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I want to show that $\text{Aut}(\mathbb{Z}_8)$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.

The group $\mathbb{Z}_8$ is cylcic and is generated by one element. The possible generators are $1,3,5,7$.

Each automorphism maps each of the element of the set $\{1,3,5,7\}$ to one of the element $\{1,3,5,7\}$, right?

But how can we show that that the automorphism group of $\mathbb{Z}_8$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ ?

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An automorphism $f:\Bbb Z_8 \to \Bbb Z_8$ is determined by $f(1)$ (prove it). Since $1$ is a generator and generators must map to generators, $f(1)\in \{1,3,5,7\}$. So there are 4 automorphisms. Since the only groups of order 4 are $\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$, you just have to show that $Aut(\Bbb Z_8)$ has no element of order 4.

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    $\begingroup$ You can also just write down an explicit isomorphism. $\endgroup$ – Qiaochu Yuan Feb 17 '16 at 22:47
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    $\begingroup$ I mean that if $f,g:\Bbb{Z}_8\to \Bbb Z_8$ are automorphisms (or homomorphisms for that matter), then $f=g$ is equivalent to $f(1)=g(1)$. To prove this just think about $f(n)=f(1+...+1)=f(1)+...+f(1)$ (with $n$ copies of 1). $\endgroup$ – Nitrogen Feb 17 '16 at 22:52
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    $\begingroup$ Yes that's exactly right. $\endgroup$ – Nitrogen Feb 17 '16 at 23:38
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    $\begingroup$ Not all elements of $\Bbb Z_4$ have order 4, however there are elements of order $4$, which doesn't happen in $\Bbb Z_2 \times \Bbb Z_2$. $\endgroup$ – Nitrogen Feb 18 '16 at 1:19
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    $\begingroup$ If $f(1)=3$ then $f(x)=3x$ not $x+3$. $\endgroup$ – Nitrogen Feb 18 '16 at 23:06

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