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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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
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.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
