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:Originally Posted by TexasGirl
For an ordered pair (G,*) to be a group we must have:
$\displaystyle (a,b,c,e \epsilon G)$
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) $\displaystyle \exists$ an element e such that for any a, a*e=e*a=a (i.e. an identity element exists)
4) For every element a, $\displaystyle \exists$ an element $\displaystyle a^{-1}$ such that $\displaystyle a*a^{-1}=a^{-1}*a=e$ (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