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I'm having trouble figuring out

Find the order of the element $\overline{8} + \langle\overline{6} \rangle$ in the quotient group $\mathbb{Z}_{60} / \langle \overline{6} \rangle$

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Maybe this is a shot in the dark.... $|\mathbb{Z}_{60}|$= 60 and $|\langle \overline{6} \rangle|=10$ so $|\mathbb{Z}_{60}/\langle \overline{6} \rangle| = 60/10 =6$? Do all elements have the same order? –  pigishpig Dec 7 '11 at 21:37
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You have asked 9 questions and accepted answers on none of them. Please accept answers on questions that people have put forth effort by clicking the check mark next to the best answer. –  Brandon Carter Dec 7 '11 at 21:38
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One way would be to just add $8$ to itself and see what happens. Have you done this? –  Dylan Moreland Dec 7 '11 at 21:41
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1 Answer

up vote 3 down vote accepted

The elements of $\mathbb{Z}_{60}/\langle \overline{6}\rangle$ are $\langle \overline{6}\rangle$, $\overline{1}+\langle \overline{6}\rangle$, $\overline{2}+\langle \overline{6}\rangle$, $\overline{3}+\langle \overline{6}\rangle$, $\overline{4}+\langle \overline{6}\rangle$, and $\overline{5}+\langle \overline{6}\rangle$. which if these is equal to $\overline{8}+\langle \overline{6}\rangle$? How many times do you have to add that one to itself to get $\langle \overline{6}\rangle$?

There’s no law against doing some actual computations to see what’s going on!

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Then the answer is 10 since you add $\langle \overline{6} \rangle$ to $\overline{8}$ 10 times to return to 8. Isn't that the same as in the group $\mathbb{Z}_{60}$. What does the quotient group have to do with it? –  pigishpig Dec 7 '11 at 22:26
    
@pigishpig: No, the answer is $3$. $\overline{8}+\langle\overline{6}\rangle=\overline{2}+\langle\overline{6}\rangle‌​$; $\left(\overline{2}+\langle\overline{6}\rangle\right)+\left(\overline{2}+\langle \overline{6} \rangle\right)=\overline{4}+\langle\overline{6}\rangle$; and $\left(\overline{4}+\langle\overline{6}\rangle\right)+\left(\overline{2}+\langle \overline{6} \rangle\right)=\langle \overline{6} \rangle$. –  Brian M. Scott Dec 7 '11 at 22:29
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