Let f be an Euler phi-function
consider the ring z(n), u(z(n)) is a multiplicative group
given any [a] in u(z(n)), gcd(a,n)=1
。(u(z(n)))=f(n) =>[a]^(f(n))=[1] in u(z(n)) (為什麼???)
=>a^(f(n)) =1 mod n
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1 則留言:
拿 a 去造一個 cyclic subgroup H of u(z(n)),
=>。(a)=。(H), 也就是說 a^。(H)=1
by Lagrange's Theorem, 。(H)|。(u(z(n)))
令。(u(z(n))) = f(n) = k*。(H), for some k
=> a^f(n) = (a^。(H))^k = 1^k = 1
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