Thread: Abstract Algebra on Quotient Algebras

Let G=(G,+,-,0) be a group.
Let \[Theta]\[Epsilon] Con(G).
Let N=[0]\[Theta]
Then For all a \[Epsilon]G prove a+N=N+a N
where a+N={n+a:n \[Epsilon] N} and N+a={n+a:n\[Epsilon]N}




Is this more clear, I don't have a program that generates the proper math symbols. Please help prove!

Originally Posted by hcourche

Let be a group.
Let .
Let
Then For all prove
where and




Is this more clear, I don't have a program that generates the proper math symbols. Please help prove!

Is that wht you mean? What's ?

Originally Posted by hcourche

Let G=(G,+,-,0) be a group.
Let \[Theta]\[Epsilon] Con(G).
Let N=[0]\[Theta]
Then For all a \[Epsilon]G prove a+N=N+a N
where a+N={n+a:n \[Epsilon] N} and N+a={n+a:n\[Epsilon]N}




Is this more clear, I don't have a program that generates the proper math symbols. Please help prove!

Either you read the directions to fast-learn to type in LaTeX in this site (very advisable!), or else you write in standard, widely-known characters, otherwise it's practically impossible to understand what you meant.

Tonio

Originally Posted by Drexel28

Is that wht you mean? What's ?

A congruence on A is an equivalence relation on A that is compatible with all operations. The set of all congruences on A is denoted by Con(A).
The line with for all a in G, prove that a+N=N+a the rest is good. By the way how did you input the proper symbols?