Exercise 1.6.1 ( $\varphi(x^n) = \varphi(x)^n$)

Proof. I write $\mathbb{N}={\mathbb{Z}}^{\text{+}}=\{0,1,2,\dots \}$.

(a)

Let $e$ be the identity of $G$ and $e^{\prime }$ the identity of $H$. Since $\phi$ is an homomorphism, $\phi (e)=\phi (e\cdot e)=\phi (e)\phi (e)$, so $\phi (e)e^{\prime }=\phi (e)\phi (e)$. By multiplying on the left by $\phi {(e)}^{-1}$, we obtain $$\phi (e)=e^{\prime }.$$

By the inductive definition of the power map, for all $x\in G$, and for all $n\in \mathbb{N}$,

$$x^0=e,x^{n+1}=x\cdot x^n.$$

Similarly for $y\in H$, $y^0=e^{\prime }$. So

$$\phi (x^0)=\phi {(x)}^0.$$

Consider the property defined for all integers $n$ by

$$𝒫(n)⟺\phi (x^n)=\phi {(x)}^n.$$

We have proved $𝒫(0)$. Suppose now that $𝒫(n)$ is true for some $n\in \mathbb{N}$, so that $\phi (x^n)=\phi {(x)}^n$. Then

$$\phi (x^{n+1})=\phi (x\cdot x^n)=\phi (x)\phi (x^n)=\phi (x)\phi {(x)}^n=\phi {(x)}^{n+1}.$$

This show $𝒫(n+1)$. The induction is done, so

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N},\phi (x^n)=\phi {(x)}^n.$$

(b)

Note that for all $x\in G$, $\phi (x)\phi (x^{-1})=\phi (x)\phi {(x)}^{-1}=e^{\prime }$, thus $\phi {(x)}^{-1}=\phi (x^{-1})$.

We define the negative powers by

$$x^{-n}={(x^n)}^{-1},(n\in \mathbb{N},x\in G).$$

Then for all $n\in \mathbb{N}$, using part (a),

$$\phi (x^{-n})=\phi ({(x^n)}^{-1})=\phi {(x^n)}^{-1}=\phi ({(x^n)}^{-1})=\phi (x^{-n}).$$

So

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