Proving Fermat's Little Theorem with algebra

(a) Let $p$ be a prime and let b be a non zero element of the field $\mathbb{Z}_{p}$. Show that $b^{p-1} = 1$. Hint Lagrange

My attempt

$|\mathbb{Z}_{p} \setminus \{0\}| = p-1$

since $b \in \mathbb{Z}_{p} \setminus \{0\}$

$\langle b\rangle \leq \mathbb{Z}_{p} \setminus \{0\}$

From Lagrange $\langle b\rangle | \mathbb{Z}_{p} \setminus \{0\}$

suppose $\langle b \rangle = n$

then $\implies n \mid p-1$

$\implies p-1 = nk$

$\implies b^{p-1}=(b^{n})^{k} = e^{k} = e = 1$

(b) Use (a) to prove that if $p$ is prime and a is an integer the p divides $a^{p}-a$.

And now I have to use the above fact to prove Fermat's Little Theorem.. any hints?

• If $(G,\cdot )$ is a group s.t. $|G|=n$, then $g^n=1$ for all $g\in G$. – Surb Apr 8 '18 at 16:38
• so?............ – user420309 Apr 8 '18 at 16:42
• As you said, $(\mathbb Z_p\backslash \{0\},\cdot )$ has cardinality $p-1$, and thus $a^{p-1}=1$ for all $a\in \mathbb Z_p\backslash \{0\}$. – Surb Apr 8 '18 at 16:46
• how can I use this to proof fermat? – user420309 Apr 8 '18 at 16:48
• I don't understand, all the proof is here ! $\mathbb Z_p\backslash \{0\}$ is a group of order $p-1$ for the multiplication. Therefore $a^{p-1}=1$ for all $a\in \mathbb Z_p\backslash \{0\}$. – Surb Apr 8 '18 at 16:50