Profinite completion

Remarks

$\hat G$ is a profinite group, and $\iota(G)$ is dense in $\hat G$. The kernel of $\iota$ is the intersection of all $N \in \mathcal{N}$, so $\iota$ is injective if and only if $G$ is residually finite. Example: for $G = \Z$, the finite-index subgroups are $n\Z$ with $n \ge 1$, and $\hat\Z = \varprojlim \Z/n\Z$; by the Chinese remainder theorem it is isomorphic to $\prod_p \Z_p$. Here $\iota$ is injective, since $\bigcap_n n\Z = \set{0}$. Non-example of injectivity: $(\Q, +)$ has no proper subgroup of finite index, so $\hat\Q$ is trivial and $\iota$ is not injective. Universal property: every homomorphism from $G$ to a profinite group $P$ factors uniquely through $\iota$ by a continuous homomorphism $\hat G \to P$.