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

• Bashyboy
If you define ##\overline{a}\cdot \overline{b} := \overline{a\cdot b}##, of course.)In summary, the conversation discusses a proved theorem that states an element in a certain set is nilpotent if and only if every prime divisor of a given number is also a prime divisor of the element. This leads to a question about whether a cyclic subgroup of a certain form consists of all nilpotent elements, which is shown to be true in the case of a specific number. The conversation also discusses the meaning of an element being nilpotent and the structure of subgroups in a ring.
Bashyboy

## 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##.

## 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?

Bashyboy said:

## 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##.

## 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?

fresh_42 said:
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?

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)).

fresh_42 said:
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.

fresh_42 said:
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.

Bashyboy said:
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.

## 1. What are nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}##?

Nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}## are elements that, when raised to a certain power, equal zero. In other words, they are elements that have a finite order and their powers "collapse" to zero.

## 2. How do you identify nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}##?

To identify nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}##, you can use the fact that an element ##a## is nilpotent if and only if ##a^{k} \equiv 0 \pmod{n}## for some positive integer ##k##. You can also use the property that the product of nilpotent elements is nilpotent.

## 3. What is the significance of nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}##?

Nilpotent elements in ##\mathbb{Z}/n\mathbb{Z}## play an important role in understanding the structure of the ring ##\mathbb{Z}/n\mathbb{Z}##. They can help identify subrings and ideals, and can also be used in solving equations and proving theorems.

## 4. Can there be more than one nilpotent element in ##\mathbb{Z}/n\mathbb{Z}##?

Yes, there can be more than one nilpotent element in ##\mathbb{Z}/n\mathbb{Z}##. In fact, any element that has a finite order and is not a unit is a nilpotent element.

## 5. What is the relationship between nilpotent elements and zero divisors in ##\mathbb{Z}/n\mathbb{Z}##?

In ##\mathbb{Z}/n\mathbb{Z}##, every nilpotent element is also a zero divisor. However, the converse is not always true. That is, there can be zero divisors that are not nilpotent elements.

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