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

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.