This question is related to exercise 1.51 from Rotman's "Introduction to the Theory of Groups".
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My question is related to exercise 1.51 from Rotman's "Introduction to the Theory of Groups".
An element a in a ring $R$ (with unit element $1$) has a left quasi-inverse if there exists an element $b\in R$ such that $a+b−ba=0$. I want to show that if every element in $R$ has a left quasi-inverse except $1$, then $R\setminus{1}$ is a group under the operation $a∗b=a+b−ba$.
If you take a ring and change its multiplication by the new, this new structure has anything interesting?
Is some reference papers or books about this operation?
First I'll answer your question about the exercise (showing that $(R\setminus \{1\},*)$ is a group).
Let's start by checking that $*$ is associative on all of $R$: \begin{align*} (a * b)*c &= (a+b-ba)*c\\ &= (a+b-ba)+c-c(a+b-ba)\\ &= a+b+c-ba-ca-cb+cba\\ &= a+(b+c-cb)-(b+c-cb)a\\ &= a*(b+c-cb)\\ &= a*(b*c). \end{align*}
Now observe that for any $a\in R$: \begin{align*} 1*a &= 1+a-a1 = 1\\ a*0 &= a+0-0a = a\\ 0*a &= 0+a-a0 = a \end{align*} And if $b$ is a left quasi-inverse of $a$, then $a*b = 0$.
Now we'd like to show that $R\setminus \{1\}$ is closed under $*$. Let $a,b\in R\setminus \{1\}$, and suppose for contradiction that $a*b = 1$. Since $b\neq 1$, $b$ has a left quasi-inverse $c$, so \begin{align*} (a*b)*c &= 1*c\\ a*(b*c) &= 1\\ a*0 &= 1\\ a &= 1 \end{align*} which is a contradiction.
So now we have an associate operation $*$ on $R\setminus \{1\}$ which has an identity element $0$, and every element has a right-inverse (for all $a$ there exists $b$ such that $a*b = 0$). So you can apply the fact that a monoid with right-inverses is a group. (Proof: Let $b$ be the right-inverse of $a$, and let $c$ be the right-inverse of $b$. Then $(a*b)*c = 0*c$, so $a*(b*c) = c$, so $a = c$, so $b*a = 0$, and $b$ is the inverse of $a$).
Ok, having solved the exercise, some comments:
An element with a left quasi-inverse is called left quasiregular. For a reference, you could start with Wikipedia and follow the references there.
An element $x$ is left quasiregular if and only if $(1-x)$ has a left inverse. Indeed, if $y(1-x) = 1$, then the left quasi-inverse of $x$ is $z = -xy$. We have $$x+z-zx = x-xy+xyx = x(1-y+yx) = x(1-y(1-x)) = x(1-1) = 0.$$ Conversely, if $x+z-zx = 0$, then $y=(1-z)$ is the left inverse of $(1-x)$. We have $$y(1-x) = (1-z)(1-x) = 1-x-z+zx = 1-(x+z-zx) = 1.$$
Suppose $R$ has the property that every element of $R\setminus \{1\}$ is left quasiregular, as in the exercise. Then for all $x\neq 1$, $(1-x)$ has a left inverse. Thus every nonzero element of $R$ has a left inverse. This means that $R\setminus \{0\}$ is a monoid with left-inverses, and hence a group (by the dual of the fact noted above). So every element of $R\setminus \{1\}$ is left quasiregular if and only if $R$ is a division ring.