## Group homomorphisms preserve exponents

Let $G$ and $H$ be groups and $\varphi : G \rightarrow H$ a group homomorphism.

1. Prove that $\varphi(x^n) = \varphi(x)^n$ for all $x \in G$ and $n \in \mathbb{Z}^+$.
2. Prove that $\varphi(x^n) = \varphi(x)^n$ for all $x \in G$ and $n \in \mathbb{Z}$.

1. We proceed by induction on $n$. For the base case, $\varphi(x^1) = \varphi(x) = \varphi(x)^1$. Suppose the statement holds for some $n \in \mathbb{Z}^+$; then $\varphi(x^{n+1})$ $= \varphi(x^n x)$ $= \varphi(x^n) \varphi(x)$ $= \varphi(x)^n \varphi(x)$ $= \varphi(x)^{n+1}$, so the statement holds for $n+1$. By induction, $\varphi(x^n) = \varphi(x)^n$ for all $n \in \mathbb{Z}^+$.
2. First, note that $\varphi(x) = \varphi(1_G \cdot x) = \varphi(1_G) \cdot \varphi(x)$. By right cancellation, we have $\varphi(1_G) = 1_H$. Thus $\varphi(x^0) = \varphi(x)^0$. Moreover, $\varphi(x) \varphi(x^{-1}) = \varphi(xx^{-1}) = \varphi(1) = 1$; thus by the uniqueness of inverses, $\varphi(x^{-1}) = \varphi(x)^{-1}$. Now suppose $n$ is a negative integer. Then $\varphi(x^n) = \varphi((x^{-n})^{-1})$ $= \varphi(x^{-n})^{-1}$ $= (\varphi(x)^{-n})^{-1}$ $= \varphi(x)^n$. Thus $\varphi(x^n) = \varphi(x)^n$ for all $x \in G$ and $n \in \mathbb{Z}$.