Aut $\mathbb Z_8$ is isomorphic to $\mathbb Z_2 \oplus\mathbb Z_2$

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

Thanks

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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.
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 :) – user42912 Apr 14 '13 at 6:51
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. – Ittay Weiss Apr 14 '13 at 6:58
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 – user42912 Apr 14 '13 at 8:11