The question is taken from Section 9, page 101 of 'A First Course in Abstract Algebra' by John B. Fraleigh, 7th edition.
"Let $G$ be a group. Prove that the permutations $\rho_{a} : G \rightarrow G$, where $\rho_{a}(x) = xa$ for $a \in G$ and $x \in G$, do form a group isomorphic to G."
My attempt at an answer: 1) Proving closure $\rho_{a}(x) = xa$, since $x \in G$ and $a \in G$, $xa \in G \Rightarrow$ closure.
2) Proving one-to-one: Let $\rho_{a}(x) = \rho_{a}(y)$ for some $x,y\in G$. Then $xa = ya$. Since $a \in G, \exists a^{-1} \in G$ such that $a.a^{-1} = a^{-1}.a = e$, where $e$ is the identity. Therefore, by multiplying on the right by $a^{-1}$, $x=y$.
Since $\rho_{a}(x) = \rho_{a}(y) \Rightarrow x=y$, this is one-to-one.
3) The last part is where I am running into problems. I think I am trying to prove $\phi: \rho_{a} \rightarrow G$, and $\phi(xy) = \phi(x)\phi(y)$. So... $\rho_{a}(xy) = \rho_{a}(x)\rho_{a}(y)$.
However, this is $xya$ on the left hand side and $xa.ya$ on the right hand side.
Any help would be appreciated.
Thanks in advance.