Problem:
Solution:
Step 1: The set $ \mathbf{F} - \{0\} $ together with multiplication is a group - all axioms can be trivially verified by virtue that $ \mathbf{F} $ is a finite field.
Step 2: The set $ a^p $, $ \forall p \in \mathbf{Z} $, $ a \neq 0 \in \mathbf F $ is the cyclic subgroup generated by $ a $. As such, the subgroup has an order $ r \mid q-1 $. Therefore $ a^{q-1} = (a^r)^k = 1^k = 1 $. $a^r = 1 $ is because it is a cyclic subgroup.
Step 3: For $ a \ne 0 $, the above show that $ a^{q} = a(a^{q-1}) = a $, therefore $ a^q - a = 0$. It is obvious that $ a^q - a = 0 $ for $ a = 0 $ as well.
Therefore we have just proved that $ a^q - a = 0 $ for all $ a \in \mathbf{F} $.
Solution:
Step 1: The set $ \mathbf{F} - \{0\} $ together with multiplication is a group - all axioms can be trivially verified by virtue that $ \mathbf{F} $ is a finite field.
Step 2: The set $ a^p $, $ \forall p \in \mathbf{Z} $, $ a \neq 0 \in \mathbf F $ is the cyclic subgroup generated by $ a $. As such, the subgroup has an order $ r \mid q-1 $. Therefore $ a^{q-1} = (a^r)^k = 1^k = 1 $. $a^r = 1 $ is because it is a cyclic subgroup.
Step 3: For $ a \ne 0 $, the above show that $ a^{q} = a(a^{q-1}) = a $, therefore $ a^q - a = 0$. It is obvious that $ a^q - a = 0 $ for $ a = 0 $ as well.
Therefore we have just proved that $ a^q - a = 0 $ for all $ a \in \mathbf{F} $.
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