3NF decomposition + database normalization
Every textbook I've ever seen gives you one or more algorithms you can use to determine every possible candidate key for R{ABCDEFG}. (It can be a fairly laborious and time-consuming job.) Based on the FDs you provided, I think R{ABCDEFG} has only one candidate key, {AC}.

Loosely speaking, for a relation to be in 3NF, it must be in 2NF and have no transitive dependencies. For a relation to be in 2NF, it must be in 1NF and have no partial key dependencies. 


The only candidate key here is {AC}.
From A->DE we know that DE is dependent on A, which is only part of the candidate key.


Therefore, R{ABCDEFG} is not in 2NF (or 3NF) based on the FDs you provided.