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. 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.

References

• Adjunction of elements to groups by B. H. Neumann, Journal of the London Mathematical Society, ISSN 14697750 (online), ISSN 00246107 (print), Volume 18, Page 12 - 20(Year 1943): ^{Official copyMore info}
• Wreath products and p-groups by Gilbert Baumslag, Proceedings of the Cambridge Philosophical Society, Volume 55, Page 224 - 231(Year 1959): ^{ungated PDFMore info}, Corollary 4.3 (Page 14 of the paper)