Proving that a metric space is a group

I'm stuck on this relatively hard problem.

• Let $G$ be a non-empty set, $d$ a distance on $G$ and $\cdot$ an associative operation on $G$

• $\cdot$ is such that $$\forall a \in G , \forall x \in G ,\forall y \in G,( a\cdot x =a \cdot y) \Rightarrow x=y $$

and $$\forall a \in G , \forall x \in G ,\forall y \in G, (x\cdot a =y \cdot a) \Rightarrow x=y $$

• $ \cdot$ is continuous

• $(G,d)$ is compact

Prove that $(G,\cdot)$ is a group, and that the inverse function ($x \rightarrow x^{-1}$) is continuous

What I need to find first is an identity element. My guess is that I should consider the minimum $m$ of the function $f_x:y \rightarrow d(x\cdot y,x) $ (which exists since $f_x$ has a compact domain, and its codomain is $\mathbb R$).

I don't know how to prove that $m=0$ though...

The fact this $G$ is a group is a theorem of Katsumi Nakamura in his paper "On bicompact semigroups", Mathematical Journal of Okayama University, 1952, see theorem 1. The proof is not hard but a bit long.

To prove that the inverse is continuous, suppose it is not. Since our group is metrizable, it suffices to work with limits of sequences (even if it were not metrizable, one can run a similar argument using nets instead). Then there exists $x\in G$ and a sequence $x_i$ converging to $x$ so that $x^{-1}\ne y=\lim_{i} x_i^{-1}$. Note that since $G$ is compact, by working with subsequences, it suffices to consider the case when the sequence $(x_i^{-1})$ converges to some $y\in G$ and $y\ne x^{-1}$. We then have, by continuity of multiplication, that: $$ yx= \lim_{i} (x_i^{-1} x)= \lim_i 1= 1. $$ For the same reason, $xy=1$. Thus, $y=x^{-1}$, which is a contradiction.