Counterexample for an alternate definition for a group

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SUMMARY

The discussion revolves around a problem from Herstein regarding the conditions under which a set G can be classified as a group. Specifically, it establishes that if G is closed under an associative product and satisfies the conditions of having an identity element e and an inverse for each element, then G is a group. However, the counterexample presented demonstrates that if the inverse condition is altered, G may not form a group, as illustrated by the integers under a defined operation where n*m = n, which fails to satisfy the group properties.

PREREQUISITES
  • Understanding of group theory concepts, specifically identity elements and inverses.
  • Familiarity with associative operations in algebra.
  • Knowledge of counterexamples in mathematical proofs.
  • Basic comprehension of Herstein's work on abstract algebra.
NEXT STEPS
  • Study the properties of groups in abstract algebra, focusing on definitions and examples.
  • Explore counterexamples in mathematical proofs to understand their significance.
  • Learn about different algebraic structures, such as rings and fields, and their properties.
  • Review Herstein's "Topics in Algebra" for deeper insights into group theory.
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, educators teaching group theory, and researchers exploring algebraic structures.

SiddharthM
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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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Take the integers and define n*m = n. This seems to work, no?
 
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.
 

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