Every group is a subgroup of a divisible group

Statement
Suppose $$G$$ is a group. Then, there exists a group $$K$$ containing $$G$$ as a subgroup such that $$K$$ is a divisible group, i.e., $$K$$ is divisible for all primes.

Related facts

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

The result can be understood in a 2 X 2 matrix of results:

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

Proof
Note that although the proof below uses primorials, we can write a similar proof that uses any sequence with the property that every prime number occurs as a divisor of infinitely many numbers in the sequence. Thus, the sequence of factorials could be used instead.

Consider the sequence of primorials: a sequence $$(a_n)_{n \in \mathbb{N}}$$ where $$a_n$$ is the product of the first $$n$$ primes. The sequence proceeds as follows:

$$2,6,30,210,\dots,$$

Now, define a sequence of groups with homomorphisms as follows:


 * $$G_0$$ is the original group.
 * For each nonnegative integer $$i$$, $$G_{i+1}$$ is defined as the external wreath product of the cyclic group $$G_i$$ with the cyclic group of order $$a_{i+1}$$ acting as a cyclic permutation group. The homomorphism from $$G_i$$ to $$G_{i+1}$$ embeds $$G_i$$ as the diagonal subgroup of a wreath product.

The direct limit of the sequence of injective homomorphisms:

$$G_0 \to G_1 \to G_2 \to \dots$$

is the desired divisible group containing $$G$$.