Question 62
Suppose $(A, +, \cdot)$ is a ring in which $x^2 = 0$ only for $x = 0$. Let $B = \{a \in A \mid a^2 = 1\}$. Show that:
a. $ab - ba = bab - a$, for all $a \in A$ and $b \in B$.
b. $(B, \cdot)$ is a group.
a. $ab - ba = bab - a$, for all $a \in A$ and $b \in B$.
b. $(B, \cdot)$ is a group.