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An abelian group $A$ is $p$-divisible when multiplication by $p$ is surjective, equivalently $A/pA=0$. This does not mean it has no $p$-torsion: $\mathbb{Q}/\mathbb{Z}$ is divisible and has elements of every finite order. The Kummer cohomology divisibility criterion relates this property to Galois cohomology.

We first give the common reduction for the three conditions, then prove the cycle from (i) to (ii), from (ii) to (iii), and from (iii) back to (i). Assume here that $\mathrm{char}K\ne p$. The Galois module of roots of unity ${\mu }_p$ fits into the Kummer sequencewhich is surjective on the right because $p$th-root polynomials are separable. Restrict it to the absolute Galois group of any algebraic extension $L/K$. The long exact sequence gives the Kummer cohomology divisibility criterionContinuous cohomology of a profinite group in positive degree with discrete coefficients is torsion: a continuous normalized cocycle becomes zero on a sufficiently small open subgroup, and restriction followed by corestriction multiplies its class by the finite index. For an abelian group, p-primary torsion consists of elements killed by powers of $p$, while p-divisibility means that multiplication by $p$ is surjective. For a torsion group $A$, $A\{p\}=0$ if and only if $A[p]=0$, because a nonzero element of $p$-power order has a multiple of exact order $p$. Consequently the two conditions on the multiplicative cohomology groups are jointly equivalent toThe coefficient module $K_s^{\times }$ is used with its restricted Galois action as in the question. Purely inseparable algebraic extensions give equivalent absolute Galois groups; the ${\mu }_p$ reduction therefore also covers those extensions.

The cohomological dimension at a prime condition ${\mathrm{cd}}_p(K)\le n$ means that $H^q(G_K,M)=0$ for every discrete $p$-primary torsion Galois module $M$ and every $q>n$. Cohomological dimension does not increase on a closed subgroup. Since absolute Galois groups of algebraic extensions identify with closed subgroups, up to the purely inseparable identification just noted, condition (i) gives $H^{n+1}(L,{\mu }_p)=0$ for every algebraic $L/K$. The Kummer reduction proves (i) implies (ii).