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). 
 A: Maybe you would like to just directly show that $\ell_g$ is both one-to-one and onto.
To see that $\ell_g$ is one-to-one, suppose that $\ell_g(x) = \ell_g(y)$.  Then $gx = gy \implies g^{-1} (gx) = g^{-1}(gy) \implies (g^{-1} g) x = (g^{-1}g) y \implies ex = ey \implies x = y$.
(Here $e$ is the identity element in $G$.)
To see that $\ell_g$ is onto, suppose that $y \in G$.  Let $x = g^{-1} y$.
Then $\ell_g(x) = g(g^{-1} y) = (g g^{-1}) y = ey = y$.  This shows that $\ell_g$ is onto.
A: If an operator is invertible, it acts bijectively. I find that intuitive and widely applicable.
A: This may provide a more direct proof of the statement.
$l_g$ is injective: Suppose $l_g(x_1) = l_g(x_2)$. Then we have $$x_1 = 1\cdot x_1 = g^{-1} g \cdot x_1 = g^{-1} \cdot gx_1 = g^{-1} \cdot gx_2 = g^{-1}g \cdot x_2 = 1 \cdot x_2 = x_2$$
using the existence of inverse in a group and the associativity of multiplication.
$l_g$ is surjective: Suppose to the contrary that there was some $x' \in G$ such that there was no $x \in G$ for which $gx = x'$. But then $x = 1 \cdot x = g^{-1} g \cdot x = g^{-1}  \cdot gx \neq g^{-1} x'$, so there is no $x \in G$ that is equal to $g^{-1} x'$, implying that the group is not closed under multiplication.
A: The other answers don't really deal with the point I wanted to address, which is that the intuition for this (at least for me) lies in the algebra in the original Arabic sense. A group is a structure where, when you're dealing with equations, you can group things together and cancel things out in a particular way.
The fact that left and right translation are bijective is actually much weaker than the axioms of associativity and inverse. The associativity implies that additionally the group is isomorphic to the group of left translation functions with composition as the operation. So really the important thing is that if
$$aba^2=gh$$
then we know for sure that
$$aba^2h^{-1}=g$$
and all the other equations you can derive from the axioms. (This is the approach of universal algebra, and it's also personally the only way I understand anything in math).
A: I agree that the proof presented here doesn't provide intuition. (To clarify for those reading: the proof goes through the Lemma: "if a composition of functions is bijective, then each function is bijective").

Your intuition should be that a bijective function is a function with an inverse, i.e. one you can "undo".
Once you have a better intuition on bijective functions (namely that they are the ones you can "undo"), the proof that translation is bijective is extremely intuitive: the defining feature of groups is that they have inverses: you can always "undo" translation by a group element by translating again by that element's inverse. 

The intuition behind the proof that the bijective functions are precisely the invertible functions is simple: "to undo my function I need for each element in the co-domain to have come from somewhere (i.e. surjectivity), so I'll have some place to send it back, and I need a unique place to send it back (i.e. injectivity) so that 'sending back' will itself be a function." And of course the observation that these are necessary conditions ("if an element came from two places, I can't use a function to send it back to both places, and if an element didn't come from anywhere I can't send it back anywhere").
Also the Lemma becomes intuitive "if a composition of functions is bijective, then each function is bijective": all this is saying is that if what you're doing involves applying a step that can't be undone, then what you're doing can't be undone.
A: 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.
A: A simple explanation that I loved is based on the "principle of the drawers" that says that if you have N drawers containing N things and no drawer contains more than one thing then each drawer has exactly one thing.
Given that in a group G xy and xz are distinct if y is distinct from z you get that all xG elements are distinct, but they are also o(G) and therefore the mapping is invertible.
This intuitive explanation is however valid only for finite groups :-)
