Solution: Let $x\in G$. Let $n\in \mathbb{N} $. We will prove that $\phi \left( x^n \right) = \phi (x)^n$ by the mathematical induction.

• For $n=1$, we have $\phi \left( x^1 \right) = \phi (x)^1$.
• Let us suppose that the result is true for $n\leq k$.
• We will prove that $\phi \left( x^{k+1} \right) = \phi \left( x \right) ^{k+1}$. Note that \begin{align*} \phi \left( x^{k+1} \right) & = \phi \left( x^{k} \cdot x\right) \\ & = \phi \left( x^k \right) \cdot \phi (x) \\ & = \phi (x)^k \cdot \phi (x) \\ & = \phi (x)^{k+1}. \end{align*} Thus, using the mathematical induction we proved that for $x\in G$ and $n\in \mathbb{N} $ \[ \phi \left( x^n \right) = \phi (x)^n. \]

Now note that \[ \phi \left( x^{-1} \right) \cdot \phi (x) = \phi \left( x^{-1} \cdot x \right) = \phi (e) = e. \] The above implies \[ \phi \left( x^{-1} \right) = \phi (x)^{-1} . \] Take $y=x^{-1} $. Then \[ \phi \left( x^{-n} \right) = \phi \left( y^n \right) = \phi (y)^n = \phi \left( x^{-1} \right) ^n = \phi (x)^{-n}. \] Hence, for every $n\in \mathbb{Z} $ and $x\in G$ we have \[ \phi \left( x^n \right) = \phi (x)^n. \]