Let $e$ be an element of group $(G,\cdot)$
$(a)$ Show that the set $G$ forms a group under the multiplication $$"*":G \times G \rightarrow G ; (x,y) \mapsto xe^{-1}y.$$ $(b)$ Show that the group $(G,*)$ with multiplication given by (a) is isomorphic to the oringinal group structure $(G, \cdot)$ on $G$.
My proof:
We're given that $(G,*)$ is closed under multiplication since $(G,\cdot)$ is a group and so $xe^{-1}y \in G$. Similarly we must show that there exists an identity element $E_{G} \in G$. Consider the following $$x*e=xe^{-1}e=xe_{G}=x$$ and so $e_{G}=e$
Next we must show $G$ contains inverses so $y=ex^{-1}e$ then, $$x*y=xe^{-1}ex^{-1}e=xe_{G}x^{-1}e=xx^{-1}e=e_{G}e=e$$ For part $(b)$ I am unsure how to show they are isomorphic, I was thinking maybe finding a suitable group homomorphism and that would imply they are isomorphic. I included my proof of part $(a)$ just to verify it is correct, any help on part $(b)$ is appreciated!
For part (b), you want to show there is an isomorphism $\varphi: G\to G$ such that for all $a,b \in G$, $$\varphi(a b) = \varphi(a) * \varphi(b) = \varphi(a)e^{-1}\varphi(b).$$
Notice that we want to define $\varphi$ in a way that "cancels out" the $e^{-1}$ in the formula. Notice if we define $\varphi(x) = xe$, then $$\varphi(ab) = (ab)e = abe$$ and $$\varphi(a) * \varphi(b) = \varphi(a)e^{-1}\varphi(b) = (ae)e^{-1}(be) = a(ee^{-1})be = abe$$ and thus, $$\varphi(a b) = \varphi(a) * \varphi(b)$$ as desired.