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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

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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. –  Arturo Magidin Mar 17 '12 at 20:46

2 Answers 2

In fact, your $A$ is not decomposed into either invariant factors or into primary divisors. I think it might be simpler using the primary divisors. You could make do with the invariant factors as well.

Since $36 = 2^2\times 3^2$, then $\mathbb{Z}_{36} = \mathbb{Z}_{2^2}\oplus\mathbb{Z}_{3^2}$.

Since $96 = 2^5\times 3$, then $\mathbb{Z}_{96} = \mathbb{Z}_{2^5}\oplus\mathbb{Z}_3$.

Since $108 = 2^2\times 3^3$, then $\mathbb{Z}_{108} = \mathbb{Z}_{2^2}\oplus\mathbb{Z}_{3^3}$.

So $$A = \mathbb{Z}_{36}\oplus\mathbb{Z}_{96}\oplus \mathbb{Z}_{108} \cong \mathbb{Z}_{2^2}\oplus\mathbb{Z}_{2^2}\oplus\mathbb{Z}_{2^5}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_{3^2}\oplus\mathbb{Z}_{3^3}.$$

Now, $18 = 2\times 3^2$; the factor of $3^2$ is irrelevant to the $2$-groups, and the factor of $2$ is irrelevant to the $3$-groups (as I noted in the Hint here). We have $$2\mathbb{Z}_{2^2}\cong \mathbb{Z}_2,\qquad 2\mathbb{Z}_{2^5}\cong\mathbb{Z}_{2^4}$$ and $$3^2\mathbb{Z}_3=\{1\},\qquad 3^2\mathbb{Z}_{3^2}=\{1\},\qquad 3^2\mathbb{Z}_{3^3}\cong\mathbb{Z}_3.$$

For the quotient, you know that $\mathbb{Z}_{2^a}/\mathbb{Z}_{2^b}$ is cyclic, and you know the order (namely, $2^a/2^b = 2^{a-b}$); so you know exactly what cyclic group it is. You can do this with each cyclic factor, and that will give you a decomposition of $A/18A$ into primary divisors. From the decomposition into primary divisors one can get the decompositioon into invariant factors in the standard way.

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First, you should clarify which canonical form you're looking for. Your original description of $A$ is not in either of the two most common canonical forms.

As for your main question, you will want to use (perhaps you need to prove) the following facts about finite(ly generated) abelian groups:

  • $(B\times C)/n(B\times C) \cong (B/nB) \times (C/nC)$
  • ${\mathbb Z}_n/m{\mathbb Z}_n = {\mathbb Z}_n /(m,n){\mathbb Z}_n = {\mathbb Z}_{(m,n)}$
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