## When is a map from ZZ/(48) to Cyc(36) a group homomorphism?

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?

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 $a = b$ mod $n$, 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 $na \equiv 0$ mod $m$, 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 $1 \equiv 3at$ mod 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.

I think that a little simpler explanation is in (http://wolfweb.unr.edu/homepage/naik/classes/731/60.9.Soln.pdf). Just checking $x^{48a}=1 \in \mathbf{Z}_{36}$ clarifies the desired homomorphisms, not need to care about order of $|x^a|$ or Lagrange’s Theorem.