Proposition a is a primitive root modulo n then it is also a primitive root modulo d, for any d dividing n.
Proof.
Claim: The map \( \mathbb{Z}_{n}^{*} \rightarrow \mathbb{Z}_{d}^* \) is surjective. \( d=\prod p_i^{r_{di}} , \ n=\prod p_i^{r_{fi}} \cdot \prod q_i^{r_{gi}} \) where \( p_i, q_i \) are all distinct prime. Set \( f=\prod p_i^{r_{fi}} , \ g=\prod q_i^{r_{gi}} \) then \( n=fg \).
Note \( (d,g)=1 \implies \exists x \in \mathbb{N} : xd\equiv 1 \bmod g \). For any \( [i]_g \in \mathbb{Z}_{d}^* \), \( i\equiv u \bmod g \) where \( u\in [0, g-1] \).
\( (g-u+1)x \equiv k \mod g \) where \( k \in [0, g-1] \). \( \implies i+kd\equiv 1 \mod g \implies (i+kd,g)=1 \). \( \forall p_i : p_i \mid f \), \( p_i \nmid (i+kd) \) since \( (i, p_i)=1 \land pi\mid kd \).
Since \( \mathbb{Z}_{n}^{*}=<[a]_n> \) and the map \( \mathbb{Z}_{n}^{*} \rightarrow \mathbb{Z}_{d}^* \) is surjective.
Therefore, a is a primitive root modulo d.
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