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Math and science::Algebra::Aluffi

# 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.

### $$\cat{C_3} \times \cat{C_2}$$ 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.

Aluffi, p69