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We know that, any abelian group G can be written as a direct sum of dG and a reduced subgroup R (where dG is the subgroup generated by all divisible subgroups of G). Is it true that R is isomorphic to G/dG?

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Yes, it is true. – Joel Cohen May 7 '11 at 14:40
up vote 4 down vote accepted

If G is written as the direct sum A⊕B, then B is always isomorphic to G/A.

Write elements of G as ordered pairs (a,b) and define the homomorphism from G to B by mapping (a,b) onto b. The kernel is precisely those (a,b) with b=0, that is, the kernel is precisely A⊕0 = A. The image is all b for b in B, that is, the image is precisely B. The first isomorphism theorem then gives that the image, B, is isomorphic to the source mod the kernel, G/A.

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