Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.1
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 $n \in \mathbb{Z}$.
Solution:
(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 $n \in \mathbb{Z}$.
