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.

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