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Let $G$ and $H$ are two divisible groups that each of which is is isomorphic to a subgroup of the other, then $G\cong H$.

What I've done is to use the injective property for both groups:

  1. $G\cong K\le H$ so we have $G\stackrel{\iota}{\hookrightarrow} H$ and $G\stackrel{id}{\longrightarrow} G$ and then there exists $H \stackrel{\phi}{\longrightarrow} G$ that $\phi\circ i=id|_G$.

  2. $H\cong S\le G$ so we have $H\stackrel{\iota}{\hookrightarrow} G$ and $H\stackrel{id}{\longrightarrow} H$ and then there exists $G \stackrel{\psi}{\longrightarrow} H$ that $\psi\circ i=id|_H$.

Is my approach right? May I ask you what will be happen if we omit the adjective divisible? Thanks.

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How do you intend to conclude from this that $G$ and $H$ are isomorphic? You need to produce two mutually inverse homomorphisms and with your argument there's no reason that $\phi$ and $\psi$ will be inverses of each other. Concerning the last question: If you omit the divisibility hypothesis then the result is wrong. For example, take two free groups of different ranks $\geq 2$ then each one is a subgroup of the other but they aren't isomorphic. – commenter Oct 13 '12 at 13:23
@commenter: I know that point and thank you. I just noted what I did. In fact 1 and 2 are my conclusions not the solution. Thanks. – Babak S. Oct 13 '12 at 13:41
up vote 3 down vote accepted

The key here is the classification of divisible groups. Every divisible group is a direct sum of copies of $\mathbb{Q}$ and $\mathbb{Z}/{p^\infty}$ for any prime $p$, where $\mathbb{Z}/{p^\infty}$ denotes the group of $p^n$-torsion elements on the unit circle.

None of these groups can map nontrivially into each other, except for $\mathbb{Q}$, which cannot inject into a sum of the others because the others are all torsion groups (check this!). Thus, if we have an injection from one divisible group to another, each summand in the domain will have its image inside summands of the codomain that are isomorphic to it.

You are now reduced to proving that the cardinality of the collection of summands of a given isomorphism type in the domain is no more than what you have in the codomain. This is easy for $\mathbb{Q}$ since you can view it as a $\mathbb{Q}$-vector space. For each $p$, you can restrict your attention to $p$-torsion elements and apply the same argument over $\mathbb{Z}/p$.

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