Example 5 :
Complete the following table so that * is commutative binary operation on the set S = {a,b,c,d}.
From the table, there is no identity and inverse element.(G1 and G2 are not satisfied).
Therefore, it is not a group.
Friday, August 13, 2010
GROUPS 6
Thursday, August 12, 2010
GROUPS 5
Finite Groups and Group Tables
Finite group - Group where the set G has finite element.
The rules :
Finite group - Group where the set G has finite element.
The rules :
- List the identity first (use definition G2).
- To find an inverse elements for each element e, a (use definition G3) a' must be either a or e.
- All axioms are satisfied except possibly the associative property. Checking associativity on a case by case basis.
GROUPS 4
Theorem 3 ( The uniqueness of identity & inverse elements in G )
In a group G with binary operation *, there is only one element e in G such that
In a group G with binary operation *, there is only one element e in G such that
e*x = x*e = x
for all x in G.
Likewise, for each a in G, there is only one element a' in G that
Likewise, for each a in G, there is only one element a' in G that
a'*a = a*a' = e
In summary, the identity & inverse of each element are unique in a group.
GROUPS 3
Elementary Properties of Groups
Theorem 1 (Left & Right Cancellation Laws)
If G is a group with binary operation *, then the left and right cancellation laws hold in G, that is
Theorem 1 (Left & Right Cancellation Laws)
If G is a group with binary operation *, then the left and right cancellation laws hold in G, that is
a*b = a*c implies b = c, and
b*a = c*a implies b = c for all a,b,c in G
Similarly, we can prove the right cancellation law.
b*a = c*a implies b = c for all a,b,c in G
Proof:
(To prove the left cancellation law)
To show if a*b = a*c then b = c
Let a*b = a*c
Since G is a group, G3 is satisfied. Then, for all a' in G such that,
(To prove the left cancellation law)
To show if a*b = a*c then b = c
Let a*b = a*c
Since G is a group, G3 is satisfied. Then, for all a' in G such that,
a'*(a*b) = a'*(a*c) --G3
(a'*a)*b = (a'*a)*c --G1,* associative
e*b = e*c --by def. of a' in G3
b = c
(a'*a)*b = (a'*a)*c --G1,* associative
e*b = e*c --by def. of a' in G3
b = c
Similarly, we can prove the right cancellation law.