Elements in a finite group of order prime

Question

I'm trying to prove the following statement:

Let $p=101$. Let $x\in (\mathbb{Z}/p\mathbb{Z})^*$ be arbitrary. Show that there exists $y\in (\mathbb{Z}/p\mathbb{Z})^*$ such that $y^3=x$.

Here is my work. Since $p$ is a prime. We know that the group of units $(\mathbb{Z}/p\mathbb{Z})^*$ is a field and cyclic. Let $\alpha\in(\mathbb{Z}/p\mathbb{Z})^*$ be a primitive root, then $x=\alpha^j$ for some $j\in \mathbb{Z}$. But I cannot go further. I also tried Fermat's little theorem as follows; \begin{equation*} x^{100} \equiv 1 \quad mod(101) \end{equation*} and \begin{equation*} (x^{3})^{33}x \equiv 1 \quad mod(101) \end{equation*} But I cannot see the exact solution. Can you help me please?

Answer 1

Hint: Fermat's little theorem also tells you that $\ x^{101}\equiv x \mod(101)\ $, and therefore that $\ x^{101+100n}\equiv x \mod(101)\ $ for any integer $\ n\ $. What happens if you choose $\ n=1\ $?

By the way, the group of units $\ (\mathbb{Z}/p\mathbb{Z})^*$ isn't a field. It's a cyclic group under multiplication, but it's not closed under addition. The entire ring $\ \mathbb{Z}/p\mathbb{Z}\ $ is what constitutes a field.

Answer 2

Hints:

• To prove existence only, prove that the map $x \mapsto x^3$ is injective.

• To exhibit a cube root, note that $201=67 \cdot 3$.

Answer 3

You almost have it, you just need to flip things around a bit:

$\begin{equation*} (x^{-1})^{100} \equiv 1 \quad mod(101) \end{equation*}$

$\begin{equation*} ((x^{-1})^{33})^{3}(x^{-1}) \equiv 1 \quad mod(101) \end{equation*}$

$\begin{equation*} ((x^{-1})^{33})^{3} \equiv x \quad mod(101) \end{equation*}$

So the cube root of $x$ is $(x^{-1})^{33}$.

Answer 4

Consider the map $f: x \mapsto x^3$. Since your group is abelian, it is easy to check that $f$ is a group homomorphism. Now, the order of your group is $\phi(101) = 100$ and so no non-identity element has an order which is either a divisor or a multiple of 3. So, $\text{ker}(f) = \{x \in (\mathbb{Z}/101 \mathbb{Z})^*| x^3 =1\} = \{1\}$ and hence $f$ is injective. Since $f$ is an injection from a finite set to itself, $f$ is also a surjection and therefore, every element has a pre-image via $f$. That is, every element is a cube.

Answer 5

There are many variations, but actually all solutions gravitate around Fermat's little theorem. Denote by $\bar a$ the class of $a$ mod $101$. Here $\bar 1=\bar x^{100} =\bar x.\bar x ^{99}$, so $\bar x=(\bar x^{-33})^3$.