The set of one-sided inverses is not a group under multiplication We proved that if $A$ is a ring and $U$ is the set of elements  of $A$ which have both a right and left inverse, then $U$ is a multiplicative group. Now I ask for an example to show that the elements of a ring which are left invertible do not necessarily form a group.
Thanks in advance   
 A: Take the ring of $A$ of all linear operators on the set of polynomials. Let $U$ be the set of all such operators with a left inverse. If $I$ and $D$ denote integration -- $I(p(x))$ is the integral with constant term $0$ -- and differentiation, respectively, then of course $DI(p(x))=p(x)$; hence $I$ is an element of $U$. Will every left inverse of $I$ be in $U$?
If $D'$ is any left inverse of $I$, then for any $n \geq 1$, we have $D'(a_nx^n)=D'(I(na_nx^{n-1})=nax^{n-1}=D(a_nx^n)$. 
Note that $D'$ is not injective -- e.g., $D'(2x+1)=2=D'(2x+2)$ -- so $D$ can't have a left inverse. If $D'$ had a left inverse, then it's a map that has a left and right inverse. This would imply that $D'$ is a bijection from the set of polynomials to itself. 
This of course implies that $U$ is not a group as it's not closed under inverses.
A: Let R be a ring and let S denote the set of all elements of R which have a left inverse.  We will prove a general statement:  
"If there is $x \in S$ such that $x$ has a left inverse, but no right inverse, then S does not form a multiplicative group." 
Since $S \subset R$, and $x$ does not have a right inverse in R, it cannot have a right inverse in S.  Elements of a multiplicative group must have two sided inverses, so S cannot be a multiplicative group. 
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So your question amounts to finding a particular ring that has (at least one) element that is left invertible, but not right invertible.  For that, I will refer you to the already answered question: A ring element with a left inverse but no right inverse?
