Let $G$ be a group. The set of all automorphisms of $G = \operatorname{Aut}(G$), with $(\operatorname{Aut}(G), \circ)$ also being a group.

Consider $C_n=\langle g:g^n=1\rangle$, the cyclic group of order $n$. For each positive integer $m$ with $1\leq m\leq n$ define a map $f_m:C_n\rightarrow C_n$ as follows: for $r\in \mathbb{Z}, f_m(g^r):=g^{rm}$

Show that if $f:C_n\rightarrow C_n$ is a homomorphism of $C_n$ then $f=f_m$ for some $m$.

Hint: show that if $f(g)=g^k$ then $f=f_k$.

I do not quite know what to do. I know when $r=1$, $f_m(g)=g^m$ but can't see if this helps.

  • $\begingroup$ To know the image of $g$ i enough since it generates $C_n$ $\endgroup$ – Ashar Tafhim Feb 22 '17 at 19:16

Recall that a homomorphims is such that $f(gh)=f(g)f(h)$. Thus by induction we have that $f(g^r)=f(g)^r$.

Now suppose $f$ is an homomorphism from $C_n\to C_n$. Since $C_n=\langle g\rangle$ we then have $f(g)= g^k$ for some $0\leq k \leq n$. Therefore $f(g^r)=f(g)^r=(g^k)^r= g^{rk}$ and $f=f_k$.

  • $\begingroup$ I know that $f(gh)=f(g)f(h)$ but I can't see exactly how you got $f(g^r)=f(g)^r$ $\endgroup$ – harry55 Feb 22 '17 at 19:29
  • $\begingroup$ The base case for induction is $f(g)=f(g)$. Now assume for some $k\in \Bbb N$, $f(g^k)= f(g)^k$. Thus we have that $f(g^{k+1})=f(g g^k)=f(g)f(g)^k=f(g)^{k+1}$. Hence by induction you have the result. $\endgroup$ – user416426 Feb 22 '17 at 19:58
  • $\begingroup$ What is your $f(h)$ in $f(g^r)=f(g)^r$? $\endgroup$ – harry55 Feb 22 '17 at 20:04
  • $\begingroup$ In my proof that that formula holds you mean? It would in that case be $g^k$. $\endgroup$ – user416426 Feb 22 '17 at 20:10
  • $\begingroup$ I understand from the induction why $f(g^r)=f(g)^r$ but I can't see how you decided to use the fact $f(g^r)=f(g)^r$ in the first place. Where did it come from? Just your own intuition? $\endgroup$ – harry55 Feb 22 '17 at 20:40

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.