Abstract Algebra on Quotient Algebras

• Apr 6th 2010, 11:06 AM
hcourche
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!
• Apr 6th 2010, 12:57 PM
Drexel28
Quote:

Originally Posted by hcourche
Let $\displaystyle G=(G,+,-,0)$ be a group.
Let $\displaystyle \Theta\in \text{Con}(G)$.
Let $\displaystyle N=[0]/\Theta$
Then For all $\displaystyle a \in G$ prove $\displaystyle a+N=N+a N$
where $\displaystyle a+N=\left\{n+a:n \in N\right\}$ and $\displaystyle N+a=\left\{n+a:n\in N\right\}$

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 $\displaystyle \text{Con}(G)$?
• Apr 6th 2010, 01:30 PM
tonio
Quote:

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
• Apr 12th 2010, 11:44 AM
hcourche
Quote:

Originally Posted by Drexel28
Is that wht you mean? What's $\displaystyle \text{Con}(G)$?

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?