Intuition behind left and right translations being bijective in a group? In my algebra class, we learn that the maps $l_g(x) = gx$ for $x \in G$ and $r_g(x) = xg$ for $x \in G$ are bijective. The proof given uses the fact that $l_g l_{g^{-1}} = l_{g^{-1}} l_g = 1_G$, so both functions are bijective since $1_G$ is and therefore $l_g$ is bijective. The proof for $r_g$ is analogous. Is there are more intuitive approach to achieving this result? The proof, while elegant, doesn't provide intuition (in my opinion). 

And no one has pointed out ...
This result is equivalent to the observation that every row and every column of the multiplication table is a permutation of any other row or column.  That is, each row and column has all the symbols in it exactly once, so is surjective and injective.
