Oreder of groups and their elements

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Homework Help Overview

The discussion revolves around proving that any group of even order contains at least one element of order 2. Participants explore the relationship between the order of a group and the orders of its elements, particularly focusing on groups with even cardinality.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the implications of having no elements of order 2 and how that would affect the group's order. They consider the pairing of elements and the identity element's role in determining orders. Questions arise about identifying specific elements of order 2 in groups of even order.

Discussion Status

The discussion is active, with participants offering insights into the implications of group order and the existence of elements of order 2. Some participants express understanding while others seek clarification on how to demonstrate the existence of such elements without specifying which they are.

Contextual Notes

Participants are navigating the constraints of proving a general statement about groups of even order without relying on specific examples or known groups. There is an emphasis on understanding the theoretical underpinnings rather than providing direct solutions.

Bellarosa
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Show that any group of even order has at least an element of order 2

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3. I know that the order of a groups tells you how many elements the group consist, but just randomly assuming that it has at least an order of 2 is what I can't really understand. For example |G| = 6, which means that G = {a,b,c,d,e,f} I know that one of its element is the identity element which is e, but the order of the other elements can all be 2, or one element can only have order 2. I just need to understand how the order of a group relates to he order of its element.
 
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If there is no element of order 2 then g is not equal to g^(-1) for every element in G except for e (the identity), right? Do you see it yet?
 
which means that the group will only have one element right...I'm understanding but it's still a bit confusing...
 
The group will be split into pairs {g,g^(-1)} which all have two elements and finally {e} which only has one element. Looks to me like that would imply the order of the group would be odd. Seem so to you?
 
yes the group will have an odd order
 
Right. So any group that has no element of order two has odd order. So if a group has even order...?
 
then how can you determine which elements are of order 2
 
No, if you know a group has order eight, then one element is it's own inverse (so has order two).
 
Bellarosa said:
then how can you determine which elements are of order 2

You can't until you know what the group is. You just know there must be one.
 
  • #10
ok I think I get it ... you're saying that for example G = {e, a, b, c} then if a has an order of two then a^2 = e, or can you give me a general example?
 
  • #11
Take Z_6. The set of integers mod 6 under addition. The order of the group is 6 which is even. So one of those integers must have order two. Which one?
 
  • #12
3...this I understand I just have trouble showing how an unspecified group of even order has an element of order 2
 
  • #13
That - is - what - you - are - supposed - to - prove. You don't have to say which one it is, you are just supposed to show it exists.
 
  • #14
ok got you
 

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