Question 69
Let $(A,+,\cdot)$ be a ring in which $x=0$ is the unique solution of $x^2=0$, $x\in A$. Let $B=\{a\in A\mid a^2=1\}$. Prove:
a. $ab-ba=bab-a$ for every $a\in A$ and $b\in B$.
b. $(B,\cdot)$ is a group.
a. $ab-ba=bab-a$ for every $a\in A$ and $b\in B$.
b. $(B,\cdot)$ is a group.