x·z) and (y + z)·x = (y·z) + (y·x). I think there is motivation to define rings in this manner: it helps explain why the multiplicative monoid of a ring can never be a group: because it is not a cancellative monoid. In fact, generally speaking, there is a fundamental difference between a cancellable monoid, and a non-cancellable monoid. A cancellable monoid may not itself be a group, such as with the nonzero integers, but it is _embeddable_ into a group, the group of nonzero rational numbers. A non-cancellable monoid is _not embeddable_ into a group. Multiplicative monoids of a ring with zero divisors excluding 0 are examples of this. If x is a zero divisor, then there exists some a such a·x = 0, so y·(a·x) = 0 = z·(a·x) = (y·a)·x = (z·a)·x, and so, if a is cancellable, then x is not cancellable. So zero divisors satisfying a certain condition are never cancellable. In fact, this answers an important question: why can a ring not be extended in such way that 0 can attain a multiplicative inverse and allow 0·x = x·0 = 0 to be false? The reason is because 0 is not multiplicatively cancellable. Extending a ring to a field can allow every cancellable element to have an inverse, but not 0 or other non-cancellable nonzero divisors.