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Consider the group $$ℤ^*_{55}$$

Is exponentiating to the 3rd power a permutation of: $$ℤ^*_{55}$$ And exponentiation to the 5th power?

I'm trying to solve this problem related to groups, but I don't know how to do it. Is there a mechanical way to find it? Something like a formula?

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    $\begingroup$ Hint: Write the modulus ($n=55$ in the question) as a product of prime powers $\displaystyle n=\prod{p_i}^{k_i}$. Use the Chinese Remainder Theorem to reduce the problem to moduli of the form $n=\displaystyle{p_i}^{k_i}$. Solve that. $\endgroup$
    – fgrieu
    Commented Mar 23, 2022 at 18:36

2 Answers 2

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A simpler approach is to prove that the map $$(\cdot)^p: \mathbb{Z}_m^{*} \rightarrow \mathbb{Z}_m^{*} \quad x \mapsto x^p$$ is a bijection. To do this determine the order of $\mathbb{Z}_{m}^{*}$, now by Lagrange's theorem you should know whether or not there are any elements of order $p$ (does $p\mid\varphi(m)$?). If there are elements of order $p$, can you see what the issue might be? If there are none, then injectivity follows with a short proof.

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You might want to begin by considering an easier problem: for $k, n \in \mathbb{N}$, whether $f(g)=g^k$ is a permutation in the cyclic group $C_n$. Then you can look into the decomposition of $\mathbb{Z}^\star_{55}$ into cyclic groups and proceed from there.

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