Lemma 8.1 Suppose that X = (X,*) is a group and suppose a E X has finite order n > 1. Ifr, s € Zn are such that r < s then a" ‡ as. If m is any integer, then there exists a unique integer r E Zn such that am = ar. 1. 2.
Lemma 8.1 Suppose that X = (X,*) is a group and suppose a E X has finite order n > 1. Ifr, s € Zn are such that r < s then a" ‡ as. If m is any integer, then there exists a unique integer r E Zn such that am = ar. 1. 2.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Lemma 8.1
Suppose that X = (X,*) is a group and suppose a E X has finite order n > 1.
Ifr, s € Zn are such that r < s then a* ‡ a³.
If m is any integer, then there exists a unique integer r € Zn such that am = ar.
1.
2.
Problem 9. Let's prove Lemma 8.1. First, suppose by way of contradiction that 0 ≤ r < s < n and a” = a³.
Part (a). Explain why as¯r = ɛ, where ɛ is the identity for X.
Part (b). Explain why the equation in Part (a) is impossible.
Part (c). Use the assumption that a" = ɛ along with the Division Algorithm to prove Claim (2).
Lemma 9.1 tells us Pow[a] = {ɛ, a,
an-¹}, and these elements are all distinct. Consequently, Pow[a] contains
the same number of elements as does Zn; hence, there are bijections from Pow[a] to Zn. Lemma 9.1 tells us one
such bijection is f : Pow[a] → Zn defined by f(a¹) = r.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1b65600-d204-4706-9da4-f66117221fbc%2F0aede41f-5369-4f08-8e95-98c342b32d47%2Fvc5qj3a_processed.png&w=3840&q=75)
Transcribed Image Text:Lemma 8.1
Suppose that X = (X,*) is a group and suppose a E X has finite order n > 1.
Ifr, s € Zn are such that r < s then a* ‡ a³.
If m is any integer, then there exists a unique integer r € Zn such that am = ar.
1.
2.
Problem 9. Let's prove Lemma 8.1. First, suppose by way of contradiction that 0 ≤ r < s < n and a” = a³.
Part (a). Explain why as¯r = ɛ, where ɛ is the identity for X.
Part (b). Explain why the equation in Part (a) is impossible.
Part (c). Use the assumption that a" = ɛ along with the Division Algorithm to prove Claim (2).
Lemma 9.1 tells us Pow[a] = {ɛ, a,
an-¹}, and these elements are all distinct. Consequently, Pow[a] contains
the same number of elements as does Zn; hence, there are bijections from Pow[a] to Zn. Lemma 9.1 tells us one
such bijection is f : Pow[a] → Zn defined by f(a¹) = r.
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