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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

  • 3
    As has been mentioned a few times, there isn't such a thing as **the** "canonical form" of an abelian group. There are *two* "standard decompositions" (which may be called canonical form), but given that there are **two** of them (into primary divisors and into invariant factors), you cannot talk about "the" canonical form, and you should always specify **which** of them you mean.2012-03-17

2 Answers 2