Mathematics Prelims

January 18, 2009

Introductory Algebra and Definitions

Filed under: Abstract Algebra,Algebra — cjohnson @ 10:21 am

If S is a set, a binary operation on S is a function \circ : S \times S \to S, where we generally write s \circ t for \circ(s, t).  We say that this binary operation is associative if for every s,t,u \in S we have that (s \circ t) \circ u = s \circ (t \circ u).  A pair (S, \circ) where S is a set and \circ an associative binary operation on S is then called a semigroup.  For example, (\mathbb{N}, +) is a semigroup.

If (S, \circ) is a semigroup and there exists an e \in S such that for every s \in S we have that s \circ e = e \circ s = s, then we say that e is an identity (of \circ) and we call (S, \circ) a monoid.  The natural numbers together with zero, which we’ll denote \mathbb{N}_0, is then a monoid under addition.

Generally we’ll let juxtaposition denote the binary operation, writing st for s \circ t.  In cases where + is the “natural” symbol to use for the binary operation, however, (e.g., the natural numbers with addition), then we’ll still use +.

When s,t \in S and st = e (where e still denotes the identity), then we say that s is the inverse of t (and vice versa).  We will normally write s^{-1} to mean the inverse of s, though when + is the operation we’ll write -s and write t - s to mean t + (-s).  A monoid where every element has an inverse is called a group.  The integers under addition form the group (\mathbb{Z}, +).  To check this, simply note that addition of integers is associative, addition of two integers yields another integer, zero is the identity element, and for any z \in \mathbb{Z} we have that -z is its inverse.

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: