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!