Normal subgroup generated by a set. If $F$ is the free group on $\{a,b\}$ and $N$ is the normal subgroup generated by $\{b^3, a^7, aba^{-2}b^{-1}\}$ I am trying to show that $F/N$ is a non abelian group on $21$ elements.
My concern is that I am not clear on how $N $ will look like. For example, it is intuitively clear that $a$ is not in $N$ but I am not sure how to prove this.
As per comments below I realize that it is somewhat difficult to understand what $N$ would look like. Is it possible to give a reference for the same?

Let $1$ be the identity element. We have the equivalences $$b^3=1,a^7=1,aba^{-2}b^{-1}=1$$ in $F$ that give us $F/N$.
Using $b^3=1$ and $b^2=aba^{-2}$ we can convert all powers of $b$ in a free product $a^{x_1}b^{x_1}...a^{x_n}b^{x_n}$ to $1$, so it looks like $a^{x_1}ba^{x_2}b...a^{x_n}b$.
Now we can manipulate the equivalence relations to get 
$$bab=a^2$$
$$ba^2b=a^2ba^2$$
$$ba^3b=a^4ba^4$$
and so on (it's a little tedious), and use this to reduce the free product down to something of the form $a^{x_1}ba^{x_2}$, and then $a^xb^y$.
The idea is to show that you can reduce everything to $a^{\{0,1,2,3,4,5,6\}}b^{\{0,1,2,3\}}$ and conversely use this to generate $F/N$.
To answer your second question, note that multiplying by generators of $N$ always makes the length longer. This is because you won't ever have an element ending with $a^{\pm 1,\pm 7},b^{\pm 1, \pm 3}$ (unless you just multiplied by the inverse generator). Thus unless you start with $a$ you won't get $a$.
