# Problem solving algebraic structures task

• January 29th 2011, 04:28 AM
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!
• January 29th 2011, 05:18 AM
Plato
Quote:

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.
$f(x \cdot y) = (x \cdot y) \cdot (x \cdot y) = x \cdot (yx) \cdot y = (x \cdot x)(y \cdot y) = f(x) \cdot f(y)$
• January 29th 2011, 05:24 AM
goroner
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 :)
• January 29th 2011, 05:31 AM
goroner
• January 29th 2011, 06:09 AM
Plato
Always start a new tread for a different question.
That is a rule of the forum.
• January 29th 2011, 06:23 AM
goroner
ok sry, ill start new one thanks!
• January 29th 2011, 10:12 AM
wonderboy1953
Quote:

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.