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The following exercise is from Rotman, An Introduction to the theory of groups, 4th ed, p324. "The following conditions on a group G are equivalent: (i) G is divisible, (ii) Every nonzero quotient of G is infinite, (iii) G has no maximal subgroups."

I can prove that (i) implies (ii) and that (ii) is equivalent to (iii) but I am having trouble with showing that (iii) implies (i) and (ii) implies (i). Any suggestions?

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  • $\begingroup$ Don't prove (ii) implies (i). It is enough to show (iii) implies (i) to show all are equivalent. $\endgroup$ – John Habert Feb 18 '14 at 21:10
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A more elementary approach to showing $(iii) \Rightarrow (i)$ is to show the contrapositive: Suppose $G$ is not divisible, then for some minimal positive integer $n$, $nG < G$ is a proper subgroup. Since $G$ is abelian, $nG$ is normal. Now consider the quotient $G/nG$; this is again an abelian group where every element of $G/nG$ has order dividing $n$.

To to find a proper maximal subgroup of $G/nG$ there are two cases to consider, if $n$ is prime then $\mathbb{Z}/n\mathbb{Z}$ is a field and $G/nG$ is in fact a $\mathbb{Z}/n\mathbb{Z}$-vector space, so by Zorn's lemma you can find a basis, and a maximal "codimension-1" subspace. If $n$ is composite, I leave the argument up to you, and I'm sure you can figure out the rest.

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  • $\begingroup$ Thanks! If n = ab for a, b > 1 then nG = abG = aG = G since a, b <n and n is minimal with nG < G. Although I think one could use the fact that G is divisible if and only if pG = G for all primes p and not consider n composite at all. $\endgroup$ – user129773 Feb 19 '14 at 20:36
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It's sufficient to show that iii) implies i).

Let $\Phi(G) $ be the Frattini subgroup of G, i.e. the intersection of all maximal subgroups of G. There is a lemma due to Dlab that states $$\Phi(G) = \bigcap_{p \in \mathbb{P}} \ pG$$ where $\mathbb{P}$ is the set of prime integers.

So $\Phi(G) = G \Leftrightarrow G$ has no maximal subgroups, and in this case $G = \bigcap_{p \in \mathbb{P}} \ pG$, i.e. G is divisible.

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