During a lecture was pointed out that one of the main feature, from a topological perspective, of normed vector spaces is the translational invariance, that is that one can study the topological structure about a fixed point and then move the open balls wherever needed.
However such a property can be stated for a simpler space than a vector space. A group $(G,\cdot)$ with a distance $d$ such that $d(x,y)=d(xy^{-1},e)$ where $e$ is the identity element should be sufficient (can be easily seen that $d(x,y)=d(xz^{-1},yz^{-1})$ follows and vice versa).
Now I wonder if
is $(G,\cdot)$ a topological group?
is $(G,d)$ complete if and only if all the absolute convergent series (with respect to $\rVert x \lVert = d (x,e)$ which obviously isn't a norm) are convergent?
I think that commutativity is sufficient (i.e. $(G,\cdot)$ an Abelian group) for the first point, is it also necessary?
It seems the following.
Let $G$ be a group which topology is determined by a two-side invariant metric $d$, that is $d(x,y)=d(zxt, zyt)$ for all $x,y,z,t\in G$. Then $G$ is topological group because the inversion and the multiplication on the $G$ are continuous. Indeed, let $x,y,x’,y'\in G$, $\varepsilon>0$, $d(x,x’)< \varepsilon$, and $d(x,x’)< \varepsilon$. Then $d(x’^{-1}, x^{-1})=d(x’^{-1}x, e)=d(x, x’)< \varepsilon $ and $d(x’y’,xy)\le d(x’y’,x’y)+d(x’y,xy)=d(y’,y)+d(x’,x)<2\varepsilon$. From the other side, I expect that there is a non-abelian group $G$ which topology is determined by a right invariant metric $d$ (that is $d(x,y)=d(xt, yt)$ for all $x,y,t\in G$) which is not a topological group.