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!