# Thread: Problem solving algebraic structures task

1. ## Problem solving algebraic structures task

Hi i got some problem solving this:

Let (G, *) be abelian group, then the function f:G->G defined with f(x) = x*x , for x E G is homomorphism, you must prove this is homomorphism.

I know what is homomorphism, one function is homomorphism if it is well defined over the operations between groups, for e.g. if we have op * in one group and # in other group than f is homomorphism defined as f(x*y) = f(x) # f(y) for every x,y E G, it is homomorphism between G and some other group H, so this task confused me , since the function is defined on the same group, as i get it the result from operating in bouth the groups for the same operands should be the same and that's how we say the function is homomorphic, if it gives same result for operating with the operations from the groups, for this reason i tought the function f is not homomorphic cause x != x*x , actualy i dont know what exactly the operation does but if in the group holds the idmepotentity law then x*x = x might be true, but that was not defined so i realy dont know how to solve this problem.

Thanks for help!

2. Originally Posted by goroner
Let (G, *) be abelian group, then the function f:G->G defined with f(x) = x*x , for x E G is homomorphism, you must prove this is homomorphism.
The key here is Abelian group.
$f(x \cdot y) = (x \cdot y) \cdot (x \cdot y) = x \cdot (yx) \cdot y = (x \cdot x)(y \cdot y) = f(x) \cdot f(y)$

3. yes i started like that and i got the same result , but i doubted about it cause i had this tought in my mind: the result x != x*x, and i never managed to see that i was doing it ok, thanks anyway ) it makes sense

4. i dont like to open one more thread, i got one more question, how do i prove that in tree there's exactly one central vertex or two but they should be neighbour vertexes, i think the terms are central and bicentral vertex.i tought for a moment that i can prove it with mathematical induction but there're problems like: the tree might be verige, or it might be rooted tree with n children per node, etc. also i tought something about the radius of a tree cause central got to do with radius as it should be the centar node like in circle there's centar point and a radius so i think of it that way but dont know if that's the case cause on faculty i havent studied nothing about central points in graphs, nor trees and they gave this as task on exams and when i asked them how do i prove it they said write what you think about it no formal prove is needed, cause as the task was given a central point in graph concept was bit explaned but not well cause was an exam task.. thanks!

5. Always start a new tread for a different question.
That is a rule of the forum.

6. ok sry, ill start new one thanks!

7. Originally Posted by goroner
Hi i got some problem solving this:

Let (G, *) be abelian group, then the function f:G->G defined with f(x) = x*x , for x E G is homomorphism, you must prove this is homomorphism.

I know what is homomorphism, one function is homomorphism if it is well defined over the operations between groups, for e.g. if we have op * in one group and # in other group than f is homomorphism defined as f(x*y) = f(x) # f(y) for every x,y E G, it is homomorphism between G and some other group H, so this task confused me , since the function is defined on the same group, as i get it the result from operating in bouth the groups for the same operands should be the same and that's how we say the function is homomorphic, if it gives same result for operating with the operations from the groups, for this reason i tought the function f is not homomorphic cause x != x*x , actualy i dont know what exactly the operation does but if in the group holds the idmepotentity law then x*x = x might be true, but that was not defined so i realy dont know how to solve this problem.

Thanks for help!
Just noting that this should have been posted in the linear and abstract algebra section.