Proving that a mapping is a group homomorphism Assume that $G$ is a group, and let $S(G)$ denote the group of bijections $G\rightarrow{G}$ under composition of maps.  Prove that the mapping ${\varphi}:G\rightarrow{S(G)}$ defined as follows is a group homomorphism. Every $a\in{G}$ is mapped to ${r_{a}}\in{S(G)}$ where ${r_{a}}(g) = ga$ for every $g\in{G}$. This mapping may be described as the right translation by $a\in{G}$. The analogous statement for left translations follows easily, but I am having trouble proving that ${\varphi}(ab) ={\varphi}(a){\varphi}(b)$. Thanks! 
In light of the comment/solution posted by Edcookie 274, the definition of right translation by a group element should be $${\varphi}(a) = {r_{a}}(g) = ga^{-1}$$ for every $g\in G.$ This definition makes everything work easily and clears up all my questions.
 A: $\varphi(ab):G\rightarrow G$ defined by $\varphi(ab)(g) = (ab)g = a(bg) = a(\varphi(b)(g)) = \varphi(a)\circ\varphi(b)(g) $
A: Assume that $G$ is a group with identity element $e$, and let $S(G)$ be the group of bijections (equipped with the group law of composition of functions) $G\rightarrow{G}.$ Then the maps $G\rightarrow{S(G)},$ $a\mapsto{l_{a}},$ and $a\mapsto{r_{a}}$ where for any $g\in{G}$ $${l_{a}}(g) = ag,$$ and  $${r_{a}}(g) = ga^{-1}$$ are both injective group homomorphisms $G\rightarrow{S(G)}.$ They are injective because ${l_{a}} = {l_{b}}$ implies that ${l_{a}}(g) ={l_{b}}(g)$ for every $g\in{G}$, and in particular ${l_{a}}(e) =ae=a=b=be={l_{b}}(e)$, and ${r_{a}}={r_{b}}$ implies that ${r_{a}}(g)={r_{b}}(g)$  for every $g\in{G}$, and in particular, ${r_{a}}(e)= ae=a=b=be={r_{b}}(e)$. We also observe that for every $g\in{G},$ $${l_{ab}}(g)=abg=a(bg)=a({l_{b}}(g))={l_{a}}({l_{b}}(g))={{l_{a}}}\circ{{l_{b}}}(g),$$ while $${r_{ab}}(g)={g(ab)^{-1}}=g{b^{-1}}{a^{-1}}={r_{a}}(gb^{-1})={r_{a}}\circ{r_{b}}(g).$$
I am grateful to Daniele Tampieri and Edcookie274 for helping me clear this up.
