Let $C_n$ be a cyclic group. Show that $C_6 ≅ C_3 × C_2$.

Question

Let be $C_n$ a cyclic group. Show that $C_6$ $\cong$ $C_3$ $\times$ $C_2$ .

Answer 1

Think about order of $(1,1) \in C_3 \times C_2$ may guide you to find the answer.

Answer 2

What about $$[k]_6\mapsto ([k]_2,[k]_3) \ \ ?$$

Answer 3

Since it is cyclic, it is abelian. By Cauchy's theorem there exist subgroups $A$ and $B$ such that $|A|= 2$ and $|B|=3$. Since $\gcd(2,3) = 1$, we have that the intersection of $A$ and $B$ is the identity, therefore we have that $AB=C_6$ ($AB$ is a group because $ C_6 $ is abelian). We have that every element $c \in C_6 $ is of the form $ab$, where $a$ is in $A$ and $b$ in $B$, now, use the homomorphism $f(c) = (a,b)$.