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How to determine the largest order of an element in the group $\mathbb Z_4 \times \mathbb Z_{18} \times \mathbb Z_{15}$.

I know the order of the element $(a, b)$ in the direct product $\mathbb Z_m \times \mathbb Z_n$ is $\operatorname{lcm}(|a|, |b|)$. How to apply this result here? Thanks for the help.

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3 Answers 3

up vote 6 down vote accepted

You are almost done. Few more thing to notice.

For any $(a, b, c) \in \mathbb Z_4 \times \mathbb Z_{18} \times \mathbb Z_{15}$, $|a|$ divides $4$, $|b|$ divides 18, and $|c|$ divides 15. We have the following possibilities $order(a) = 1, 2, 4$, $order(b) = 1, 2, 3, 6, 9, 18$, and $order(c)= 1, 3, 5, 15$. The largest least common multiple among the possibilities is that of $4$, $18$ $15$, which is $180$, which is the order of $(1, 1, 1)$, and this is the largest order an element of $ \mathbb Z_4 \times \mathbb Z_{18} \times \mathbb Z_{15}$.

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thanks a lot for quick reply –  monalisa May 5 '13 at 17:14

Using the Fundamental Theorem of Finite Abelian Groups you may write $$\mathbb{Z}_4\times \mathbb{Z}_{18}\times \mathbb{Z}_{15}\cong\mathbb{Z}_4\times (\mathbb{Z}_9\times\mathbb{Z}_2)\times(\mathbb{Z}_3\times\mathbb{Z}_5)\cong (\mathbb{Z}_3\times \mathbb{Z}_9)\times (\mathbb{Z}_2\times\mathbb{Z}_4)\times (\mathbb{Z}_5)\cong \mathbb{Z}_6\times \mathbb{Z}_{180}$$

The last is written in canonical form, where each order divides the next higher one. Once this is done the largest order of any element (called the exponent of the group) is always the largest order, in this case $180$.

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sir, can u please elaborate what a canonical form is? I am unaware of the term. Thanks –  monalisa May 5 '13 at 17:33
2  

You do the same thing. The order of an element $(a,b,c)\in \mathbb{Z}_n\times \mathbb{Z}_{m}\times \mathbb{Z}_{k}$ is $\text{lcm}(\lvert a\rvert, \lvert b\rvert , \lvert c\rvert )$.

Example: The order of $(0,0,3)\in \mathbb{Z}_4\times \mathbb{Z}_{18}\times \mathbb{Z}_{15}$ is $\text{lcm}(1,1,5) = 5$.

Example: The order of $(2,9,3)$ is $\text{lcm}(2, 2, 5) = 10$.

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thanks for your timely reply. –  monalisa May 5 '13 at 17:14

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