A subset $S\subseteq G$ <topologically generates> $G$ exactly when
$$
\langle p_j(S)\rangle=p_j(G)
$$
for every $j$. In the usual presentation by surjective finite quotients this reads $\langle p_j(S)\rangle=G_j$. Indeed, a subgroup is dense exactly when its image in every finite discrete quotient is the whole quotient. This is the <finite-quotient criterion for topological generation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 2 / a / i / Solution

For a <profinite group> $G=\varprojlim_jG_j$, a subset $S\subseteq G$ is a <topological generating set> exactly when $\langle p_j(S)\rangle=p_j(G)$ in every finite quotient. Indeed, a subgroup of a profinite group is dense exactly when its image in every finite continuous quotient is surjective.

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