This is Exercise 2.1.4(a) of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by W. Magnus et al.
The Question:
Let $F=\langle a, b\rangle$. If $N$ is the normal subgroup of $F$ generated by each of the following sets of words, find the index of $N$ in $F$:
(a) $b^2, ab$.
Thoughts:
I'm stumped. Here $N=\langle b^2, ab\rangle$, so is composed of words of the form $$b^{2n_1}\underbrace{a^{\varepsilon_2}b^{n_2}\dots a^{\varepsilon_m}b^{n_m}}_{m-1\text{ times } a^{\varepsilon_i}b^{n_i}}$$ for $n_1\in\Bbb Z$ and $n_2, \dots, n_m\in \{2\nu+1\mid \nu\in\Bbb Z\}$ for each $m\in\Bbb N$, $\varepsilon_i\in\{\pm 1\}$. I don't know if that helps (but at least it's a start and I have something to say about the problem); my guess is that the index would be a multiple of $4$ since $b^2, ab$ each have length $2$. Do you see what I mean?
There are other questions like this one - it goes up to (f) - so I'd like to know how to get started on them.
Please help :)
Note that $N$ is defined to be the normal subgroup generated by certain elements (here, $ab$ and $b^2$). That means we want to consider the smallest normal subgroup containing these elements and not just the subgroup generated by these elements.
Let us note some things about the quotient $F/N$. It is clearly generated by $a' = aN$ and $b' = bN$. You have $a'b' = 1$ since $ab \in N$. It follows that $a' = b'^{-1}$ so $F/N$ cyclic being generated by $b'$. You also have $b'^2 = 1$ since $b^2 \in N$, so $F/N$ is cyclic of order at most $2$.
Now consider the homomorphism $f: F \to \mathbb{Z}/2\mathbb{Z}$ defined by $f(a) = f(b) = 1 + 2\mathbb{Z}$. Then $f$ is surjective and $ab$ and $b^2$ lie in $\ker(f)$. It follows that $N \subseteq \ker(f)$ and since $\ker(f)$ has index $2$ in $F$, it follows that we actually have $N = \ker(f)$ (as the index of $N$ was at most $2$). We conclude that the index of $N$ in $F$ is $2$.