I'm trying to prove that Aut $\mathbb Z_8$ is isomorphic to $\mathbb Z_2 \oplus\mathbb Z_2$, but I have no idea how to prove it. First of all, I'm trying to prove that Aut $\mathbb Z_8$ has four elements. Can I argue that because $\mathbb Z_8$ has four possibilities of generators, say $\bar 1$, $\bar 3$ $\bar 5$, $\bar 7$, since each isomorphism is compleated determined by the image of its generators, then $\mathbb Z_8$ has four elements?

I need help



You're on to a good start. An automorphism of a cyclic group is uniquely and completely determined by the image of any fixed generator, and that image must itself be a generator. That proves to you indeed that $Aut(\mathbb Z_8)$ has four elements.

Now, to go on, you just need to distinguish between the two possibilities for a group of order $4$. It is either isomorphic to $\mathbb Z_4$ or to $\mathbb Z_2 \times \mathbb Z_2$. There are four possibilities for generators of $Aut(\mathbb Z_8)$, so you can try each of them and see if it generates the whole group or not. You'll quickly find the correct answer, proving the result.

  • $\begingroup$ why a group of order 4 has only two possibilities: isomorphic to $\mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2$? thank you for your answer :) $\endgroup$ – user42912 Apr 14 '13 at 6:51
  • $\begingroup$ It requires a bit of thought. It's easy enough to just brute force construct the multiplication table of a group of order $4$ and see that there are only two ways to form such a table. Remembering that in each row and columns all elements of the group must appear (necessarily only once) makes it very easy to complete the multiplication table. $\endgroup$ – Ittay Weiss Apr 14 '13 at 6:58
  • $\begingroup$ I solved the question, but I've been thinking, I explicitly wrote down the elements of Aut $\mathbb Z_8$ and easily saw that this group is isomorphic to the Klein group, do you know a more elegant approach? thank you again $\endgroup$ – user42912 Apr 14 '13 at 8:11
  • $\begingroup$ what is not elegant in what you did? $\endgroup$ – Ittay Weiss Apr 14 '13 at 8:19
  • $\begingroup$ yes, you're right! the problem is I don't like brute force because in bigger automorphisms maybe I will not be able to solve the question using this procedure, but I understood your point. $\endgroup$ – user42912 Apr 14 '13 at 8:23

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.