Is the infinite dihedral group an inverse limit of the finite dihedral groups? The p-adic numbers are the inverse limit of the rings $\mathbb Z / p^n \mathbb Z$. Can the infinite dihedral groups be construed as some sort of inverse limit of finite dihedral groups?
 A: For $n \geq 1$ we have $D_n = \langle x,y : x^n = y^2 = (xy)^2 = 1 \rangle = \mathbb{Z}/n\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$. For $n|m$ there is a canonical surjective homomorphism $D_m \to D_n$, $x \mapsto x$, $y \mapsto y$. The limit of this diagram is not $D_{\infty} = \langle x,y : y^2 = (xy)^2 = 1 \rangle = \mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$. Instead, it is $\widehat{\mathbb{Z}} \rtimes \mathbb{Z}/2\mathbb{Z}$, where $\widehat{\mathbb{Z}} = \varprojlim_n \,\mathbb{Z}/n\mathbb{Z}$. This is a profinite group which contains $D_{\infty}$ as a dense subgroup.
A: If "infinite dihedral group" has subgroup $\mathbb Z$, then the answer to the literal question is "no", as is the answer to the modified question (with limit replaced by colimit). In any case, we can see what the plausible limit and colimit really give.
The (projective) limit of $\mathbb Z/n$ with "inverse" adjoined gives $\widehat{\mathbb Z}$ (the product of all $p$-adic integers) with an inverse adjoined.
For the colimit, treat the cyclic case first, map $\mathbb Z/m$ to $\mathbb Z/mn$ by $\ell\mod n\to m\ell \mod mn$. Then take the ascending union of all the $\mathbb Z/n$'s, and we get $\mathbb Q/\mathbb Z$.
The dihedral case is similar, with an extra "reflection" element tagging along.
