Exercise 2.3.18 (Universal property of $Z_n$)

Proof. Let $H$ be a group and $h\in H$ such that $h^n=1$

• Unicity. Suppose that $f:Z_n\to H$ is an homomorphism such that $f(x)=h$. Then for all $k\in \mathbb{Z}$, $f(x^k)=h^k$. Indeed $f(x^0)=f(1)=1=h^0$. If $f(x^k)=h^k$ for some integer $k\ge 0$, then $f(x^{k+1})=f(x^k\cdot x)=f(x^k)f(x)=h^{k}h=h^{k+1}$. The induction is done, which proves that for all $k\in \mathbb{N}$, $f(x^k)=h^k$. Moreover, for $k\ge 0$, $f(x^{-k})=f({(x^k)}^{-1})=f{(x^k)}^{-1}={(h^k)}^{-1}=h^{-k}$, therefore

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},f(x^k)=h^k.$$

  Let $g$ be another homomorphism such that $g(x)=h$. If $y$ is any element of $Z_n$, then $y=x^k$ for some integer $k$, therefore

  $$g(y)=g(x^k)=h^k=h(x^k)=f(y)(y\in Z_n).$$

  This shows that $f=g$, so there is at most one homomorphism $f$ such that $f(x)=h$.

• Existence. Consider the map

  $$f\{\begin{array}{ccc}\hfill Z_n\hfill & \hfill \to \hfill & H\hfill \\ \hfill x^k\hfill & \hfill \mapsto \hfill & h^k\hfill \end{array}$$

  • $f$ is well defined: Every element $y\in Z_n$ is of the form $y=x^k$ for some $k\in \mathbb{Z}$. If $y=x^k=x^l$, then $x^{k-l}=1$, where $x$ has order $n$, thus $n\mid k-l$, so $l=k+\mathit{\lambda n}$ for some integer $\lambda$. Since $h^n=1$,

    $$h^l=h^{k+\mathit{\lambda n}}=h^k{(h^n)}^{\lambda }=h^k,$$

    so $f$ is well defined.

  • $f$ is a homomorphism: If $y,z\in H$, then $y=x^k$ and $z=x^l$ for some integers $k,l$. Then

    $$\begin{array}{llll}\hfill f(y)f(z)& =f(x^k)f(x^l)=h^k{h}^l=h^{k+l}=f(x^{k+l})=f(x^k{x}^l)=f(\mathit{yz}).& \hfill & \end{array}$$
  • Finally $f(x)=f(x^1)=h^1=h$.

  So $f:Z_n\to H$ is a homomorphism such that $f(x)=h$.

In conclusion, if $h$ is an element of $H$ with $h^n=1$, then there is a unique homomorphism from $Z_n=⟨x⟩$ to $H$ such that $x$ maps to $h$. □