Let $M$ be an abelian monoid. For sake of simplicity we shall assume that in $M$ the cancellation law holds true. With this last assumption we define the group completion $G$ of $M$ as $$G:=M\times M/\sim$$ where $(a,b)\sim (a',b')$ if and only if $a+b'=a'+b$. It has been quite surprizing find out, reading the paper Completions and Fibrations for Topological Monoids  Paulo LimaFilho, that in general, if $M$ is a topological abelian monoid then its group completion $G$, endowed with the quotient topology, is not a topological group. Can anyone help me to provide an example of this fact?
A suitable counterexample can be constructed as follows.
Let $(e_n)_{n\in\omega}$ be the standard orthonormal basis of the Hilbert space $\ell_2$. For every $n\in\mathbb N$ consider the linear hull $L_n$ of the vectors $e_1,\dots,e_{n}$ in $\ell_2$. On the union $L^\infty:=\bigcup_{n=1}^\infty L_n$ consider the strongest topology that induces the Euclidean topology on each space $L_n$. It is wellknown (and easy to see) that the space $L^\infty$ is a nonmetrizable $k_\omega$space (i.e., the direct limit of a sequence of compact Hausdorff spaces). Observe that the vector $e_0$ is orthogonal to the subset $L^\infty$ in $\ell_2$.
In the Hilbert space $\ell_2$ consider the submonoid $S:=\bigcup_{n=0}^\infty(n e_0+L_n)$. Now consider the product $S\times S$ and the map $$q_S:S\times S\to SS=\mathbb Z+\bigcup_{n=1}^\infty L_n,\;\;q_S:(x,t)\mapsto xy.$$ It is easy to see that the quotient topology on $SS$ is not metrizable  it coincides with the topology of the direct limit of the sequence $(\mathbb Z+L_n)_{n=1}^\infty$.
Finally consider the topological monoid $M:=\ell_2\times S$. We claim that $M$ is a required counterexample. Indeed, consider the subgroup $MM\subset \ell_2\times\ell_2$ and the map $q:M\times M\to MM$, $q:(x,y)\mapsto xy$.
Since the space $M\times M$ is secondcountable, the set $MM$ endowed with the quotient topology is a sequential $\aleph_0$space (see Theorem 11.3 on p.494 here) and hence $MM$ has countable $cs^*$character. Assuming that $MM$ is a topological group, we can apply Theorem 1 from this paper of Banakh and Zdomskyy and conclude that $MM$ is either metrizable or contains an open $k_\omega$subgroup. But none of these two conditions applies to $MM$: this space contains a topological copy of the $k_\omega$space $L^\infty$ and hence is not metrizable and contains a copy of the Hilbert space $\ell_2$, so cannot contain an open (and hence closed) $k_\omega$subgroup.

$\begingroup$ Is it difficult to describe the universal continuous homomorphism of your $M$ to a topological group? $\endgroup$ – მამუკა ჯიბლაძე May 20 '18 at 8:57

3$\begingroup$ The universal homomorphism will be the natural map of $M$ to the topological group $\ell_2\times\mathbb Z\times L^\infty$, but the latter topological group is not sequential, so, not a quotient space of any metrizable space. $\endgroup$ – Taras Banakh May 20 '18 at 9:00