All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Commutative Group (Posted on 2006-09-18) Difficulty: 2 of 5
Here is a simple problem from abstract algebra.

Prove that a group with exactly five elements is commutative.

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Counting elements | Comment 4 of 8 |
(In reply to Counting elements by vswitchs)

My earlier post overlooked that a or b may be their own inverses.  If this is the case only for one of them, we still have at least six elements by my previous arguments. If both equal their inverses, I show that the products aba and bab will be yet different elements:

Take a b a.  It cannot equal e, as then both ab and ba would equal a_ (the inverse of a).  It cannot equal a, because then ba would equal e (and therefore ab, see my last post). And it cannot equal b, because then (by multiplication with a) ab=ba.  It also cannot equal ab or ba because then a would be the unit element.

Likewise for b a b.  So this again gives us at least six elements (six because aba might equal bab).



  Posted by vswitchs on 2006-09-18 13:28:40

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information