= Solution
If $\gcd(m,|H|)=1$, choose $r$ with $mr\equiv1\pmod{|H|}$ 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 $x\ne1$ 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>.
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