Is $\langle abab^{-1}\rangle$ a normal subgroup of $\langle a,b|\varnothing\rangle$? Let $G=\langle a,b | \varnothing\rangle$ and let $H\leq G$ s.t $H=\langle abab^{-1}\rangle$. Is $H\triangleleft G$?
I'm asking this question in order to understand the fundamental group of the Klein bottle which is $\langle a,b|abab^{-1}\rangle$.
 A: $H$ is not normal. Consider conjugation by $a$: $a(abab^{-1})a^{-1}=a^2bab^{-1}a^{-1}$, which is not a power of the original element.
So in order to find out what that fundamental group is, you must quotient by the normal closure of the subgroup generated by $abab^{-1}$. In general, this is a hard thing to get your hands on. 
A: You want to know if the subgroup $\langle abab^{-1}\rangle$ is a normal subgroup of $F(a, b)$. As has been pointed out already, this is false. However, it is actually "very false", in the sense that $\langle abab^{-1}\rangle$ is a malnormal subgroup of $F(x, y)$.
A subgroup $H\leq G$ is called malnormal if $H\cap H^g\neq 1\Rightarrow g\in H$.
The reason that $\langle abab^{-1}\rangle$ is malnormal is because it is a maximal cyclic subgroup of a non-abelian free group. A subgroup $\langle w\rangle$ is maximal cyclic if it is not contained in any other cyclic subgroup, so $w\neq v^k$ for $k\neq 0$. It is easy to verify that $abab^{-1}$ is not a proper power of any other element, and so the subgroup is maximal cyclic.
To see why $\langle abab^{-1}\rangle$ is malnormal, we have the following:
Theorem: Let $C=\langle w\rangle$ be some cyclic subgroup of a finitely generated free group $F(X)$, $|X|\geq 2$, such that $C$ is a maximal cyclic subgroup. Then $C$ is malnormal.
Proof: Assume otherwise and look for a contradiction. Then, there exists some $i>0$ and some $u\in F(X)$ such that $u^{-1}w^iu=w^{j}$. Note that $|i|=|j|$ (why?). If $i=j$ then we have $u\in \langle w\rangle$, as elements in a free group commute if and only if they are in the same maximal cyclic subgroup. Thus, $i=-j$. Therefore, $u^{-1}w^iu=w^{-i}$, so $u^{-1}w^{-i}u=w^{i}$. Then, $u^{-2}w^iu^2=u^{-1}(u^{-1}w^iu)u=u^{-1}w^{-i}u=w^i$, and so $u^2\in C$. As $C$ is maximal cyclic, $u\in C$. Thus, $C$ is malnormal in $F(X)$, as required.
A: Eric Auld has given a fine answer to the original problem. More generally, since we're talking about fundamental groups, there is a nice way to get at the normal subgroups of $\langle a, b\rangle$. The fundamental group of the one-point union of circles $\mathbb{S}^1\vee\mathbb{S}^1$ is isomorphic to the free group on 2 letters. If you can classify the $n$-fold covering spaces of $\mathbb{S}^1\vee\mathbb{S}^1$, then you will be able to classify the index $n$ subgroups of $\langle a,b\rangle$. At the very least, you can try to draw just a few of them for each $n$.
