Free Group Norms 
Hello everyone, I'm trying to solve this problem, but I'm stuck... i don't quite understand the definition of the norm, If you guys can give me a better explanation, I would appreciate it, Thanks

You have to calculate how many groups elements are represented by words at length at most $N$ in the group generators.
Let $G$ be free abelian with generators $a,b$. How many elements have shortest representative of length $n$? For $n=0$ there is just $1$. For $n>0$, we have $a^n,b^n,a^{-n},b^{-n}$, which is $4$. We also have $a^kb^{n-k}$ for $k=1,2,\ldots,n-1$, which makes $n-1$, and we get three more sets of size $n-1$ by replacing $a$ by $a^{-1}$ and/or $b$ by $b^{-1}$. So that makes $4+4(n-1) = 4n$ in total of length $n$. Summing these from $n=1$ to $N$ and adding on $1$ for length $0$, gives the required answer.
For the free group, the first generator in a reduced word can be any one of $a^{\pm 1}$ or $b^{\pm 1}$, making four possibilities, but there are only $3$ for subsequent generators in the word, since they must be the inverse of the preceding genertor. So, for $n >0$, there are $4.3^{n-1}$ group elements whose shortst representatives have length $n$. Again you need to some from $1$ to $N$ and add on $1$ for length $0$. 
