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Nilpotent Elements of ##\mathbb{Z}/n\mathbb{Z}##

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    I just finished proving that ##\overline{a} \in \mathbb{Z}/n\mathbb{Z}## is nilpotent if and only if every prime divisor of ##n## is a prime divisor ##a##, and this lead me to wonder whether the cyclic subgroup ##\langle \overline{p_1 p_2 ... p_k} \rangle## be consist of all nilpotent elements, where ##p_1,...p_k## appear in the prime factorization of ##n##. This seems to be the case when ##n = 72##.

    2. Relevant equations


    3. The attempt at a solution
    It seems that this would be an immediate corollary of the theorem to which I alluded. Is this right? Every element of ##\langle \overline{p_1 p_2 ... p_k} \rangle## is of the form ##\overline{x(p_1 p_2 ... p_k)}##, and it seems that every prime divisor of ##n## would be a divisor of ##x(p_1 p_2 ... p_k)##. Does this sound correct?
     
  2. jcsd
  3. Dec 1, 2016 #2

    fresh_42

    Staff: Mentor

    It sounds a little bit confused. What is ##\overline{x(p)}##? And what does it mean for an element ##a\in\mathbb{Z}_n## to be nilpotent?
     
  4. Dec 1, 2016 #3
    I would take ##\overline{x(p)}## to be a congruence class in ##\mathbb{Z}/n \mathbb{Z}##, where ##x(p) = xp## is some multiple of ##p##. From my understanding, ##\overline{a} \in \mathbb{Z}/n \mathbb{Z}## if there is some ##m \in \mathbb{N}## such that ##\overline{a}^m = \overline{0}##, where ##\overline{a}^m = \overline{a^m}##. Is this wrong?
     
  5. Dec 1, 2016 #4

    fresh_42

    Staff: Mentor

    Yes, that's right. So you have ##n=p_1^{c_1}\cdot \ldots p_k^{c_k}##.
    Now if you defined ##c:= max\{c_1, \ldots ,c_k\}##.
    What is ##(x(p_1\cdot \ldots \cdot p_k))^c##? And what does it mean for the elements of your subgroup.

    I don't really understand the rest under 3.) Since we are talking about nilpotent elements, I assume we are talking about rings. ##\mathbb{Z}_n## is a ring, so far so good. The only guaranteed group in a ring is the additive group. (Because e.g. ##3\cdot 24=0## in ##\mathbb{Z}_{72}##, multiplication is in general no group.)
    So if you talk about subgroups, you have to mean according to addition. These groups however don't automatically inherit the multiplication structure of the ring they are from. Structures which do that are either subrings or ideals. This is why I don't understand, what you actually mean, esp. by ##x\cdot p##. Do you mean ##x## is a natural number and ##xp## is an abbreviation for ##p+p+\ldots +p## (##x## times)?

    So the short answer to your question is probably, yes. (I just don't follow your argument. Try the one above (first 3 lines)).
     
  6. Dec 5, 2016 #5
    I don't think this is true. I believe the units of a group form a multiplicative group.

    Yes, I am taking ##x \cdot p = p + ... + p## (##|x|##-times). So would ##\langle \overline{p_1 p_2 ...,p_k} \rangle## form a subring? I will try to verify this in the meantime.
     
  7. Dec 5, 2016 #6

    fresh_42

    Staff: Mentor

    O.k., this is true. The units of a ring form a group. But units are rather boring elements with respect to the ring structure.
    Yes, it has to be. I simply had difficulties to understand what you meant.
    Say ##m:=p_1\cdot \ldots \cdot p_k \,\vert \, n##. Then ##n\mathbb{Z} \subseteq m\mathbb{Z}## is an ideal and thus ##m\mathbb{Z} / n\mathbb{Z} = \langle \overline{m} \rangle = \langle \overline{p_1 p_2 ...,p_k} \rangle## a ring.
     
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