Remark 2 on Groups

i)

According to Remark 1 on groups iv), there exists exactly one inverse element for $a^{-1}$. Since $a \circ a^{-1} = e$, $a$ is the unique inverse element for $a^{-1}$. It follows that $(a^{-1})^{-1} = a$.

ii)

Let $a, b \in G$ be elements of the group $(G, \circ)$. It is shown that the inverse element of $a \circ b$ consists of $b^{-1} \circ a^{-1}$ by demonstrating that the combination of the two elements results in the neutral element $e$.