Order of Group Elements in Abelian and Non-Abelian Groups

  • Context: Graduate 
  • Thread starter Thread starter astronut24
  • Start date Start date
  • Tags Tags
    Elements Group Groups
Click For Summary

Discussion Overview

The discussion revolves around the properties of group elements in both abelian and non-abelian groups, particularly focusing on groups of order 2p (where p is prime) and groups where every element has order 2. Participants explore the implications of these properties on the existence of elements of specific orders and the structure of subgroups.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether a group of order 2p that is abelian has exactly one element of order 2, while noting that in the non-abelian case, there could be p elements of order 2.
  • Another participant suggests that if G is abelian and has two elements of order 2, then the set {e, a, b, ab} forms a subgroup, but points out that 4 does not divide 2p unless p=2.
  • A different participant poses a question about showing that if every element in G satisfies a^2=e, then the order of G must be 2^n for some n ≥ 0, and wonders if induction is the method to use.
  • One participant notes that any abelian group of order 2p for p an odd prime is cyclic, which relates to the earlier discussion about the order of elements.
  • Another participant states that if p divides |G|, then there is an element of order p in G, which is relevant to the discussion of groups of order p^a.
  • One participant confirms they have resolved the question regarding groups where a^2=e for all a in G.
  • Another participant suggests that the problem about the group of non-singular upper triangular matrices could be approached by finding a counterexample.
  • A later reply mentions a useful result that if all elements of a group have order 2, then the group is abelian, hinting at a proof involving the inverse of products of elements.

Areas of Agreement / Disagreement

Participants express various viewpoints on the properties of groups of order 2p and the implications of elements having order 2. There is no consensus on the specific claims regarding the number of elements of certain orders in abelian versus non-abelian groups, and the discussion remains unresolved on several points.

Contextual Notes

Some arguments depend on the definitions of group order and element order, and the implications of group structure are not fully resolved. The discussion includes assumptions about the nature of primes and subgroup properties that may not be universally applicable.

astronut24
Messages
10
Reaction score
0
if a group of order 2p ( p prime) is abelian...then does it have exactly one element of order 2 ?? if a group is non abelian...i could figure out that there are p elements of order 2. but the abelian case is a bit confusing...
also..is it like...any group of order 2p has an element of order p?

if a group has orer p^a , a>=1 where p is prime...then I've got to show that G has an element of order p.
can i say that any non-identity element in G can have order p or p^2 or p^3...or p^a. then if x in G is of the form x^(p^i) =e ...we can say, (x^(p^(i-1))^p= e and we've found an element x^(p^(i-1)) that is of order p...it seems too simple to be correct.
 
Physics news on Phys.org
suppose G is abelian and has 2 elements of order 2, say a and b...then {e,a,b,ab} becomes a subgroup of G. but 4 does not divide 2p unless p=2. if p=2 then anyway {e,a} is a subgroup of G of order 2( p in this case).
 
thanks for the help!
here's another question on the same lines: if G is a group where a^2=e for each a in G. show that order of G is 2^n for some n >=0. it's clear that the group is abelian and also clear that 2 divides O(G). how do you proceed further? by any chance, is it induction that we're supposed to use?
 
The first answer also follows by the observation that any abelian group of order 2p for p an odd prime is cyclic.

The second follows from the fact that if p is a prime and p divides |G| then there is an element of order p in G.
 
if G is a group such that a^2=e for each a in G. show that order of G is 2^n for some n>=0.
please help...
is the group of 3 non-singular upper triangular matrices a normal subgroup of
GL(3,R), the group of 3 cross 3 non singular matrices over R.
 
I've answered the a^2=e for all a in G one.

As for the p^r one, all elements have order dividing p^r say the order of g is p^s, then p^{s-1] has order p.
 
And why don't you just do the matrix one? it simple boils down to multiplying matrices together.

Hint: it's probably easier to find a counter example.
 
Just to put here another useful result: if G is any group and all of its elements have order 2 then it is abelian. proof an exercise (hint what is (xy)^{-1})
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
10K
  • · Replies 8 ·
Replies
8
Views
3K