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

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Homework Help Overview

The problem involves a group (G, *) defined by a condition related to the operation * and the exponentiation law. The task is to demonstrate that this group is Abelian based on the given condition that for all integers 'i' in the set {2, 3, 4}, the equation (x * y) ^ i = (x^i) * (y^i) holds for all elements (x, y) in G.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the given condition, questioning how the distributive property of exponentiation over the group operation relates to the commutativity of the group. Some participants suggest starting with the assumption that the group is not Abelian and examining the consequences.

Discussion Status

The discussion includes various attempts to understand the problem, with some participants providing specific cases (e.g., when i = 2) to illustrate their reasoning. There is recognition of the complexity introduced by the different values of i and the potential for confusion regarding the operations involved. Guidance has been offered, but no consensus has been reached on a complete solution.

Contextual Notes

Participants note that the operation * is not limited to multiplication, which adds complexity to the problem. There is also mention of the need to clarify the question further if considering cases where i takes on specific values.

JPC
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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 :)
 
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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:
 
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
 
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.
 
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
 

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