# Counterexample for an alternate definition for a group

1. Dec 12, 2007

### SiddharthM

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?

2. Dec 12, 2007

### morphism

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

3. Dec 12, 2007

### SiddharthM

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.