How to show that $ U(3^{n})= \langle [2] \rangle $ for all natural number. I think 1st I have to find out the order of $[2]$ in $ U(3^{n})$ and then find the size of $ U(3^{n})$. But i can't do, please someone help me.
Multiplication group of integers modulo
0
$\begingroup$
abstract-algebra
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0$ [2] $ will not even be in the multiplicative group if $ n $ is even, and even if it is, it might not generate the entire group. Did you mean to say $ 3^n $ and not $ 3n $? – 2017-02-10
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0yes, my mistake – 2017-02-10
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0Do you see how it suffices to show that $ 2^{2 \cdot 3^n} \neq 1 \pmod{3^{n+2}} $? (Use induction on $ n $ to get the claim from this.) – 2017-02-10
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0can you respond to proof , please – 2017-02-10