Every pi-group is a subgroup of a divisible pi-group

Statement
Suppose $$G$$ is a periodic group with $$\pi$$ the set of primes that occur as divisors of the orders of elements of $$G$$. Then, there exists a group $$K$$ containing $$G$$ satisfying the following conditions:


 * 1) $$K$$ is a periodic group and the set of primes dividing the orders of elements of $$K$$ is $$\pi$$. Thus, $$K$$ is a $$\pi$$-group.
 * 2) $$K$$ is a divisible group for all primes. Note that divisibility by primes outside $$\pi$$ is automatic, so the interesting claim is that $$K$$ is $$\pi$$-divisible.

Related facts

 * Every group is a subgroup of a divisible group

Facts used

 * 1) uses::Diagonal subgroup of a wreath product with a cyclic permutation group is divisible by all primes dividing the order of the cyclic group
 * 2) When we do such a wreath product, we do not introduce any new prime divisors of orders of elements

Proof
The proof method is similar to that for every group is a subgroup of a divisible group.