Inverse of product of two elements of a group
Dependencies:
$$(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = e$$
Therefore, $(ab)^{-1} = b^{-1}a^{-1}$.
Dependency for:
Info:
- Depth: 1
- Number of transitive dependencies: 1
$$(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = e$$
Therefore, $(ab)^{-1} = b^{-1}a^{-1}$.