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}.
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