This is another problem from Herstein. The first problem 12 asks to prove that if G is closed under associative product and(adsbygoogle = window.adsbygoogle || []).push({});

(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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# Counterexample for an alternate definition for a group

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