RE: Random Thoughts
July 16, 2018 at 11:25 pm
(This post was last modified: July 16, 2018 at 11:27 pm by Kernel Sohcahtoa.)
My proof was actually via a direct approach. The specific definition dealt with the order of an element in a group G. Specifically, if an element x in a group G has finite order, then "there is a positive integer n such that x^n=e," where e is the unique identity element of the group G.
The specific exercise asked me to prove that the order of x is equal to the order of x^-1 (assuming that G is a group and x is in G), so I considered the cases where x is of finite order and x is not of finite order (in other words, x has order infinity). For the latter case (order of x equals infinity), taking the negation of the definition of finite order gives "for every positive integer n, x^n does not equal e." Making use of this definition helped me understand that since x^n does not equal e for every positive integer n, then (x^-1)^n = x^-n = 1/x^n does not equal 1/e = e^-1 = e for every positive integer n , and so, the order of x^-1 must be infinity which is equal to the order of x.
The specific exercise asked me to prove that the order of x is equal to the order of x^-1 (assuming that G is a group and x is in G), so I considered the cases where x is of finite order and x is not of finite order (in other words, x has order infinity). For the latter case (order of x equals infinity), taking the negation of the definition of finite order gives "for every positive integer n, x^n does not equal e." Making use of this definition helped me understand that since x^n does not equal e for every positive integer n, then (x^-1)^n = x^-n = 1/x^n does not equal 1/e = e^-1 = e for every positive integer n , and so, the order of x^-1 must be infinity which is equal to the order of x.
