9

I at a loss here. Showing that $G$ has an identity is the difficult part here. Obviously for each $a\in G$ there are $b,c\in G$ such that $ab=a$ and $ca=a$, but I need a hint as to how to prove these are all the same element.

In the finite case, the pigeonhole principle certainly shows that they all can't be different.

Tim Raczkowski
  • 20,046
  • See also http://math.stackexchange.com/questions/505302/g-is-a-group-if-and-only-if-for-all-a-b%E2%88%88g-ax-b-has-solution-true-or-false and http://math.stackexchange.com/questions/656881/proving-that-g-is-a-group-if-ax-b-and-ya-b-have-solutions – Martin Sleziak Dec 01 '15 at 15:18

1 Answers1

7

You can use the given property to check directly that $ab = a$ for some $a,b$ forces $b$ to be a right identity. For any $d$, write $d = ea$ and then $db = eab = ea = d$. From here it is straightforward to show that $ca = a$ similarly implies that $c$ is a left identity, and then $c = cb = b$ shows that there exists a two-sided identity.

Rolf Hoyer
  • 12,399