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I am trying to prove this:

If a divisible group $E$ containining $A$ is minimal divisible then $A$ is an essential subgroup of $E$.

Let $ < c > =C, \ C\cap A = 0$. Without loss of generality we can say that $ o (c) = \infty$ or $o (c) = p$.

Now I want to say that $C$ can be embedded in a subgroup $B \subset E$, such that $B \simeq \mathbb{Q} $, or $B \simeq \mathbb{Z}_{p^\infty }$. But I can't figure out how to proove it.

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I guess $E$ is commutative. Right? What is "minimal divisible"? – Mark Sapir Feb 9 '12 at 15:35

Just got it. Stupid question.

Groups are abelian.

Given $A$, call the divisible group $Ε$ containing $A$ minimal divisible if no proper divisible subgroup of $Ε$ contains $A$.

Every divisible group that contains $A$, also contains the minimal divisible of $A$. So there is divisible subgroup $B$, such that $C\subset B\subset E$.

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Your $E$ is what would be called the divisible hull (or envelope) of $A$. Essentially by definition, $E$ is essential over $A$. This is the result of several equivalent conditions that are summarized in Section 4.2 of Lambek's "Lectures on Rings and Modules." For abelian groups, or modules over a PID, injective is equivalent to divisible. – Chris Leary Feb 10 '12 at 13:32

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