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


by JPC
Tags: abelian, condition, definied
JPC
JPC is offline
#1
Jan3-09, 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 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


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 :)
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
tiny-tim
tiny-tim is offline
#2
Jan3-09, 12:31 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
Quote Quote by JPC View Post
(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!

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 …
JPC
JPC is offline
#3
Jan4-09, 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

Sisplat
Sisplat is offline
#4
Jan5-09, 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.
JPC
JPC is offline
#5
Jan7-09, 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


Register to reply

Related Discussions
Abelian group Calculus & Beyond Homework 9
Abelian Group; what to do if the set is G=R-{1/3}? Calculus & Beyond Homework 3
Abelian group where a*a=e Calculus & Beyond Homework 13
What is this abelian group? Calculus & Beyond Homework 14
Abelian group Introductory Physics Homework 8