Let $g \in \text{ group } G $ and $n \in N$. Let $\phi : \mathbb{Z_n} \rightarrow G$ be defined by $\phi(i) = g^i$ for $0 \le i \le n$. Give a necessary and sufficient condition (in terms of g and n) for $\phi$ to be a homomorphism. Prove your assertion. My $g =$ solution's $h$. I think $h$ is mismatched for group G.
(1.) Where does the necessary and sufficient condition spring up from? How do you preordain it?
(2.) How do you preordain to calculate $h^n$? I know homomorphism maps $0 \to id_G$.
(3.) I know if $|a| = n < \infty$ then $a^i = a^j \iff n | (i - j)$. However why not $<h> \simeq \mathbb{Z_n}$ ?
(4.) In the same line as the purple underline, the solution used the Division Algorithm to induce $q_i, r_i$ for $i = 1,2$ such that $i = q_1m + r_1$ and $j = q_2m + r_2$ for all $0 \le r_1,r_2 < m$.
Add these two equations: $i + j = (q_1 + q_2)m + r_1 + r_2$ for all $0 \le r_1 + r_2 < 2m$.
Hence why does the solution apply the Division Algorithm to $i + j$ to induce $i + j = qm +r, 0 \le r < m$? Isn't this redundant?