Group Inverse of a Product - kipawaa/Proof-Tree GitHub Wiki
Statement
The inverse of a product of elements $ab$ of a Group $G$ is $b^{-1} a^{-1}$.
Explanation
Proof(s)
Let $a, b \in G$ and hence $ab \in G$ by closure.
By definition of inverse, we have that $(ab)(ab)^{-1} = e$.
Notice that
$$\begin{align*} (ab)(b^{-1} a^{-1}) &= a(bb^{-1})a^{-1}\ &= a e a^{-1} & \text{by definition of inverse}\ &= a a^{-1} & \text{by definition of identity}\ &= e & \text{by definition of inverse} \end{align*}$$
Since Group Inverse is Unique we have that $(ab)^{-1} = b^{-1} a^{-1}$, as wanted.