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I really appreciate all of you who helped me! Please take a look at my argument below and feel free to give me some comments:

Write $G = \{ e, a, a^2, a^3, a^4 \}$. Since any element $f$ in Aut$(G)$ is an isomorphism that maps $G$ to itself, we notice that $f(e)=e$ and a generator of $G$ is always mapped to a generator of $G$ under $f$. But we also observe that $a, a^2, a^3$ and $a^3$ are generators of $G$. Then we can find $f_1, f_2, f_3, f_4 \in$Aut$(G)$ such that

  • $f_1(a) = a$
  • $f_2(a)=a^2$
  • $f_3(a)=a^3$
  • $f_4(a)=a^4$

Note that the above exhaust all elements in Aut$(G)$. If not, then we have $f(a)=e$, but $e$ is not a generator of $G$. Then we define $\phi:$Aut$(G) \rightarrow \mathbb{Z}_4$ as follow: For any $f \in$ Aut$(G)$, if $f(a)=a^i$, then $\phi(f) = \overline{i-1}$. By the above, we observe that the only possible values of $\overline{i-1}$ are $\bar{0}, \bar{1}, \bar{2}, \bar{3}$, which are exactly the elements of $\mathbb{Z}_4$. Then the following can be checked:

1. $\phi$ is a homomorphism.

  1. $\phi$ is bijective.

This completes the proof.

To a certain extent, I can complete the whole proof, but I am not sure about the two bold sentences. 1. Do I make the correct claim about exhausting all elements in Aut$(G)$?

  1. How to show that $\phi$ is a homomorphism? I verify that it is a homomorphism by some testing. But in general case, I get the following: $$(f \circ g)(a) = f(a^j) = a^{ij}$$ but $$f(a)g(a)=a^ia^j=a^{i+j}.$$ What happen here?

It is a question from my abstract algebra test! I am completely lost here. I have spent some time after the test but still have no clues.

Let $G$ be a cyclic group of order $5$. Show that Aut$(G) \cong \mathbb{Z}_4$.

Aut$(G)$ : the set containing all the isomorphisms that map $G$ to itself

The only thing I get is $G = \{ e, a, a^2, a^3, a^4 \}$. I try to define explicitly an isomorphism between Aut$(G)$ and $\mathbb{Z}_4$. But I cannot move on: if you give me any $f \in$Aut$(G)$, how can I convert it to an element of $\mathbb{Z}_4$? I also notice that the order of $G$ is $5$, which is a prime number, but no other things can I tell.

Please don't give me the whole proof! It would be great if you can leave me some hints so that I can work out the proof on my own! Thanks in advance.

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  • $\begingroup$ First observe that if you know $f(a)$ then you know $f(a^2)$ et cetera. In other words you know all of $f$. What choices of $f(a)$ give rise to a well defined homomorphism that also happens to be a bijection. $\endgroup$ – Jyrki Lahtonen Oct 3 '14 at 6:56
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To show $Aut(G)\cong \mathbb{Z_4}$, pick any $f \in Aut(G)$. Now your trouble is how you will associate it with an element of $\mathbb{Z_4}$. I guess the reason you are getting confused is because you are not putting enough concentration on elements of $Aut(G)$ i.e. $f$, think like this , $f:G \to G$ where $G=<a>$, so write $f$ as $f:<a> \to$ $<a>$. You muse know under $f$ where will $e$ go? Now $a$ is the generator of <$a$>, so next thing you need to do is think where can $a$ possibly go? as soon as you will figure it out, you will see the result right in front of your eyes. TRY!!

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Write down possible images of $a\in G$..

The very next line would turn out to be the one you are searching for....

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    $\begingroup$ (+1) ... and also observe that if $f(a)=\ldots$ and $g(a)=\ldots$, then $(g\circ f)(a)=\ldots$, where $\ldots$ $\endgroup$ – Jyrki Lahtonen Oct 3 '14 at 6:54
  • $\begingroup$ You mean if $f \in$Aut$(G)$, find out all $f(a) \in G$? $\endgroup$ – Nighty Oct 3 '14 at 6:54
  • $\begingroup$ @LeeKM : Yes... try recalling what is definition of homomorphism of a group and look at the order of $a$... $\endgroup$ – user87543 Oct 3 '14 at 6:56
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Hint: Any homomorphism $\Phi$ of a group $G$ is characterized by its values on a set of generators. Your group $\mathbb{Z}_5$ is cyclic, and so generated by a single element $a$, so any homomorphism $\Phi: G \to G$ is determined by $\Phi(a)$. There are five possibilities: $\Phi(a) = e$, $\Phi(a) = a$, $\Phi(a) = a^2$, etc. Which of these define automorphisms?

(NB that in general not all assignments of generators to elements of the codomain need define homomorphisms, but in this case they do, as you can readily check.)

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