Given the abelian group : A=\mathbb{Z}_{36} ×\mathbb{Z}_{96}×\mathbb{Z}_{108}
I need to write the canonical form of $18A$ and $A / 18A$
Here is my calculation ,using the followings:
n(B\times C)=nB×nC
$m\mathbb{Z}_n=(m,n) \mathbb{Z}_n\cong\mathbb{Z}_{n / (m,n)}$
$18A=18\mathbb{Z}_{36}\times 18\mathbb{Z}_{96}\times 18\mathbb{Z}_{108} \cong \mathbb{Z}_{2} \times 18\mathbb{Z}_{96}\times \mathbb{Z}_{6}$
Since 96/18 is not an integer, we take care of the $18\mathbb{Z}_{96}$ element using:
$m\mathbb{Z}_n=(m,n) \mathbb{Z}_n\cong\mathbb{Z}_{n / (m,n)}$ (By the way , is there any other way ???!)
$18A=Z_2\times 18Z_{96}\times Z_6=Z_{2}\times Z_{16}\times Z_{6}$
The problem starts here , when I want to calculate $A / 18A$:
$\begin{align*}A / 18A&=\mathbb{Z}_{36}\times \mathbb{Z}_{96}\times \mathbb{Z}_{108} / (18\mathbb{Z}_{36}\times 18\mathbb{Z}_{96}\times 18\mathbb{Z}_{108} )\\ &=\mathbb{Z}_{36} / 18\mathbb{Z}_{36} \times \mathbb{Z}_{96} / 18\mathbb{Z}_{96} \times \mathbb{Z}_{108} / 18\mathbb{Z}_{108} = \;??? \end{align*}$ How do I continue from here ?
Regards