Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.9
Let $Z_{36} = \langle x \rangle$. For which integers $a$ does the map $\psi_a$ defined by $\overline{1} \mapsto x^a$ extend to a well defined homomorphism $\mathbb{Z}/(48) \rightarrow Z_{36}$? Can $\psi_a$ be a surjective homomorphism?
Solution: We begin with a couple of lemmas.
Lemma 1: Let $G = \langle x \rangle$ be a cyclic group, $H$ a group, and $\varphi : G \rightarrow H$ a group homomorphism. Then $\varphi$ is uniquely determined by $\varphi(x)$.
Proof: Suppose we have two homomorphisms $\varphi$, $\psi$ : $G \rightarrow H$, such that $\varphi(x) = \psi(x)$. Since $G$ is cyclic, an arbitrary $y \in G$ can be written as $x^a$ for some $a$. Then $$\varphi(y) = \varphi(x^a) = \varphi(x)^a = \psi(x)^a = \psi(x^a) = \psi(y).$$ So $\varphi = \psi.$ $\square$
Lemma 2: Let $G$ be a group, $S = \{ x \}$, $Z_n = \langle x \rangle$, and $\overline{\varphi} : S \rightarrow G$ a set mapping. Then $\overline{\varphi}$ extends to a homomorphism $\varphi : Z_n \rightarrow G$ if and only if $\overline{\varphi}(x)^n = 1$.
Proof: ($\Rightarrow$) We have $x^n = 1$, so that $$\varphi(x^n) = \varphi(x)^n = \overline{\varphi}(x)^n = 1. $$($\Leftarrow$) Let $a,b$ be integers such that $x^a = x^b$. Then $a = b \pmod n$, so that $b = nk+a$ for some $k$. Thus $$\varphi(x^b) = \varphi(x^{kn+a}) = (\varphi(x)^n)^k \varphi(x^a) = \varphi(x^a),$$ so that $\varphi$ is well defined. Now suppose $z,w \in G$ with $z = x^a$, $w = x^b$. Then $$\varphi(zw) = \overline{\varphi}(x^{a+b}) = \overline{\varphi}(x)^{a+b} = \overline{\varphi}(x)^a \overline{\varphi}^b = \varphi(z) \varphi(w).$$ Thus $\varphi$ is a homomorphism. $\square$
Lemma 3: A map $\overline{\varphi} : \mathbb{Z}/(n) \rightarrow \mathbb{Z}/(m)$ given by $\overline{1} \mapsto \overline{a}$ extends to a group homomorphism $\varphi : \mathbb{Z}/(n) \rightarrow \mathbb{Z}/(m)$ if and only if $\frac{m}{\mathsf{gcd}(m,n)}$ divides $a$.
Proof: If $\overline{\varphi}$ extends, then by Lemma 2, we have $na \equiv 0 \pmod m$, so that $m$ divides $na$. Let $d = \mathsf{gcd}(m,n)$ and write $m = dm_1$ and $n = dn_1$. Then $m_1$ divides $n_1a$, and since $m_1$ and $n_1$ are relatively prime, $m_1 = \frac{m}{\mathsf{gcd}(m,n)}$ divides a by Euclid’s lemma.
Conversely, if $\frac{m}{\mathsf{gcd}(m,n)}$ divides $a$, then $m$ divides $na$, so that $\overline{\varphi}$ extends to a homomorphism by Lemma 2. $\square$
Here, $\mathsf{gcd}(36,48) = 12$ and $\frac{36}{12} = 3$. So $\overline{\varphi}$ extends if and only if $3|a$. There are 12 such $a$.
Suppose some such $\varphi$ is surjective; then we have $1 \equiv 3 \pmod{36}$, a contradiction.
Note that since $\mathsf{im}\ \varphi_a = \langle x^a \rangle$, $|\mathsf{im}\ \varphi_a| < 36$. Hence $\varphi_a$ cannot be surjective.
See also here https://wolfweb.unr.edu/homepage/naik/classes/731/60.9.Soln.pdf
