How to embed $U(1)$ (or other groups) into a bigger group, using Dynkin diagrams I am trying to find the embedding and the branching rules for some group decompositions. For example, I consider $E_7$ and its maximally compact subgroup $SU(8)$ and I want to "see" how the Dynkin diagram of $E_7$ is modified to get $A_7$. I tried to take a linear combination of roots to produce $A_7$.
The problem is for an embedding like $SU(8) \supset SU(6) \times U(1)$. How do I get this?
 A: Since $SU(8)$ is a maximal regular subalgebra of $E_7$, you can identify it by looking at the extended Dynkin diagram.  This is the diagram you get by appending the lowest root to the simple roots that you have.  The $E_7$ Dynkin diagram looks like this: 

The extended diagram you get by appending the lowest root has an extra circle on the left side of the diagram: 

Now you obtain the subalgebras by erasing one of the circles in the diagram, so to get $A_7$, you just erase the circle that is sticking up.  This means to get the roots of the $A_7$ subalgebra from the $E_7$ roots, you just take the six original simple roots that don't include the one sticking up, and add to it the lowest $E_7$ root.  
This procedure will give you all the maximal regular subalgebras of a given Lie algebra.  Each diagram has a unique extended diagram from which you can construct these maximal regular subalgebras.  You can find the extended diagrams here.
The other example you asked about, $SU(8) \supset SU(7)\times U(1) $ (I'm doing $SU(7)$ instead of $SU(6)$, since it is the bigger subalgebra) simply comes from erasing one of the nodes of the $A_7$ diagram.  The resulting Dynkin diagram is $A_6$ which describes the $SU(7)$ factor, and then the node you erased is the $U(1)$ generator, which commutes with all elements of the $A_6$ algebra.  
