Counterexample for an alternate definition for a group

In summary, problem 12 from Herstein states that if a set G is closed under associative product and satisfies conditions (a) and (b), then G is a group. The proof involves showing that y(a)a=ay(a)=e and then ea=ae=a. However, problem 13 shows that this conclusion is false if condition (b) is replaced with a slightly different condition, (b'), and G is still closed under associative product. A counterexample is given by taking the integers with n*m=n, which is closed and associative, but not a group.
  • #1
SiddharthM
176
0
This is another problem from Herstein. The first problem 12 asks to prove that if G is closed under associative product and
(a) there is a e so that for any element a in G we have ae=a
(b) for each a in G there is a y(a) so that ay(a)=e
then G is a group. Although it took me some time to do, it comes down to showing that y(a)a=ay(a)=e and from there show ea=ae=a, which is straightforward.

then problem 13 asks to show that the conclusion of problem 12 is false if we assume instead
(a)same
(b')given a in G there is a y(a) so that y(a)a=e.

I presume that this G is closed under an associative product as well, because if not a trivial example is;

positive integers with 0=e under subtraction. Then for any a, a is our y(a) so a-a=e=0 and ae=a-e=a-0=a.

What counterexample is there if we assume G is closed under an associative product?
 
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  • #2
Take the integers and define n*m = n. This seems to work, no?
 
  • #3
it is closed and associative. and e=any integer so that (a) is satisfied, say 1, then for any a in G, 1a=1=e so that (b) is satisfied. However

1*9=1 which is not 9*1=9 which means G cannot be a group.
 

1. What is a counterexample for an alternate definition of a group?

A counterexample for an alternate definition of a group would be a set with a binary operation that does not satisfy the four group axioms: closure, associativity, identity, and invertibility.

2. Can you give an example of a counterexample for an alternate definition of a group?

Yes, an example of a counterexample for an alternate definition of a group is the set of all positive integers under the operation of multiplication. This set does not have an identity element (0 is not a positive integer) and therefore does not satisfy the group axioms.

3. Why is a counterexample important in understanding alternative definitions of a group?

A counterexample is important because it shows that a certain definition of a group is not valid. It helps to clarify the necessary and sufficient conditions for a set to be considered a group, and can lead to a better understanding of the concept.

4. Are there different types of counterexamples for alternate definitions of a group?

Yes, there are different types of counterexamples depending on which group axiom is not satisfied. For example, a set with a binary operation that is not associative would be a counterexample for the associativity axiom, while a set with no identity element would be a counterexample for the identity axiom.

5. How can a counterexample be used in mathematical proofs about groups?

A counterexample can be used in mathematical proofs to disprove a statement or claim about groups. By providing a counterexample, it shows that the statement is not always true and therefore cannot be used to make generalizations about groups. It can also be used to identify and correct errors in proofs that rely on alternative definitions of a group.

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