# Some Problems Involving Homomorphisms of Abelian Groups

EDIT: G(p) = {x in G : |x| = p^k for some k greater than or equal to 0}

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#### Solution Preview

1. Let g be an element of G(p). Then |g| = p^k for some nonnegative integer k. Now |alpha(g)| must divide |g| = p^k, so we have |alpha(g)| = p^j for some nonnegative integer j <= k. Therefore, alpha(g) must be an element of H(p), so alpha[G(p)] must be a subset of H(p).

2. First we show that if G and H are isomorphic finite abelian groups and p ...

#### Solution Summary

In this solution, we solve some problems involving homomorphisms an isomorphisms of finite abelian groups.

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