The Gateway to Computer Science Excellence
0 votes
28 views

Let $G$ be a group with identity element $e$. If $x$ and $y$ are elements in $G$ satisfying $x^5y^3=x^8y^5=e$, then which of the following conditions is true?

  1. $x=e, \: y=e$
  2. $x\neq e, \: y=e$
  3. $x=e, \: y \neq e$
  4. $x\neq e, \: y \neq e$
in Set Theory & Algebra by Veteran (431k points)
recategorized by | 28 views

1 Answer

0 votes
Start with x5y3=x8y5x5y3=x8y5 and multiply it with x−5x−5 to the left and with y−3y−3 to the right. We get e=x3y2e=x3y2 and therefore x3y2=x5y3=ex3y2=x5y3=e By the same way we get to e=x2y=x3ye=x2y=x3y and again this yields x=ex=e Replacing in the initial identity we get y3=y5=ey3=y5=e and through the same approach we get to y=e
by (17 points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,365 answers
198,493 comments
105,260 users