# Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topological groups

An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a monomorphism of topological semigroups) into an abelian Hausdorff topological group? And what if the embedding is required to be a homeomorphism when corestricted to the image (endowed with the relative topology induced by the overlying topological group)?

I'm aware of work of E. Schieferdecker (Math. Ann. 131, 1956), N. J. Rothman (Math. Ann. 139, 1960), and a few more authors on the subject, but couldn't find any counterexample in print.

Clearly, there are examples for the second question. Each Hausdorff abelian paratopological group (that is, a group endowed with a topology making the multiplication continuous) which is not a topological group (for instance, the Sorgenfrey arrow, that is the real line endowed with the Sorgenfrey topology generated by the base consisting of half-intervals $[a,b)$, $a<b$) is a counterexample. Less trivial and locally compact counterexample (which necessarily is not a group, because each locally compact paratopological group is a topological group) should be the additive semigroup of non-negative real numbers, endowed with the standard topology and then with isolated zero. (Maybe I even wrote a proof somewhere).
Update. A counterexample to your first question. For our purpose it is important that each Hausdorff space which admits a continuous injective map into a Hausdorff (para)topological group is functionally Hausdorff, because each Hausdorff topological group is Tychonoff (and each Hausdorff paratopological group is functionally Hausdorff). I recall that a space is functionally Hausdorff if for each distinct points $x,y\in X$ there exists a continuous function $f:X\to[0;1]$ such that $f(x)=0$ and $f(y)=1$. Clearly, each functionally Hausdorff space is Hausdorff. Let $X$ be an arbitrary Hausdorff space which is not functionally Hausdorff (see, for instance, quotations from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989) added at the end of the answer). Let $S(X)$ be a free abelian semigroup over a the space $X$, that is $S(X)$ is a set of formal sums $x=x_1+\cdots+x_n$ of elements of the set $X$. Clearly, $S(X)$ is a cancellative abelian semigroup. Determine a topology on the set $S(X)$ by its base consisting of all sets of the form $U_1+\cdots+U_n$, where $U_i$ are arbitrary open subsets of the space $X$ and $U_1+\cdots+U_n$ is a set of formal sums $x=x_1+\cdots+x_n$, where $x_i\in U_i$ for each $i$. Clearly, $S(X)$ is a topological semigroup. Since the space $X$ is naturally homeomorphic to a subspace of $S(X)$, the space $X$ is not functionally Hausdorff. It rests to show that the space $S(X)$ is Hausdorff. Let $x=x_1+\cdots+x_n$ and $y=y_1+\cdots+y_m$ be arbitrary distinct points of the set $S(X)$. Let $\{z_1,\dots, z_k\}=\{x_1,\dots, z_n\}\cup\{y_1,\dots, y_m\}$. Since the space $X$ is Hausdorff, there exist mutually disjoint neighborhoods $U(z_i)\ni z_i$. Put $U(x)=U(x_1)+\cdots+U(x_n)$ and $U(y)=U(y_1)+\cdots+U(y_m)$. Define a (not necessarily continuous) homomorphism $h:S(X)\to S^0(X)$ (where $S^0(X)=S(X)\cup \{0\}$ is the semigroup $S(X)$ with an added zero element $0$ such that $0+x=x+0=x$ for each element $x\in S(X)$) on the set $X$ of generators of the semigroup $S(X)$ by putting $h(z)=z_i$ if $z\in U(z_i)$ for some $i$ and $h(z)=0$, otherwise. Since the map $h$ is homomorhism, $h(U(x))=\{x\}$ and $h(U(y))=\{y\}$, so the neighborhoods $U(x)$ and $U(y)$ are disjoint.
• @SalvoTringali Suppose that the additive semigroup $S$ of non-negative real numbers endowed with the topology described above is a subgroup of a topological group $G$. Then there exists a neighborhood $U$ of the common zero $0$ of the semigroup $S$ and the group $G$ such that $U\cap S=\{0\}$. Since $1-1=0$, the continuity the group operation on $G$ implies that there exists a neighborhood $V$ of the zero such that $(1+V)-1\subset U$. Then there exists a natural number $n$ such that $1/n\in V$. Then $1/n=(1+1/n)-1\subset ((1+V)-1)\cap S\subset U\cap S=\{0\}$, a contradiction. – Alex Ravsky Jan 30 '16 at 7:17