This is the problem I've been working on. I have some ideas on parts (a) and (b), but I'm not sure if that's correct or not. Please double check my work. Also, any hints on part (c) will be appreciated. Thank you in advance!
Let $ n \in \mathbb{N}$, let G be a group, and let $C_n =<x>$ be the cyclic group of order n generated by x.\
(a) (An attempt) Suppose $\tilde{\varphi} : \{x\} \rightarrow G$ is defined by $\tilde{\varphi}(x) = g$. The following is a naive series of steps to prove that $\tilde{\varphi}$ extends to a group homomorphism $\varphi: C_n \rightarrow G$. State in which step the proof fails:
Define $\varphi: C_n \rightarrow G $ by \ $\varphi(x^k) = \tilde{\varphi}(x)^k$
for all $k \in \{0,... ,n - 1\}$. We will show that $\varphi $ is a homomorphism. Let $x^k,x^l \in C_n$ with $k,l \in \{0,...,n-1\}$. Then
$\varphi(x^kx^l) = \varphi(x^{k+l})$ by the group operation \ $= \tilde{\varphi}(x)^{k+l}$ by definition of $\varphi$ \ $= \tilde{\varphi}(x)^k\tilde{\varphi}(x)^l $ by the group operation \ $= \varphi(x^k)\varphi(x^l)$ by definition of $\varphi $ \ so that $\varphi $ is a homomorphism.
Here's what I have come up with for part (a):
The proof fails when $k+l >n - 1$ since our definition of $\varphi(x)$ is only for some $k \in \{0,1,...,n-1\}$.\
(b) (When an extension can be made) What condition on $\tilde{\varphi}(x)$ must we impose? Use your answer to rewrite the proof above into a correct one. \ Solution that I have for part (b):
The extension must be that $\tilde(\varphi(x)) = g$ since we know that \ $\varphi(x)^0 = \tilde{\varphi(x)}^0 = g^0$ which means that $g^0 = e_G$ by homomorphism properties.
(c) (Uniqueness of extensions) Let $\tilde{\varphi} : \{x\} \rightarrow G$ be a function that extends to two homomorphisms $\varphi $ and $\psi$ . Prove that $\varphi = \psi$.
What you have for part (a) is the right idea. Since elements of $C_n$ have multiple representations in terms of $x^k$ for some $k \in \mathbb{Z}$, you need to check that the defined morphism is well defined. Since for any $k,a \in \mathbb{Z}$ we have $x^k = x^{k+an}$ so we need to check $\phi (x^k) = \phi (x^{k+an})$. A counter example is when $G = \mathbb{Z}$, $n=2$, and $g=1$.
Now that we know that we need to consider well-definedness, let us think about what conditions we need to ensure that the extension is well defined. Let $m$ be the order of $g$. Since we must have $\phi (x^k) = \phi (x^{k+an})$, it follows that $\phi (x)^k = \phi (x)^{k+an}$ and therefore $g^k = g^{k+an}$. Now we must have that $id_G = g^{an}$ and if we take $a=1$, then $g^n = id_G$. Therefore, $m$ must divide $n$ and I claim that this condition is sufficient for the morphism to be well defined. I'll leave it to you to finish this since I presume that this is homework, feel free to comment if you need further help.
For part (c), it follows from $\varphi (x^k) = \varphi (x) ^ k = \tilde{\varphi } (x) ^ k = \psi (x^k)$ so the two extensions are the same.