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

  • Thread starter Bashyboy
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  • #1
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Homework Statement


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##.

Homework Equations




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?
 

Answers and Replies

  • #2
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12,838

Homework Statement


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##.

Homework Equations




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?
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?
 
  • #3
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It sounds a little bit confused. What is ¯¯¯¯¯¯¯¯¯¯x(p)x(p)¯\overline{x(p)}? And what does it mean for an element a∈Zna∈Zna\in\mathbb{Z}_n to be nilpotent?

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?
 
  • #4
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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)).
 
  • #5
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The only guaranteed group in a ring is the additive group.

I don't think this is true. I believe the units of a group form a multiplicative group.

This is why I don't understand, what you actually mean, esp. by x⋅px⋅px\cdot p. Do you mean xxx is a natural number and xpxpxp is an abbreviation for p+p+…+pp+p+…+pp+p+\ldots +p (xxx times)?

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.
 
  • #6
15,129
12,838
I don't think this is true. I believe the units of a group form a multiplicative group.
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, 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.
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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