A criterion for the existence of unique group homomorphisms from an abelian group

Let A = \langle x_1 \rangle \times \cdots \langle x_t \rangle be a finite abelian group with |x_i| = n_i for each i.

  1. Find a presentation for A.
  2. Prove that if G is a group containing commuting elements g_1,\ldots,g_t such that g_i^{n_i} = 1 for each i, then there is a unique group homomorphism \theta : A \rightarrow G such that \theta(x_i) = g_i.

  1. We claim that A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle, where 1 \leq i,j \leq t. Indeed, the “standard basis” elements e_i, consisting of x_i in the i-th coordinate and 1 in all other coordinates, satisfy these relations.
  2. Note that every element of A can be written uniquely as (x_i^{b_i})_{i=1}^t. Define \theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}, where the product on the right hand side is as computed inside G. It is clear that \theta(x_i) = g_i for each i. Moreover, \theta is a homomorphism because the g_i commute pairwise.

    Finally, suppose \sigma is a homomorphism A \rightarrow G such that \sigma(x_i) = g_i for each i. Then \sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i} = \prod \theta(x_i)^{b_i} = \theta((x_i^{b_i})). Thus \sigma = \theta (products computed inside G), so that \theta is unique.

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