
#1
Jan309, 12:22 PM

P: 230

1. The problem statement, all variables and given/known data
(G, *) is a group (where * is a law) And for all 'i' belonging to {2, 3, 4}, for all (x, y) belonging to G^{2} (x * y) ^ i = (x^i) * (y^i) (where ^ is the law : to the power of) Question : Show that G is an Abelian (commutative) group 2. Relevant equations 3. 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 :) 



#2
Jan309, 12:31 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

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 … 



#3
Jan409, 09:52 AM

P: 230

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
Jan509, 03:44 PM

P: 9

Show that a Group (G, *) definied by a condition is Abelian
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
Jan709, 01:06 PM

P: 230

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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