# Do profinite completions commute with direct products?

I am trying to prove that

$$\widehat{\mathbb{Q}}_p^\times\cong\widehat{\mathbb{Z}}\times\mathbb{Z}_p^\times,$$

where $$\widehat{G}$$ denotes the profinite completion of a group $$G$$. Using $$\mathbb{Q}_p^\times\cong\mathbb{Z}\times\mathbb{Z}_p^\times$$, this would be easy if profinite completion commuted with (at least finite) direct products, but is this true?

I have not been able to find a reference for the validity/falsity of that fact and have trouble proving it mainly because there is not really an easy description of the subgroups of $$G\times H$$ in terms of the subgroups of $$G$$ and $$H$$.

EDIT: Here is my attempt at a proof of this at least for abelian groups $$G, H$$ (is it correct?):

We claim that $$\widehat{G}\times\widehat{H}$$ satisfies the universal property of the profinite completion of $$G\times H$$. Namely, let $$N$$ be any normal subgroup of $$G\times H$$ of finite index. By Goursat's lemma, there are subgroups $$G_2\unlhd G_1\leq G$$ and $$H_2\unlhd H_1\leq H$$ and an isomorphism $$\phi: G_1/G_2\rightarrow H_1/H_2$$ such that $$\begin{equation*} N=\{(g, h)\in G_1\times H_1: \phi(gG_2)=hH_2\}. \end{equation*}$$ Notice that $$N$$ has index $$|G_1/G_2|=|H_1/H_2|$$ in $$G_1\times H_1$$ and $$G_1\times H_1$$ has index $$[G:G_1][H:H_1]$$ in $$G\times H$$; hence, as $$[G\times H: N]$$ is finite, we must have that $$[G:G_2]$$ and $$[H:H_2]$$ are finite. In particular, $$N$$ contains $$G_2\times H_2$$. Using the projections from $$\widehat{G}$$ and $$\widehat{H}$$ to $$G/G_2$$ and $$H/H_2$$, respectively, we obtain a morphism $$\widehat{G}\times\widehat{H}\rightarrow (G\times H)/(G_2\times H_2)$$ and by composition with the canonical projection a morphism $$\phi_N$$ into $$(G\times H)/N$$. It is easy to see that $$\phi_N$$ does not depend on the choice of subgroups $$G_2, H_2$$ with finite index in $$G, H$$ such that $$G_2\times H_2\leq N$$ and hence it follows that the $$\phi_N$$ are compatible with the projections among the $$(G\times H)/N$$.

Now let $$Z$$ be a group with morphisms $$f_N: Z\rightarrow (G\times H)/N$$ for any normal subgroup $$N$$ of finite index such that the $$f_N$$ are compatible with the projections between the groups $$(G\times H)/N$$. Then any morphism $$f: Z\rightarrow \widehat{G}\times\widehat{H}$$ such that $$\phi_N\circ f=f_N$$ induces a morphism $$f_G: Z\rightarrow\widehat{G}$$ compatible with the maps $$Z\rightarrow (G\times H)/(N_G\times H)\cong G/N_G$$ and $$\widehat{G}\rightarrow G/N_G$$ for $$N_G$$ normal in $$G$$ with finite index and thus $$f_G$$ is unique by the universal property of the profinite completion. Analogously, $$f_H$$ is unique, so $$f$$ is unique.

Conversely, again by the universal property of the profinite completion, we obtain a morphism $$f_G: Z\rightarrow\widehat{G}$$ compatible with the maps $$Z\rightarrow (G\times H)/(N_G\times H)\cong G/N_G$$ and $$\widehat{G}\rightarrow G/N_G$$ for $$N_G$$ normal in $$G$$ with finite index and analogously a morphism $$f_H$$. We claim that the morphism $$f: Z\rightarrow\widehat{G}\times\widehat{H}$$ obtained by putting them together satisfies $$\phi_N\circ f=f_N$$ for $$N$$ normal in $$G\times H$$ of finite index. Namely, by the above, any such $$N$$ contains a subgroup $$N_G\times N_H$$ for $$N_G, N_H$$ normal in $$G, H$$ of finite index and we know that $$\phi_{N_G\times N_H}\circ f=f_{N_G\times N_H}$$ by construction of $$f$$. The equality $$\phi_N\circ f=f_N$$ then follows by composing both sides with the projection $$(G\times H)/(N_G\times N_H)\rightarrow (G\times H)/N$$ on the left.

• Related: math.stackexchange.com/q/2751883/96384, but as you say here the question becomes more intricate because we are looking at a different kind of completion, involving a bigger class of subgroups. Commented Jun 1, 2022 at 15:16
• @reuns Am I right in assuming that this follows from the fact that if $G$ is abelian, then any subgroup $N$ of finite index contains some $G^n$? Commented Jun 2, 2022 at 4:14
• Oops yes but $G^n$ doesn't have to be finite index, forget it. (Of course it holds for $\Bbb{Q}_p^\times$) Commented Jun 2, 2022 at 13:40

$$\mathrm{Hom}_{\mathbf{Grp}}(G \times H,K)\\ \cong \{(f,g) \in \mathrm{Hom}_{\mathbf{Grp}}(G,K) \times \mathrm{Hom}_{\mathbf{Grp}}(H,K) \mid \forall x \in G\ \forall y \in H: f(x)g(y)=g(y)f(x) \}$$ The same also holds in the category of topological groups.
Using this, we prove that $$\widehat{G} \times \widehat{H}$$ satisfies the universal property of the profinite completion. Let $$K$$ be a profinite group and let $$f:G \times H \to K$$ be a group homomorphism. Then $$f$$ corresponds to a pair $$(g,h) \in \mathrm{Hom}_{\mathbf{Grp}}(G,K) \times \mathrm{Hom}_{\mathbf{Grp}}(H,K)$$ as above. By the universal property of profinite completions, we obtain group homomorphisms $$\widehat{g}:\widehat{G} \to K$$ and $$\widehat{h}:\widehat{H} \to K$$. Note that these homomorphisms satisfy $$\forall \widehat{x} \in \widehat{G}\ \forall \widehat{y} \in \widehat{H}:\widehat{g}(\widehat{x})\widehat{h}(\widehat{y})=\widehat{h}(\widehat{y})\widehat{g}(\widehat{x})$$ Indeed, this may be proved by taking nets $$(x_i)_{i \in I}, (y_j)_{j \in J}$$ in $$G$$ and $$H$$ respectively such that $$x_i$$ converges to $$\widehat{x}$$ and $$y_j$$ converges to $$\widehat{y}$$. Then $$(g(x_i)h(y_j))_{(i,j) \in I \times J}$$ converges to $$\widehat{g}(\widehat{x})\widehat{h}(\widehat{y})$$. But we have $$g(x_i)h(y_j)=h(y_j)g(x_i)$$ for all $$i,j$$, so it also converges to $$\widehat{h}(\widehat{y})\widehat{g}(\widehat{x})$$. As limits of nets in a Hausdorff space are unique, we get $$\widehat{g}(\widehat{x})\widehat{h}(\widehat{y})=\widehat{h}(\widehat{y})\widehat{g}(\widehat{x})$$ as claimed.
Thus we obtain a morphism of topological groups $$\widehat{G} \times \widehat{H} \to K$$ that makes the necessary triangle commute. Because the image of $$G \times H$$ in $$\widehat{G} \times \widehat{H}$$ is dense and $$K$$ is Hausdorff, this homomorphism is unique. Thus $$\widehat{G} \times \widehat{H}$$ satisfies the universal property of profinite completion and we get
$$\widehat{G} \times \widehat{H} \cong \widehat{G \times H}$$ as desired.