# Power maps of contractible topological groups

I'm little bit puzzled with the following question, I was wondering if someone could help with a suggestion or a hint.

Let start with some notations. Suppose that $$G$$ is a Hausdorff topological group such that

1. As a space $$G$$ is contractible.
2. As a group $$G$$ is isomorphic to the free product $$\mathrm{Q}\ast F(X)$$ where $$\mathrm{Q}$$ is the additive abelian group of rational numbers and $$F(X)$$ is a free group generated by a set $$X$$.
3. We assume that $$\mathrm{Q}$$ is a topological subgroup of $$G$$ in an obvious way and that $$H=\{1\}\ast_{\mathrm{Q}} G$$ is a retract of a free topological group (generated by a CW-complex).

Let $$\mathbf{Ab}: \mathsf{Tgr}\rightarrow \mathsf{AbTgr}$$ be the abelianization functor from the category of topological groups to abelian topological groups.

Since $$G$$ is a contractible group, for any natural number $$n>0$$ the power maps $$p^{n}: G\rightarrow G$$, $$t\mapsto t^n$$ are weak homotopy equivalences of underlying topological spaces.

My question is the following: is there a hope that the power maps: $$p^{n}: \mathbf{Ab}(G)\rightarrow \mathbf{Ab}(G)$$ induce weak homotopy equivalences of underlying topological spaces? The same question of $$H$$ and power maps $$p^{n}: H\rightarrow H$$, are they weak homotopy equivalences ?

Thank you in advance for any help!

• I'm having a hard time picturing such a $G$. For one thing, a Hausdorff contractible space is uncountable, so I guess the set $X$ must be uncountable (and have an interesting topology). For another, you say that $G$ is an amalgamated product of $Q$ and $F(X)$ but do not mention what subgroup it's amalgamated over. Could you give an example of such a $G$? – Kevin Casto May 1 at 23:07
• This a list of questions rather than a helpful comment. (1) Are there example of such $G$ for some specific examples of $X$? So, what is $G$ if $X=\{a,b\}$ or $X=\mathbb{R}$. (2) As to the above comment, do the above desired properties of $G$ depends on whether or not $X$ is countable or not? (3) If you choose $X$ to be a finite set then is there any reason to believe that the truth of `hope' depends on the divisibility of $|X|$ by $p$ or $n$? – user51223 May 2 at 5:48
• @KevinCasto roughly speaking, an ideal example is homotopy theoretical, take $\mathrm{Q}\rightarrow \ast$ the obvious homomorphism of groups. There is a factorization $Q\rightarrow G\rightarrow \ast$ where the first map is a cofibration (of topological groups) and the second map is a trivial fibration. – Ilias A. May 2 at 17:35
• @user51223 I just wrote a comment of an ideal example. in that case my question has a positive answer. the set X has not to be countable in general. – Ilias A. May 2 at 17:38
• @IliasAmrani What is the set $X$ in this case, and what topology does it have? Why is $G$ the (amalgamated? over what?) free product of $Q$ and $F(X)$? – Kevin Casto May 2 at 19:07