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Let $I = \lbrace \overline{0}, \overline{8}, \overline{16} \rbrace$ be an ideal in $\mathbb{Z}_{24}$. Find all elements of quotient ring $\mathbb{Z}_{24}/I$.

The answer is $\mathbb{Z}_{24}/I = \lbrace I, \overline{1} + I, \overline{2}+I, \dots, \overline{7} + I \rbrace$. But, I still can't understand how to obtain it. Anyone can explain, please? Thanks in advance.

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It results directly from the Third isomorphism theorem: $I$ is simply the quotient $\:8\mathbf Z/24\mathbf Z$, so $$(\mathbf Z/24\mathbf Z)\big/I=(\mathbf Z/24\mathbf Z)\big/(8\mathbf Z/24\mathbf Z)\simeq \mathbf Z/8\mathbf Z.$$

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  • $\begingroup$ Why it is $Z/24Z$ ? $\endgroup$
    – kuah mi
    Aug 5 '20 at 22:36
  • $\begingroup$ and how can $I = 8Z/24Z$ ? $\endgroup$
    – kuah mi
    Aug 5 '20 at 22:37
  • $\begingroup$ what if like this: $Z_{24} = \lbrace \overline{0}, \overline{1}, \dots, \overline{23} \rbrace$. Then, $Z_{24}/I = \lbrace \overline{0}+I, \overline{1}+I, \dots, \overline{7}+I \rbrace$ since for $\overline{8}$, it would be $\overline{8}+I = I+I = I$, for $\overline{9}: \overline{9} + I = \overline{1+8} + I = \overline{1} + I$, and so on. $\endgroup$
    – kuah mi
    Aug 5 '20 at 23:15
  • $\begingroup$ Please answer my question Sir @Bernard $\endgroup$
    – kuah mi
    Aug 5 '20 at 23:29
  • $\begingroup$ Sorry, I had not seen the above commentsYes, I think you can add these details. $I=8\mathbf Z/24\mathbf Z$ because $I$ is made up of the multiples of $8\bmod 24$ (since $4\cdot\bar 8=\bar 0$). $\endgroup$
    – Bernard
    Aug 5 '20 at 23:47

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