Isomorphisms of cyclic group products
Isomorphisms to a product of cyclic groups
If \( n \) and \( m \) are positive integers with \( \operatorname{gcd}(n, m) =
1\),
then \( \cat{C_{nm}} \cong \cat{C_n} \cat{C_m} \).
Proof on back side.
The proposition refers to cyclic groups, defined below:
Cyclic group
A group is cyclic iff it is isomorphic to \( \mathbb{Z} \) or \( \mathbb{Z}/n\mathbb{Z}\).
Proof essense
Consider \( nm = 6 \), \( n = 3 \) and \( m = 2 \). \( [1]_{6} \) generates \( \cat{C_6} \). \( [1]_{3} \) generates \( \cat{C_3} \). We can map \( \cat{C_6} \) to \( \cat{C_3} \) by mapping \( [1]^2_{6} \) to \( [1]_{3} \) to create a morphism \( \cat{C_6} \to \cat{C_3} \). Similarly, we can map \( [1]^3_{6} \) to \( [1]_{2} \) to create a morphism \( \cat{C_6} \to \cat{C_2} \).
Because 3 and 2 have no common divisor, incrementing \( [1]_3 \) and \( [1]_2 \) vicariously through \( [1]_{6} \) must continue for the full 6 elements of \( \cat{C_6} \) before both \( [1]_3 \) and \( [1]_2 \) loop back together on the identity, \( [0]_6 \). This means the morphism from \( \cat{C_6} \) to \( \cat{C_3} \times \cat{C_2} \) must be injective. Furthermore, \( \cat{C_3} \times \cat{C_2} \) only has 6 elements, so the morphism is also surjective. The morphism is thus a bijection.
TODO: actual proof
\( \cat{C_3} \times \cat{C_2} \) is a cyclic group
The cyclic group \( \cat{C_6} \) is isomorphic to \( \cat{C_3} \times \cat{C_2} \), so the latter is a cyclic group too.
3-4 example
