Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 3 / c

If $\gcd (m,|H|)=1$, choose $r$ with $mr\equiv 1(\mathrm{mod}|H|)$ by the Bezout identity. The Lagrange theorem gives $x^{|H|}=1$ for every $x\in H$, so $(x^m{)}^r=x$ and $(x^r{)}^m=x$. Thus $x\mapsto x^m$ is bijective, even though it need not be a homomorphism.

Conversely, if a prime $p$ divides both $m$ and $|H|$, the Cauchy theorem for groups gives $x\ne 1$ with $x^p=1$. Then $x^m=1$, so the power map sends both $x$ and the identity to the identity and is not injective. This proves the power-map criterion for a finite group.

Necessity follows by applying $p_j$ to an equality $x^m=g$. Conversely, suppose every finite quotient contains an $m$th root of $p_j(g)$ and defineThese are nonempty finite sets, and the transition maps preserve them. The nonemptiness theorem for inverse limits of finite sets supplies a compatible tuple $x\in {\underleftarrow{\lim }}_j{X}_j=G$, for which $x^m=g$. This is the finite-quotient criterion for roots in a profinite group.

If $m$ is coprime to every $|G_j|$, the power-map criterion for a finite group says that every finite-quotient power map is bijective. Part ii gives surjectivity of $f_m:G\to G$. If $x^m=y^m$, then $p_j(x)=p_j(y)$ for every $j$ by injectivity in $G_j$, and hence $x=y$. Thus the continuous power map is bijective.