I’m currently going through Lang’s Algebra, I’ve only recently started it, and I’m trying to do every proof left as an exercise. One such question is the following:

Let $G$ be a cyclic group, and let $a,b$ be two generators. Then there exists an automorphism of $G$ mapping $a$ onto $b$. Conversely, any automorphism of $G$ maps $a$ on some generator of $G$.

I think I was able to prove the second part of the statement (see below), but I am struggling with the first part.

For the second part, assume $f$ is an automorphism of a cyclic group $G$ with generator $a$, then there is a unique $x\in G$ such that $f(x)=a$. Since $G=\langle a\rangle$, we have for all $y\in G, y=f(x)^k=f(x^k)$. Since $f$ is an automorphism it has an inverse, then $\forall y\in G, \; f^{-1}(y)=x^k$ but $f^{-1}$ is an automorphism as well so $f^{-1}(y)=g\in G$, so $\forall g\in G, g=x^k$ with $f(x)=a$ hence $x$ is a generator. Hopefully that’s correct…?

Now for the first part of the statement, any endomorphism of a cyclic group $G$ such that $f(a)=b$ (where $a,b$ are generators of $G$) is surjective since $f(a^n)=f(a)^n=b^n$. But I’m not sure how to show that it can also be injective. I feel there’s something simple I’m missing here. Any help is appreciated.

  • $\begingroup$ If $a^k$ is in the kernel of $f$, then $e = b^k = f(a)^k = f(a^k)$, so the order of $b$ divides $k$. How are the orders of $b$ and $a$ related? $\endgroup$ May 10, 2020 at 23:13
  • $\begingroup$ @ArturoMagidin I think that since both $a$ and $b$ generate $G$ they both have the same order right ? $\endgroup$
    – t_kln
    May 10, 2020 at 23:19
  • $\begingroup$ @T_Kln That's correct. $\endgroup$ May 10, 2020 at 23:38
  • $\begingroup$ Oh and that would suffice to have $f$ injective because $\mathrm{ord}(a)=\mathrm{ord}(f(a))$, if so, is that argument still valid in the case where $G$ is infinite because then the orders of the generators are not finite if I'm not mistaken. $\endgroup$
    – t_kln
    May 10, 2020 at 23:44

1 Answer 1


We use the result that the homomorphic image of a cyclic group is cyclic to prove part 2.

For part 1, we are supposed to prove the existence of an automorphism on the group $G$ mapping any two given generators of $G$, $a$ and $b$.

We will first prove that if a group homomorphism mapping the generator $a$ to another generator $b$ exists, it has to be bijective. If $\varphi\,:G\rightarrow G$ is a homomorphism (or an endomorphism, if you are specific about terminology) such that $\varphi(a)=b$ then $\varphi(a^n)=\varphi(a)^n=b^n$, and it follows that $\varphi$ is surjective. Moreover, if $a^k$ is in the kernel of $\varphi$, then $e=b^k=\varphi(a)^k=\varphi(a^k)$, so that the order of $b$ divides $k$. Finally, since the order of $a$ equals the order of $b$, by virtue of both of them being generators of $G$, it follows that $\varphi$ is injective, since $|a|=|\varphi(a)|$. Thus, $\varphi$ is an automorphism on $G$.

We now show the existence of the automorphism mapping the generator $a$ to another generator $b$ by explicit construction. If $x$ is a generator of $G$, then any element in the group is of the form $nx$ for some positive integer $n$. In particular, since $a$ is a generator, the other generator $b$ in question, also being a member of the group, is of the form $ma$ for some positive integer $m$. The mapping $x\mapsto mx\,\forall\,x\,\in\,G$ is the required homomorphism.

  • $\begingroup$ @MarianoSuárez-Álvarez Thank you for pointing that out. I will edit the answer. $\endgroup$ Dec 29, 2022 at 10:37

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