群论问题：$(g^{-1})^{-1}=g$是否始终成立？

Great question—your intuition is spot-on: yes, this equality holds for every element $g$ in any group $G$, no exceptions.

Let’s break this down using core group theory definitions to confirm it:

• By definition, the inverse of $g$ (written $g^{-1}$) is the unique element in $G$ such that $g \cdot g^{-1} = e$ and $g^{-1} \cdot g = e$, where $e$ is the group’s identity element.
• To find $(g^{-1}){-1}$, we need the element that acts as the inverse of $g^{-1}$—meaning an element $h$ where $g^{-1} \cdot h = e$ and $h \cdot g^{-1} = e$.
• Looking back at the first point, $g$ satisfies exactly those conditions! Multiplying $g^{-1}$ by $g$ (in either order) gives the identity.
• Since group inverses are unique, there’s no other element that can be the inverse of $g^{-1}$—so $(g^{-1}){-1}$ must be $g$.

As for why your textbook writes $(g^{-1}){-1}$ explicitly instead of simplifying to $g$ right away, that’s all about rigor, especially for beginners. Textbooks often prioritize showing the exact operation being performed rather than skipping to the conclusion, to make sure you connect the result back to the definition of inverses. In non-group algebraic structures (like monoids where not every element has an inverse, or semigroups), this property might not hold—but in groups, it’s a rock-solid rule.

内容的提问来源于stack exchange，提问作者Stephen