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?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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.