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.

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Chapter2: Second-order Linear Odes
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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.
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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