Nilpotent elements in $\mathbb Z_n$

Question

How to prove that an element $a$ is a nilpotent element of $\mathbb Z_n$ if and only if each prime divisor of $n$ is a divisor of $a$?

Answer 1

Suppose $a$ is nilpotent in $Z_n$, then some $m \in \mathbb{N}$, we have $n | a^m$. Now if $p$ is a prime divisor of $n$, by transitivity of divisibility we have $p | a^m$. As $p$ is prime, this implies that $p|a$. (Use induction and $a^m = a \cdot a^{m-1}$)

Conversely, suppose every prime divisor $p_i$ of $n$, $p_i | a$. Let $v_{p_i}(n)$ be the exponent of the largest power of $p_i$ that divides $n$. This means that $n = \prod_i p_i^{v_{p_i}(n)}$ Now let $m = \max_i\{v_{p_i}(n)\}$ where the maximum is taken over all prime divisors $p_i$ of $n$. Then we have $ n = \prod_i p_i^{v_{p_i}(n)} | a^m$, so $a$ is nilpotent in $Z_n$