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  "body": "\n## Proof\n\nWe compute with powers of a generator, and we use Bézout's identity to invert $\\sigma_a$.\n\n1. Let $G$ be a cyclic group of order $n$ with identity $e$, and let $g$ be a [[generator|0B85Q41RP6]]. By the [[definition of $\\langle g \\rangle$|0YT586TKXF]], each $x \\in G$ is $g^k$ for some integer $k$.\n2. If $g$ had infinite order, $\\langle g \\rangle$ would be infinite by [[the order of a cyclic subgroup|031M8HX7DA]]. So $g$ has finite order $m$, and $|\\langle g \\rangle| = m$. Hence $g$ has order $n$.\n3. Let $x = g^k \\in G$. By the [[laws of exponents|00DRK90Z6K]], $x^n = g^{kn} = (g^n)^k = e$.\n4. For an integer $a$, let $\\sigma_a \\colon G \\to G$, $\\sigma_a(x) = x^a$. Let $x = g^i$ and $y = g^j$. The laws of exponents give $(xy)^a = (g^{i+j})^a = g^{ia + ja} = g^{ia} g^{ja} = x^a y^a$.\n5. By step 4, each $\\sigma_a$ is a [[group homomorphism|0HV9Y7WG5R]] from $G$ to $G$.\n6. Let $a \\equiv b$ [[modulo $n$|0TPW25CBN6]], say $a = b + tn$ with $t \\in \\mathbb{Z}$. For $x \\in G$, step 3 gives $x^a = x^b (x^n)^t = x^b$. So $\\sigma_a = \\sigma_b$.\n7. For integers $a, b$ and $x \\in G$, $\\sigma_a(\\sigma_b(x)) = (x^b)^a = x^{ab}$. So $\\sigma_a \\circ \\sigma_b = \\sigma_{ab}$.\n8. Let $\\gcd(a, n) = 1$. By [[Bézout's identity|0DQVHMYAGA]], there are integers $u, v$ with $au + nv = 1$. So $au \\equiv 1$ and $ua \\equiv 1$ modulo $n$.\n9. By steps 6 to 8, $\\sigma_a \\circ \\sigma_u = \\sigma_u \\circ \\sigma_a = \\sigma_1$, which is the identity of $G$. So $\\sigma_a$ is [[bijective|01VG2A0BHW]].\n10. By steps 5 and 9, $\\sigma_a$ is an automorphism of $G$ when $\\gcd(a, n) = 1$. This is the first claim, and $\\sigma_a$ lies in the [[automorphism group|0CWXCWP0Y0]] $\\operatorname{Aut}(G)$.\n11. By [[the units of Z/nZ are the residues coprime to n|0GXNHEHQNA]], the [[group of units|08FHNR8Z5D]] $(\\mathbb{Z}/n\\mathbb{Z})^\\times$ is the set of classes $a \\bmod n$ with $\\gcd(a, n) = 1$.\n12. By steps 6, 10 and 11, the rule $\\Phi(a \\bmod n) = \\sigma_a$ defines a map $\\Phi \\colon (\\mathbb{Z}/n\\mathbb{Z})^\\times \\to \\operatorname{Aut}(G)$.\n13. The product of the classes of $a$ and $b$ is the class of $ab$. By step 7, $\\Phi$ of this product is $\\sigma_{ab} = \\sigma_a \\circ \\sigma_b$. So $\\Phi$ is a homomorphism.\n14. Let $\\Phi(a \\bmod n) = \\Phi(b \\bmod n)$. Then $g^a = g^b$, so $g^{a-b} = e$. By step 2 and [[the criterion for $g^n = e$|0F5QC101Q7]], $n$ divides $a - b$. So $\\Phi$ is injective.\n15. Let $\\alpha \\in \\operatorname{Aut}(G)$. By step 1, $\\alpha(g) = g^a$ for some integer $a$.\n16. For $k \\ge 0$, $\\alpha(g^k) = \\alpha(g)^k$ by induction on $k$. The case $k = 0$ holds since [[homomorphisms preserve the identity|08FKQD7NJ3]], and $\\alpha(g^{k+1}) = \\alpha(g^k) \\alpha(g)$.\n17. For $k \u003e 0$, $\\alpha(g^{-k}) = \\alpha((g^{-1})^k) = \\alpha(g^{-1})^k = (\\alpha(g)^{-1})^k = \\alpha(g)^{-k}$, since homomorphisms preserve inverses.\n18. By steps 16 and 17 and the laws of exponents, $\\alpha(g^k) = (g^a)^k = g^{ak} = \\sigma_a(g^k)$ for each integer $k$. By step 1, $\\alpha = \\sigma_a$.\n19. Since $\\alpha$ is surjective, $g = \\alpha(g^k) = g^{ak}$ for some integer $k$. So $g^{ak - 1} = e$, and by the criterion used in step 14, $ak - 1 = nt$ for some integer $t$.\n20. Thus $ak - nt = 1$. Each common divisor of $a$ and $n$ divides $1$, so the [[greatest common divisor|02WF878BH1]] of $a$ and $n$ is $1$.\n21. By steps 18 to 20, $\\alpha = \\Phi(a \\bmod n)$. So $\\Phi$ is surjective.\n22. By steps 13, 14 and 21, $\\Phi$ is a bijective homomorphism, that is, an [[isomorphism|085PND4KNT]] from $(\\mathbb{Z}/n\\mathbb{Z})^\\times$ to $\\operatorname{Aut}(G)$.\n\n## Idea\n\nEvery automorphism is fixed by where it sends the generator $g$, and it must send $g$ to some power $g^a$. Then it is the power map $x \\mapsto x^a$, which only depends on $a$ modulo $n$. The power map is invertible exactly when $a$ is invertible modulo $n$, and composing power maps multiplies the exponents.\n",
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