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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Карпатський, when $G$ is finite, injectivity follows from surjectivity. $\endgroup$ – lhf Sep 24 '17 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.