Let $q$ be a prime. $G$ is a cyclic group of order $q^2$. Show that for solving the DLP in $G$ it's enough to solve two distinct DLPs in two groups of order $q$.
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We are looking for an $x$ such that $\alpha^x=\beta$ in this group $G$.
By the CRT $G \cong C_q \times C_q$ alright? So taking the CRT-isomorphism yields
$\phi(\alpha^x)=(\alpha^x \mod q, \alpha^x \mod q)\overset{!}{=} (\beta \mod q,\beta \mod q)=\phi(\beta)$
So we have to solve $\alpha^x=\beta$ in $C_q$ now? But is this correct? I doubt that because in the exercise they want two distinct DLPs ???
Write $x=x_1+x_2q$. Solve $a^{x_1} = b \bmod{q}$ to find $x_1$. Let $a^q \bmod{q} = c$, $b a^{-x_1} \bmod{q}=d$. Find $x_2$ by solving $c^{x_2} = d \bmod{q}$.