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Problem:

If $a, b \in G$ are generators of a group $G$, then there is an automorphism $f:G\to G$ such that $f(a)=b$.

Since $G=\langle a \rangle$ and $b\in G$, I proposed that $f$ be given by: $$ f(g)=g^t, \text{where } t\in\mathbb{Z} \text{ is the integer such that }a^t=b$$

This is an homomorphism, but I'm not sure if is injective and surjective since I have not been able to prove it so far.

Is this really an isomorphism? If it is, can you give me some hints to prove it please?

Edit#1: I already managed to solve the surjectivity part. Here is my attempt to prove injectivity.

Let $x \in \ker f$, this is $e=f(x)=x^t$

Since $G=\langle a \rangle$ and $x\in G$, there is a $k\in \mathbb Z$ such that $x=a^k$. Then, $e=x^t=(a^k)^t=(a^t)^k=b^k$.

Here I consider two different cases. If $G$ is infinite, since $G=\langle b \rangle$ all powers of $b$ must be different, therefore given that $e=b^k$ we have $k=0$. Consequently $x=a^k=a^0=e$.

If $G$ is finite, let's suppose $|G|=n$. Since $a$ and $b$ are generators, the order of both elements is $n$. From $b^k=e$ we have that $n$ divides $k$, this is, $k=nk'$. Therefore, $x=a^k=a^{nk'}=(a^n)^{k'}=e^{k'}=e$.

Is this correct? If so, do you think there is a more direct argument to prove it?

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  • $\begingroup$ Well, if it weren't injective, say, then there would be a non-trivial element with $f(h)=e$. What could you then say about $h$? $\endgroup$ – lulu Sep 23 '17 at 13:42
  • $\begingroup$ $h$ has finite order? $\endgroup$ – Карпатський Sep 23 '17 at 13:44
  • $\begingroup$ Should I treat this in cases, when $G$ is finite and when is infinite? $\endgroup$ – Карпатський Sep 23 '17 at 13:46
  • $\begingroup$ Well, what can you say about the order of $h$? Is it a possible order? $\endgroup$ – lulu Sep 23 '17 at 13:48
  • $\begingroup$ I see that the order of $h$ must divide $t$, but I don't see contradictions with that... $\endgroup$ – Карпатський Sep 23 '17 at 13:55
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Here is a roadmap:

  • If $G=\langle a \rangle$ and $b \in G$, then there is an homomorphism $f:G\to G$ such that $f(a)=b$.

  • The image of $f$ is $\langle b \rangle$.

  • If $G=\langle b \rangle$, then $f$ is injective.

For the last step, you need to consider the two cases: $G$ finite and $G$ infinite.

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  • $\begingroup$ Thanks. I edited the post and the solution I proposed is based on your answer. Can you check if is correct, please? $\endgroup$ – Карпатський Sep 24 '17 at 13:30
  • $\begingroup$ @Карпатський, add your solution as a separate answer. $\endgroup$ – lhf Sep 24 '17 at 15:37
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    $\begingroup$ @Карпатський, when $G$ is finite, injectivity follows from surjectivity. $\endgroup$ – lhf Sep 24 '17 at 15:38

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