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Proposition 3. Let G be an arbitrary group, x E G and let m, n e Z. If x" = 1 and x = 1, then x = 1, where d = (m, n). In particular, if x = 1 for some m € Z, then |x| divides m. Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers r and s such that d = mr + ns, where d is the g.c.d. of m and n. Thus xd = xmr+ns = (xm)²(x") = 1'1$ = 1. This proves the first assertion. If x= 1, let n = |x|. If m = 0, certainly n|m, so we may assume m #0. Since some nonzero power of x is the identity, n < ∞o. Let d = (m, n) so by the preceding result x = 1. Since 0 < d < n and n is the smallest positive power of x which gives the identity, we must have d = n, that is, n | m, as asserted.

Proposition 3. Let G be an arbitrary group, x E G and let m, n e Z. If x" = 1 and x = 1, then x = 1, where d = (m, n). In particular, if x = 1 for some m € Z, then |x| divides m. Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers r and s such that d = mr + ns, where d is the g.c.d. of m and n. Thus xd = xmr+ns = (xm)²(x") = 1'1$ = 1. This proves the first assertion. If x= 1, let n = |x|. If m = 0, certainly n|m, so we may assume m #0. Since some nonzero power of x is the identity, n < ∞o. Let d = (m, n) so by the preceding result x = 1. Since 0 < d < n and n is the smallest positive power of x which gives the identity, we must have d = n, that is, n | m, as asserted.

Advanced Engineering Mathematics
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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please explain the following proposition in more detail, I am straight up confused at what we are doing here, and why we are proving this. Thank you in advance

Proposition 3. Let G be an arbitrary group, x € G and let m, n € Z. If x" = 1 and x = 1, then x = 1, where d = (m, n). In particular, if x = 1 for some m € Z, then |x| divides m. Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers r and s such that d = mr + ns, where d is the g.c.d. of m and n. Thus xd = xmr+ns = (xm)' (x") = 1'15 = 1. This proves the first assertion. If x= 1, let n = |x|. If m = 0, certainly n|m, so we may assume m #0. Since some nonzero power of x is the identity, n<∞o. Let d = (m, n) so by the preceding result x = 1. Since 0 < d ≤ n and n is the smallest positive power of x which gives the identity, we must have d = n, that is, nm, as asserted.
Transcribed Image Text:Proposition 3. Let G be an arbitrary group, x € G and let m, n € Z. If x" = 1 and x = 1, then x = 1, where d = (m, n). In particular, if x = 1 for some m € Z, then |x| divides m. Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers r and s such that d = mr + ns, where d is the g.c.d. of m and n. Thus xd = xmr+ns = (xm)' (x") = 1'15 = 1. This proves the first assertion. If x= 1, let n = |x|. If m = 0, certainly n|m, so we may assume m #0. Since some nonzero power of x is the identity, n<∞o. Let d = (m, n) so by the preceding result x = 1. Since 0 < d ≤ n and n is the smallest positive power of x which gives the identity, we must have d = n, that is, nm, as asserted.
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