There is a theorem that states that $\mathbb Z_{m} \times \mathbb Z_n$ is isomorphic to $\mathbb Z_{mn}$ if the $gcd(m,n) = 1$ because otherwise $\mathbb Z_{m} \times \mathbb Z_n$ would not be cyclic. Would deriving that proof and saying that $\mathbb Z_{12} \times \mathbb Z_5$ is isomorphic to $\mathbb Z_{60}$ by stating that $gdc(12,5) = 1$ be a sufficient proof?
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1Yes, the proof is more general. – ZirconCode Nov 05 '16 at 23:19
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1You're "specializing" a theorem, i.e. giving values to $n,m$ for which the theorem is valid, whence the conclusion is correct. – Pedro Nov 05 '16 at 23:19
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Yes, that's an application of said theorem. There is a more thorough explanation about when, and when not (and why not), the product of cyclic groups is cyclic. – amWhy Nov 05 '16 at 23:26