Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 1

For an inverse system of sets $(X_j,f_{ij})$ indexed by a directed set $J$, the inverse limit is the set of compatible tuplesIts projection to $X_j$ sends $(x_i{)}_i$ to $x_j$.

Give each nonempty finite set $X_j$ the discrete topology. The product space $X={\prod }_j{X}_j$ is compact by the Tychonoff theorem. For each $i\le j$, the compatibility condition $f_{ij}(x_j)=x_i$ defines a closed subset $C_{ij}\subseteq X$.

These sets have the finite intersection property. Indeed, for finitely many conditions choose an index $k$ above every index occurring in them, choose any $x_k\in X_k$, and use the transition maps from $k$ to define all required coordinates; choose the remaining coordinates arbitrarily. Compactness therefore givesThis is the nonemptiness theorem for inverse limits of finite sets.

The product ${\prod }_j{G}_j$ of the finite groups with their discrete topology is a topological group under coordinatewise multiplication and inversion. The compatibility equations defining $G={\underleftarrow{\lim }}_j{G}_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the subspace topology on $G$ are continuous. Hence $G$ with its standard inverse-limit topology is a topological group.

For fixed $h\in G$, the conjugation map is the compositefollowed by multiplication. Inversion, constant maps, diagonal maps, and multiplication are continuous in a topological group, so $c_h(g)=gh{g}^{-1}$ is continuous.

The profinite group $G$ is compact, and the preceding part shows thatis its image. By the continuous image of a compact space theorem, the conjugacy class is compact. Since a profinite group is Hausdorff, every compact subset is closed, so ${\mathrm{Cl}}_G(h)$ is closed.

If $g=xh{x}^{-1}$ in $G$, applying each projection map gives $p_j(g)=p_j(x)p_j(h)p_j(x{)}^{-1}$ in $G_j$.

Conversely, suppose $p_j(g)$ and $p_j(h)$ are conjugate for every $j$. Define the nonempty finite setEvery transition map $G_j\to G_i$ carries $X_j$ into $X_i$, so the $X_j$ form an inverse system. By the nonemptiness theorem for inverse limits of finite sets, there is a compatible tuple $(x_j)\in {\underleftarrow{\lim }}_j{X}_j\subseteq G$. Coordinatewise equality then gives $xh{x}^{-1}=g$. This proves the finite-quotient criterion for conjugacy in a profinite group.

The canonical image of $\Gamma$ is dense in its profinite completion $\hat{\Gamma }$. Therefore every $x\in \hat{\Gamma }$ is a limit of a net $({\delta }_{\lambda })$ in $\Gamma$. Continuity of conjugation givesso every element of ${\mathrm{Cl}}_{\hat{\Gamma }}(\gamma )$ lies in the closure of ${\mathrm{Cl}}_{\Gamma }(\gamma )$. The reverse inclusion follows because the larger conjugacy class contains the smaller one and is closed by part b(iii). Hence

Suppose first that $\Gamma$ is conjugacy separable. If $h\in \Gamma$ is not conjugate to $\gamma$ in $\Gamma$, some homomorphism to a finite group sends them to nonconjugate elements. This homomorphism factors through a finite quotient of $\hat{\Gamma }$, so $h$ cannot be conjugate to $\gamma$ in $\hat{\Gamma }$. Thus

Conversely, suppose this equality holds and $h$ is not conjugate to $\gamma$ in $\Gamma$. Then they are not conjugate in $\hat{\Gamma }$. By the finite-quotient criterion for conjugacy in a profinite group, their images fail to be conjugate in some finite quotient. This is precisely conjugacy separability.