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Let $p$ be a prime, and $\newcommand{\Z}{\mathbb{Z}} (\Z/p\Z)^*$ the multiplicative group $(\Z/p\Z) \setminus \{0\}$. Define $$ f \colon (\Z/p\Z)^* \to (\Z/p\Z)^*, \quad f(x)=x^5. $$

  • Show that if $5\mid p-1$ then $|\ker(f)| = 5$.
  • Show that if $5\nmid p-1$ then $|\ker(f)| = 1$.

I already know that $|\ker(f)|$ = {$ x \in (\Z/p\Z)^*\mid x^5=1$} and $|\ker(f)| \leq 5$ I was thinking of using that if $x \in (\Z/p\Z)^*, x^{p+1}=1 $ combine to $x \in ker(f)$ then $x^{5}=1$ but $5\nmid p-1$ so $f(\Z/p\Z)$ should be of order $p-1$ ?
And using isomorphism theorem ?
I don't know how to finish

1 Answers1

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Note that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic. Let $x$ be a generator. If $5 \mid p-1$ let $5n = p-1$ and consider the elements $$A = \{x^{n},x^{2n},x^{3n},x^{4n},1\}.$$ Notice that $A\subset \ker(f)$ but $|\ker(f)|\leq 5$ so $A=\ker(f)$.

On the other hand suppose that $5\nmid p-1$ and let $1\ne y\in (\mathbb{Z}/p\mathbb{Z})^{\times}$. Since $x$ is a generator there is an $1\leq m < p-1$ with $x^{m}=y$. If $y^{5}=1$ then also $x^{5m}=1$ which entails that $p-1 \mid 5m$, a contradiction. Hence $|\ker(f)|=1$.

Manifoldski
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