I know the following statement is false: A semigroup with left identity and right inverse is group. I have a counterexample. But I somehow unable to see problem in my proof that stating the statement is true.
My proof goes follow: Let $G$ be a semigroup. Then we know $G$ is a group if and only if it has a left identity and left inverse. We claim a semigroup with left identity and right inverse has also left inverse.
Suppose $g\in G$ and let $g^{-1}$ be a right inverse of $g$. Then \begin{equation*} (g^{-1}g)(g^{-1}g)=g^{-1}(gg^{-1})g=g^{-1}g. \end{equation*} Therefore $g^{-1}g=e$. Hence $g^{-1}$ is also a left inverse and we can apply the fact that if $G$ has a left identity and left inverse then it is a group. What is wrong with my proof?
When you have $$(g^{-1}g)(g^{-1}g)=g^{-1}g,$$ you immediately go from this to $$g^{-1}g=e.$$ I'm guessing your logic is to multiply by the right inverse $(g^{-1}g)^{-1}$ on the right hand side. Unfortunately, the right and left identities may not coincide in a semigroup, in other words you actually have $$(g^{-1}g)e=e.$$ But because $e$ is a left identity, and not a right identity, you cannot get from this to $$g^{-1}g=e.$$