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This result was used in a proof of a theorem, i am not sure if it's true. can someone tell the proof idea.

Can it be generalized to additive order of any zero divisor in $Z_{p^k}$, is there any formula to calculate the additive order of zero divisors of $Z_{p^k}$ in general ?

Rising Star
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    Why don't you try generalizing it? What are the zero divisors of $\Bbb Z_8$? What are their additive orders? How about $\Bbb Z_{27}$ or $\Bbb Z_{16}$? – Arthur May 03 '19 at 08:29
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    See this question and apply it for $n=p^k$. – Dietrich Burde May 03 '19 at 08:32
  • zero divisors of $Z_8$ are 2,4,6. O(2) =4, O(4)=2, O(6) =4 – Rising Star May 03 '19 at 08:33
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    @RisingStar And if you look at the prime factorisations of the zero divisors $2, 4$ and $6$, do you see a connection between that and their respective additive orders? If you don't see it right away, then looking at the other two rings I suggested is probably a good idea. – Arthur May 03 '19 at 08:39

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Well, the zero divisors of ${\Bbb Z}_{p^k}$ are exactly the nonzero elements $a$ with $\gcd(a,p^k)\ne 1$, i.e., the zero divisors are exactly the nonzero multiples of $p$. Then the $p^{k-1}$-multiple of a zero divisor $a$ written as $a=pb$ is $p^{k-1}\cdot a = p^{k-1}pb = p^kb=0$.

Wuestenfux
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