Let $R$ be a non empty set and $T$ be a subset of functions $R\to R$ (we call them translations) which satisfies the following:
- the identity map $t(x)=x$ is in $T$
- given $x,y\in R$ there exists one and only one function $t_{xy}\in T$ such that $t_{xy}(x)=y$.
- if $t,s\in T$ then $t\circ s\in T$.
Now fix any element $0\in R$ and define a addition operation as $$ x + y = t_{0y}(x) $$ where $t_{0y}$ is the only map in $T$ such that $t_{0y}(0)=y$.
I'm able to prove that $(R,0,+)$ is a group. Is it commutative?
I don’t think so.
Let $G$ be an nonabelian group and denote the group operation by $\bullet$. The set of left-multiplications $T=\{L_g:G\rightarrow G, h \mapsto g\bullet h \mid g \in G\}$ is a set of translations.
Indeed $\operatorname{id} = L_e \in T$, $L_g,L_h \in T \implies L_g\circ L_h= L_{g\bullet h} \in T$ and given $x,y\in G$ we find that $L_{y\bullet x^{-1}}$ defines a (unique by evaluating at $e$) translation.
Now for $x,y\in G$ we find that $x+y = L_y(x) = y\bullet x$ and $y+x = L_x(y) = x \bullet y$ won’t coincide in general, as we chose $G$ to be nonabelian.