How many group homomorphism from $\Bbb Z_3$ to $\text{Aut}(\Bbb Z_7)$?
My attempt: Since $\Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $\phi(\overline{1})$. On the other hand $\text{Aut}(\Bbb Z_7)\cong \Bbb Z_6$
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Since $o(\phi(\overline{1}))\mid |\Bbb Z_3|$, so $o(\phi(\overline{1}))$ can only be $1,~3$. If the order is $1$, it means $\phi(\overline{1})=\overline{1}$. It is indeed a homo. If the order is $3$, it means $\phi(\overline{1})=\overline{2}$ or $\overline{4}$. So the total number of group homomorphisms from $\Bbb Z_3$ to $\text{Aut}(\Bbb Z_7)$ is three, as I show above. Am I correct?
Best Answer
$$\operatorname{Hom}(\Bbb Z/3\Bbb Z, \operatorname{Aut}(\Bbb Z/7\Bbb Z)) = \operatorname{Hom}(\Bbb Z/3\Bbb Z, C_6) = C_6[3] = \{e, g^2, g^4\}$$