Assume $X$ is a Tychonoff space. Then $A(X)$ is the free topological abelian group over $X$. I know that $A(X)$ is Hausdorff and the canonical embedding from $X$ to $A(X)$ is a topological embedding.


Now consider the subgroup $N:=2A(X)=\left\{g+g\ \colon\ g\in A(X)\right\}$.
The quotient $B(X):=A(X)/N$ is the free topological Boolean group. I want to show that $B(X)$ is Hausdorff which is equivalent to say that $N$ is closed in $A(X)$. Unfortunately, I do not know how show the closedness of $N$ since I can only describe the topology on $A(X)$ by the universal property and I do not know how that helps me here... Or is this maybe not true for a general Tychonoff space $X$ and one needs additional properties on $X$ like hereditarily disconnected?


I have found a reference on Free Boolean Groups


Genze, L.V. Free Boolean topological groups. Vestn. Tomsk. Gos. Univ. 2006, 290, 11–13


which maybe could answer my question but I am unable to find this text on math.sci.net (or anywhere else).


Thank you!



Why are free Boolean topological groups Hausdorff?
You can use the universality property with the following Boolean group as codomain: $B$ is the measure algebra over the unit interval (the quotient of the $\sigma$-algebra of Lebesgue measurable sets by the ideal of sets of measure zero), with symmetric difference as operation this is a Boolean group and $d(A,B)=\lambda(A\Delta B)$ defines a metric that will turn it into a topological group.
This group contains a copy of the unit interval, namely the set $\{[0,t]:0\le t\le 1\}$.
Since $X$ is Tychonoff you now have enough continuous functions to separate all non-trivial words from the empty word.
This survey by Sipacheva gives more information: [*Free Boolean Topological Groups*, Axioms 4 (2015) 492-517](https://doi.org/10.3390/axioms4040492)


