A group is an algebraic structure that is closed under binary operation, associative, each of its element has an inverse and there is a common identity element.
Problem
Let be a finite group of even order. Show that has an element such tha
Proof
Let be the group of even order.
Suppose that each element of is paired with its inverse ( Note that the identity element is paired to itself)
Then consists of the union of all of the pairs
Now,
Suppose that no element other than identity, is paired to itself
Let the number of pairs excluding the identity be , then (since there are pairs + the identity)
But, this makes the order of odd, a contradiction, and thus there must be at least one element that is paired with its inverse (this makes the order of even)
Let be paired with its inverse, then , and hence
-
- Active Topics
-
-
- by Eli 1 day ago Iran Launches Retaliatory Attack Against Israel, and Israel Retaliates by Attacking Iranian Isfahan Millitary Base View the latest post Replies 28 Views 882
- by Eli 2 days ago All in One: YouTube, TED, X, Facebook and Instagram Reels, Videos, Images and Text Posts View the latest post Replies 319 Views 8940
- by Eli 3 days ago Re: What is in Your Mind? View the latest post Replies 685 Views 274054
- by Eli 5 days ago Python Packages for Scientific Computing View the latest post Replies 8 Views 3000
- by Eli 5 days ago Dunia Yetu: Building Tanzania's Digital Future Together View the latest post Replies 5 Views 1839
- by Eli 1 week ago Russia Invades Ukraine View the latest post Replies 646 Views 210446
- by Eli 1 week ago Programmatically Move Files from One Folder to Another View the latest post Replies 6 Views 1426
- by Eli 1 week ago Collection of Greatest Christian Hymns of all Times View the latest post Replies 33 Views 43733
- by Eli 2 weeks ago What is Retrieval-Augmented Generation (RAG)? View the latest post Replies 2 Views 358
- by Eli 2 weeks ago Chat With ChatGPT - An Interactive Conversational AI View the latest post Replies 22 Views 24353
-
A Group of Even Order
-
- Information
-
Who is online
Users browsing this forum: No registered users and 0 guests