Show that a Group (G, *) definied by a condition is Abelian

In summary, the conversation discusses a group (G, *) where * is a law, and a condition where (x * y)^i = (x^i) * (y^i) for all i belonging to {2, 3, 4} and (x, y) belonging to G2. The question is to show that G is an Abelian (commutative) group. The solution is to show that if the group is not Abelian, there are elements a and b with ab - ba ≠ 0, and then using the given condition, prove that * is commutative. This can be done for all i values, showing that G is indeed an Abelian group.
  • #1
JPC
206
1

Homework Statement



(G, *) is a group (where * is a law)
And for all 'i' belonging to {2, 3, 4}, for all (x, y) belonging to G2
(x * y) ^ i = (x^i) * (y^i)
(where ^ is the law : to the power of)

Question : Show that G is an Abelian (commutative) group


Homework Equations





The Attempt at a Solution



we have never done any questions of that sort yet, all i can say is that
"(x * y) ^ i = (x^i) * (y^i)" shows that the law '^' (to the power of) is distributive over the law '*' for 'i' belonging to {2, 3, 4}
But then i don't know where to go

Any help or directions would be appreciated, thank you :)
 
Physics news on Phys.org
  • #2
JPC said:
(G, *) is a group (where * is a law)
And for all 'i' belonging to {2, 3, 4}, for all (x, y) belonging to G2
(x * y) ^ i = (x^i) * (y^i)
(where ^ is the law : to the power of)

Question : Show that G is an Abelian (commutative) group

Hi JPC! :smile:

My inclination would be to start by saying …

if it's not Abelian, then there are a and b with ab - ba ≠ 0,

so let ab - ba = c, and … :smile:
 
  • #3
oh, i should have precised, here "*" is not necessarily the multiply law, it can be any law that respects the given conditions

i tried with your method, putting to the square after, but then i end up working with 4 laws at the same time :D

Thank you for the indication anyways, i have to find the little trick inside that exercise now
 
  • #4
Hello Jpc,

In the case that i = 2,

(x*y)*(x*y) = x*x*y*y

by associativity:

x*(y*x)*y = x*(x*y)*y

and then just left and right multiply by x inverse and y inverse respectively.

This proves that G is Abelian, since we have shown * to be commutative.

If you want to show this for the more complicated cases where i is only allowed to take the value 3 or only allowed to take the value 4, then you would need to rephrase your question.
 
  • #5
yes thank you Sisplat, that works very well :)
I was looking for very complicated things, and i did a little confusion, thinking that the "^i" was associated to the multiply law and not the * law

And yes, you were right, there were too many data given, to confuse you probably

Thanks again
 

FAQ: Show that a Group (G, *) definied by a condition is Abelian

What is a group defined by a condition?

A group defined by a condition is a mathematical structure that consists of a set of elements and a binary operation that satisfies certain conditions. These conditions include closure, associativity, identity element, and inverse element.

How is the binary operation defined in a group?

The binary operation, denoted by *, is a rule that combines any two elements in the group to produce a third element. It is defined such that for any two elements a and b in the group, a * b is also an element of the group.

What does it mean for a group to be Abelian?

A group is considered Abelian if its binary operation is commutative, meaning that for any two elements a and b in the group, a * b = b * a. In other words, the order in which the elements are combined does not affect the result.

How do you show that a group defined by a condition is Abelian?

To show that a group (G, *) defined by a condition is Abelian, we must prove that for any two elements a and b in the group, a * b = b * a. This can be done by using algebraic manipulations and the properties of the group, such as associativity and identity element.

Can a group be both defined by a condition and Abelian?

Yes, a group can be defined by a condition and also be Abelian. In fact, many commonly studied groups, such as the integers under addition or the real numbers under multiplication, are both defined by a condition and Abelian.

Similar threads

Commutator group in the center of a group
Replies
5
Views
2K
Proving Subgroups in Abelian Groups
Replies
7
Views
2K
Proving that an Abelian group of order pq is isomorphic to Z_pq
Replies
5
Views
2K
Group Theory: Finite Abelian Groups - An element of order
Replies
3
Views
2K
Show injectivity, surjectivity and kernel of groups
Replies
1
Views
1K
Prove if ##a\cdot c = b \cdot c## then ##a = b## using Peano postulates
Replies
2
Views
844
Show that given conditions, element is in center of group G
Replies
1
Views
1K
Show that ##G\simeq \mathbb{Z}/2p\mathbb{Z}##
Replies
2
Views
1K
Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates
Replies
3
Views
1K
Determining a group, by checking the group axioms
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top