# Group Theory

• February 5th 2006, 08:18 AM
TexasGirl
Group Theory
Can someone explain why the following are groups:

(R, +)
({0}, +)

and the following are not:
(N, +)
(R, *)
(Z\{0}, *)

* is denoting multiplication
• February 5th 2006, 08:44 AM
topsquark
Quote:

Originally Posted by TexasGirl
Can someone explain why the following are groups:

(R, +)
({0}, +)

and the following are not:
(N, +)
(R, *)
(Z\{0}, *)

* is denoting multiplication

For a set and binary operation "ordered pair" to be a group we must meet several properties:

For an ordered pair (G,*) to be a group we must have:
$(a,b,c,e \epsilon G)$
1) The set G must be closed under *: For any a, b we must have that a*b is also a member of G.
2) a*(b*c) = (a*b)*c (i.e. * is associative)
3) $\exists$ an element e such that for any a, a*e=e*a=a (i.e. an identity element exists)
4) For every element a, $\exists$ an element $a^{-1}$ such that $a*a^{-1}=a^{-1}*a=e$ (i.e. each element has an inverse.)

So.
(R,+) is a group because it meets all these requirements.
({0}, +) is a group because it meets these requirements...I would say "vacuously" because + doesn't actually do anything. Either way, e=0 and the inverse of 0 is 0. 0+(0+0)=(0+0)+0, etc.

(N, +): Why is this not a group? What is the inverse of, say, 6? Is it an element of the set N?

(R, *): Again, look at inverses. We may easily find the inverse of any element but 0. What is the inverse of 0?

(Z\{0}, *): I'm not clear on the notation here. I take it that Z\{0} is the set {...,-3, -2, -1, 1, 2, 3, ...}? If so then what is the inverse of 2?

-Dan