Can someone explain why the following are groups:

(R, +)

({0}, +)

and the following are not:

(N, +)

(R, *)

(Z\{0}, *)

* is denoting multiplication

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- Feb 5th 2006, 08:18 AMTexasGirlGroup Theory
Can someone explain why the following are groups:

(R, +)

({0}, +)

and the following are not:

(N, +)

(R, *)

(Z\{0}, *)

* is denoting multiplication - Feb 5th 2006, 08:44 AMtopsquarkQuote:

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