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Let $\varphi:G\to G$ . and $\varphi(g) = g^{-1}$ which I said as inverse map.

First, I proof $\varphi$ is an isomorphism by checking bijection and homomorphism.

and I know the theorem which said the inverse of isomorphic map is also isomorphic. Thus I said identity map is isomorphic.

I noted that $G$ is infinite cyclic group, so it is isomorphic to $\Bbb Z$.

So I thought the operation on $G$ should be addition "+".

Then I conclude the theorem that there is no other function that can goes from $G$ to $G$.

But I need some exact proof and want to know it's wrong or not.

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Because the group is cyclic, $G=\langle g\rangle$, we have $f(G) = \langle f(g)\rangle$. Since it is a homomorphism, if $f(g) = kg$ then

$$f(ng) = nf(g) =nkg.$$

In order for this to be surjective, $g\in f(G)$ hence $nk=1$ for some $n$ and therefore $k|1$ by definition, but then $k=\pm 1$ is necessarily the case.

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  • $\begingroup$ Thank you. I really appreciate your exact proof. I also want to ask you if my proof about Identity map is isomorphic. $\endgroup$ – 최선웅 Mar 29 '17 at 16:39
  • $\begingroup$ @최선웅 Your proof of that is not correct since the inverse of the inversion map is itself (not the identity). Instead use the fact that the composition of isomorphisms is an isomorphism, then you can use the fact that $\varphi\circ\varphi =\text{id}_G$ is the composition proves it is an isomorphism. $\endgroup$ – Adam Hughes Mar 29 '17 at 16:43
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Let's start with a group $G=\langle g\rangle$. Assuming that you prove that the inverse map (in this case, using addition it would be $f(g)=-g$), as you've stated, what remains is to prove that there are no other such maps.

A typical way to do this is to assume that you have an isomorphism and then prove that it must be one of the two functions that you've already identified.

Let $h:G\to G$ be an isomorphism. Then $h(g)=ng$ for some $n\in \mathbb{Z}$ (because $G$ is generated by $g$). Furthermore, $h^{-1}(g)=mg$ for some $m\in\mathbb{Z}$ (for the same reason). Then \begin{align} g&=h^{-1}(h(g))\\&=h^{-1}(ng)\\&=nh^{-1}(g)\\&=nmg \end{align}

Since $G$ is infinite, $nm=1$, which immediately implies that $n=m=1$ or $n=m=-1$ and $h$ is one of the previously identified isomorphisms.

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Let $u$ be a generator of $G$. Then the only other generator is $-u$.

If $\phi$ is an automorphism of $G$, then $\phi(u)$ must be a generator and so $\phi(u)=\pm u$.

Therefore, $\phi$ is the identity map or the inverse map.

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