Homomorphisms $\mathbb Z/p\mathbb Z\to G$ and the mapping property of quotient groups Let $p$ be a prime. Let $g$ be an element of a group $G$ of order 1 or $p$. Let $$\phi':\mathbb Z/p\mathbb Z\to G\\ \bar n\mapsto g$$
What's the most general way (so to speak) to prove that this map is well-defined?
I thought there may be some relationship to the following theorem:

Is it right that this theorem also implies that any group homomorphism $\psi': G'/N\to G$ "comes from" a homomorphism $\psi:G'\to G$ where $N\subset \ker \psi$? So is there a bijection between homomorphisms $G'/N\to G$ and homomorphisms $G'\to G$ whose kernel contains $N$?
If so, then our $\phi': \mathbb Z/p\mathbb Z\to G$ has to come from $\phi: \mathbb Z\to G$ such that $p\mathbb Z\subset \ker \phi$ and if $\phi'([n])=g$, then $\phi(n)=g$. Then it would follow that if $[n]=[n']$, then $n'=n+pl$, so $\phi(n')=\phi(n)+\phi(pl)=\phi(n)$; and then the above proposition would imply that $\phi([n'])=g$, showing well-definiteness.
Is that right? Are there more general (or simpler) ways to see this (or say what I said)?
Also, I don't see where I used that $p$ is prime. Is this assumption needed?
 A: I hope my answer below does a decent job of answering the questions from the original post. If anything is unclear, or if you have any questions, please leave a comment. Also, if you would like help with the exercises below, please let know.
(1) The original post starts out like this:

Let $p$ be a prime. Let $g$ be an element of a group $G$ of order 1 or $p$. Let $$\phi':\mathbb Z/p\mathbb Z\to G\\ \bar n\mapsto g$$

So we were given that $\phi'$ is a constant function. The only way a constant function can be a group homomorphism is if it maps each element to the identity. It follows that $\vert g\vert=1$.
(2) The original post also asks about the mapping property of quotient groups. Let $G'$ and $G$ be groups, and let $N$ be a normal subgroup of $G'$. Indeed there is a bijection
$$f:\left\{\begin{matrix}\text{homomorphisms }G'\to G\\\text{whose kernel contains }N\end{matrix}\right\}\to\left\{\begin{matrix}\text{homomorphisms}\\G'/N\to G\end{matrix}\right\}.$$
If $\phi:G'\to G$ with $N\le\ker\phi$, then $f(\phi)=\overline{\phi}$, the homomorphism you get from the mapping property of quotient groups.
Proof: First, let's show that $f$ is well-defined. Given $\phi:G'\to G$ with $N\le\ker\phi$, the mapping property of quotient groups gives you a homomorphism $\overline{\phi}:G'/N\to G$ with $\overline{\phi}\circ\pi=\phi$. Suppose we had another homomorphism $\psi:G'/N\to G$ with $\psi\circ\pi=\phi$. In this case we can show that $\psi=\overline{\phi}$. This is a good exercise, so I'll hide the proof below:

 Let $x\in G'/N$. Then $x=\pi(y)$ for some $y\in G'$. Then $$\psi(x)=\psi\circ\pi(y)=\overline{\phi}\circ\pi(y)=\overline{\phi}(x).$$ Hence $\psi=\overline{\phi}$.

So the map $f$ sends each homomorphism $\phi:G'\to G$ with $N\le\ker\phi$ to the unique homomomorphism $\overline{\phi}:G'\to G$ with $\overline{\phi}\circ\pi=\phi$. Hidden below is a proof that $f$ is injective:

 Let $\phi_1,\phi_2:G'\to G$ with $N\le\ker\phi_1,\ker\phi_2$, and suppose that $\overline{\phi_1}\ne\overline{\phi_2}$. Then there is an $a\in G'/N$ with $\overline{\phi_1}(a)\ne\overline{\phi_2}(a)$. Note that $a=\pi(b)$ for some $b\in G'$. Since $$\phi_1(b)=\overline{\phi_1}(a)\ne\overline{\phi_2}(a)=\phi_2(b),$$ it must be that $\phi_1\ne\phi_2$.

Hidden below is a proof that $f$ is surjective:

 Let $\psi:G'/N\to G$ be a homomorphism, and let $\phi=\psi\circ\pi$. It will follow that $\psi=\overline{\phi}$ if we can show that $N\le\ker\phi$. Let $x\in N$. Then $\pi(x)=1_{G/N}$. Hence $$\phi(x)=\psi\circ\pi(x)=\psi\left(1_{G/N}\right)=1_G.$$ Hence $x\in\ker\phi$.

Note: an alternative way to show that $f$ is a bijection would be to explicitly define the its inverse:
$$g:\left\{\begin{matrix}\text{homomorphisms}\\G'/N\to G\end{matrix}\right\}\to\left\{\begin{matrix}\text{homomorphisms }G'\to G\\\text{whose kernel contains }N\end{matrix}\right\}.$$
For each $\psi:G'/N\to G$, $g(\psi)=\psi\circ\pi$. $\;\Box$
(3) Let $G$ be a group, and let $k\in\Bbb{N}$ with $k\ge1$. It follows that there is a bijection
$$\left\{\begin{matrix}\text{homomorphisms}\\\Bbb{Z}/k\Bbb{Z}\to G\end{matrix}\right\}\leftrightarrow\left\{\begin{matrix}\text{homomorphisms }\Bbb{Z}\to G\\\text{whose kernel contains }k\Bbb{Z}\end{matrix}\right\}.$$
(4) Here is a characterization of homomorphisms $\Bbb{Z}\to G$ for an arbitrary group $G$:
Exercise 1: Let $G$ be a group, and let $g\in G$. Show that the map $\phi:\Bbb{Z}\to G$ which sends $n\mapsto g^n$ is a group homomorphism.
Exercise 2: Let $\phi:\Bbb{Z}\to G$ be a group homomorphism. Then there is a $g\in G$ such that for all $n\in\Bbb{Z}$, $\phi(n)=g^n$.
Here's a hint for the second exercise:

 Let $g=\phi(1)$.

(5) Let $k\in\Bbb{N}$ with $n\ge1$. Putting (3) and (4) together we can get a characterization of homomorphisms $\Bbb{Z}/k\Bbb{Z}\to G$ for an arbitrary group $G$:
We can now put everything together to get a description of the homomorphisms $\phi:\Bbb{Z}/k\Bbb{Z}\to G$, where $G$ is an arbitrary group:
Exercise 1: Let $G$ be a group, and let $g\in G$ with $g^k=1_G$. Show that the map $\phi:\Bbb{Z}/k\Bbb{Z}\to G$ which sends $n\mapsto g^n$ is a group homomorphism.
Exercise 2: Let $\phi:\Bbb{Z}/k\Bbb{Z}\to G$ be a group homomorphism. Then there is a $g\in G$ such that $g^k=1_G$ and for all $n\in\Bbb{Z}$, $\phi(n)=g^n$.
