**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.18**

We write $Z_n = \langle x \rangle$. Show that if $H$ is any group and $h \in H$ such that $h^n = 1$, then there is a unique homomorphism $\varphi : Z_n \rightarrow H$ with $x \mapsto h$.

Solution:

Existence: Define $\varphi by \varphi(x^k) = h^k$.

Suppose $x^a = x^b$; then $a = b \pmod n$, so that $a = qn + b$ for some $q$. Now $$\varphi(x^a) = h^a = h^{qn+b} = h^b = \varphi(x^b),$$ so that $\varphi$ is well defined.

Clearly now $$\varphi(x^ax^b) = \varphi(x^{a+b}) = h^{a+b} = h^ah^b = \varphi(x^a)\varphi(x^b),$$ so $\varphi$ is a homomorphism.

Uniqueness: Suppose we have another homomorphism $\psi : Z_n \rightarrow H$ with $\psi(x) = h$. Then for all $x^a \in Z_n$, $$\varphi(x^a) = \varphi(x)^a = \psi(x)^a = \psi(x^a),$$ so that $\varphi = \psi$. Thus $\varphi$ is unique.

## Jimmy

29 Jun 2022Minor typo in the first sentence where it says "Define $\varphi by\varphi(x^{k})$..."