91Ó°ÊÓ

Let's consider the left-hand side (LHS) of the equation, which is \(\left(a^{-1}\right)^{-1}\). By the property of group, we know that any element multiplied by its inverse produces the identity. Therefore, multiplying \(\left(a^{-1}\right)^{-1}\) with \(a^{-1}\) should give us the identity element. So \(\left(a^{-1}\right)^{-1} a^{-1} = e\), where \(e\) is the identity element. Now let's consider multiplying both sides of the equation \(a a^{-1} = e\) from left-hand side by \(\left(a^{-1}\right)^{-1}\). It would give \(\left(a^{-1}\right)^{-1} a a^{-1} = \left(a^{-1}\right)^{-1} e\). Simplifying the LHS by associativity property \( (\left(a^{-1}\right)^{-1} a) a^{-1} = \left(a^{-1}\right)^{-1}\), hence \(a^{-1} a^{-1} = \left(a^{-1}\right)^{-1}\) or \(e = \left(a^{-1}\right)^{-1}\). This implies \(\left(a^{-1}\right)^{-1}=a\). Hence, the first statement is proved.

We first calculate the product of the two elements on the right-hand side (RHS), i.e., \(b^{-1} a^{-1}\). Now consider the multiplication from left to right of \(a b b^{-1} a^{-1}\), where \(a b\) is considered as one element. This can be rearranged using associativity as \((a b) (b^{-1} a^{-1})\). We know that any element multiplied by its inverse results in the identity. Hence this reduces to \((a b) (e) = a b\), where \(e\) is the identity element. This means that the element \(b^{-1} a^{-1}\) is the inverse of \(a b\). Hence, \((a b)^{-1}=b^{-1} a^{-1}\) as required by the problem statement. Hence, the second statement is proved.