Question

7. Let $r \in U(n)$. Prove that the mapping $\alpha: \mathbb{Z}_n \to \mathbb{Z}_n$ defined by $\alpha(s) = sr \mod n$ for all $s \in \mathbb{Z}_n$ is an automorphism (an isomorphism from a group to itself) of $\mathbb{Z}_n$.

Name: 7. Let r ∈ U(n). Prove that the mapping α: ℤn →ℤn defined by α(s) = sr n for all s ∈ℤn is an automorphism (an isomorphism from a group to itself) of ℤn.
Uploaded: 2022-06-23T13:31:41-08:00
Duration: 2 min 49 s
Channel: Madhur L
Description: 7. Let r ∈ U(n). Prove that the mapping α: ℤn →ℤn defined by α(s) = sr n for all s ∈ℤn is an automorphism (an isomorphism from a group to itself) of ℤn.

          7. Let $r \in U(n)$. Prove that the mapping $\alpha: \mathbb{Z}_n \to \mathbb{Z}_n$ defined by $\alpha(s) = sr \mod n$ for all $s \in \mathbb{Z}_n$ is an automorphism (an isomorphism from a group to itself) of $\mathbb{Z}_n$.

Added by Jacob R.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Madhur L

06/23/2022

Step 1

To show that α is a homomorphism, we need to prove that for any two elements s, t in Zn, α(s + t) = α(s) + α(t) mod n. Let's compute: α(s + t) = (s + t)r mod n = (sr + tr) mod n = (sr mod n + tr mod n) mod n = α(s) + α(t) mod n Therefore, α is a homomorphism. Show more…

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Let rinU(n). Prove that the mapping \\alpha :Z_(n)->Z_(n) defined by \\alpha (s)=srmodn for all sinZ_(n) is an automorphism (an isomorphism from a group to itself) of Z_(n). order. c6r0ap an automorphism (an isomorphism from a group to itself) of Zn.