If this is obvious and a silly question, I'm sorry. As a part of an exercise, I had to see that the inverse of $(g\cdot h)$, with $g$ and $h$ in a group $G$ is $h^{-1}\cdot g^{-1}$. I proved it by this way:
Let $e$ be the neutral element on $G$:
\begin{equation} (g\cdot h)^{-1} \cdot (g\cdot h)= e \\ \Downarrow \\(g\cdot h)^{-1} \cdot (g\cdot h)\cdot h^{-1}= e \cdot h^{-1}\\ \Downarrow\\ (g\cdot h)^{-1} \cdot g\cdot (h\cdot h^{-1})= e \cdot h^{-1} \\ \Downarrow\\ (g\cdot h)^{-1} \cdot g\cdot e= h^{-1} \\ \Downarrow \\ (g\cdot h)^{-1} \cdot g\cdot g^{-1}= h^{-1} \cdot g^{-1} \\ \Downarrow \\ (g\cdot h)^{-1} = h^{-1}\cdot g^{-1} \end{equation}
Is that proof correct? Can I multiply the equation with elements of $G$ on both sides? Thank you!
$\endgroup$ 51 Answer
$\begingroup$A simpler proof is:$(gh)(h^{-1}g^{-1}) = g(h h^{-1})g^{-1} = g g^{-1}=e$, and similarly, $(h^{-1}g^{-1})(gh)=e$. Hence $(gh)^{-1}=h^{-1}g^{-1}$.
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