Question

In Your solutions for Pinters a Book of Abstract Algebra Chapter 10 Exercise 6D. I cannot follow the argumentation that ord(a)=ord(bab^-1) for all b implies a=bab^-1 for all b.Also i cannot see how the prerequisite that a is the only element of order k comes into the argumentation?? Please help and explain?

Answer 1

Explanation for where you've stuck = Since a is the only element of order k in G. and ord(a)=ord(bab^-1) says that order of (bab^-1) is same as of. i.e. ord(bab^-1)=k

but a is only element of order k in G. and we've found (bab^-1) to be of order k. so it must be equal to a.

I've also written fresh solution with every argument proved. And also explanation of where hypothesis is used.

First let's show that ord(a) = ord (bab) P if ord (a) =P => 9'-1 (babtje: pabt. I babt, hole: bb?=). e times P bant ball- babaye since alal. onth= ord (bab")./ ord (a). ? _o bbt 9P = b²b =7 a ap=1. & if ord (bab") = p =>. (babyjP=1 balb = 1 =) =7 ord (a) / ord (baby) - O & b = ord (a) = ord (baby) We're using 6D. It's given that for 'a' is only element in G with order k. since for all beG ord(baby) = ord(a) => order of babt is also k. #bEG this where < * (but since a is only element of order ke in o) hypothesis. 7 babt=a. for all bEG multiply by b => => a is commutative will all elements of G bas ab on right side 27 as belongs to sentre