10) View the group \( G \) in Problem 9 as a \( \mathbb{Z} \)-module. For this module (i) find its invariant factor
decomposition; (i) find its elementary divisor decomposition.
Proof. We have \( \mathbb{Z}_{1000}^\times = \mathbb{Z}_{2^3}^\times \times \mathbb{Z}_{5^3}^\times, \), where
\( \mathbb{Z} / k \mathbb{Z} = \mathbb{Z}_k, \) and \( \mathbb{Z}_k^\times \) is the multiplicative group of units in \( \mathbb{Z}_k \).
Now \( \mathbb{Z}_{2^3}^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \) and \( \mathbb{Z}_{5^3}^\times \cong \mathbb{Z}_{100}. \)
Therefore \( \mathbb{Z}_{1000}^\times \cong {\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{100}}. \) This is the invariant
factor decomposition with \( n_1 = {2^2 \cdot 5^2} = 100, \; n_2 = 2, \; n_3 = 2. \)
Given the basic principle that \( \mathbb{Z}_{mn} \cong {\mathbb{Z}_{m} \times \mathbb{Z}_{n}} \) when \( \text{gcd} (m, n) = 1, \)
there is the further decomposition \( \mathbb{Z}_{1000}^\times \cong {\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{4} \times \mathbb{Z}_{25}}. \)
Given the prime powers involved, this last must be the unique elementary divisor decomposition. QED.
Note: The breakdown of \( \mathbb{Z}_n^\times \) in terms of the prime power factorization of \( n \), and further characterization
of \( \mathbb{Z}_{p^\alpha}^\times, \) is given in §9.5, Corollary 20, of Dummit & Foote.