# 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 $\langle a,b|abab^{-1}\rangle$.

• What is $\phi$? – Skatche Jul 13 '13 at 4:06
• @Skatche i dont understand what you mean – Ronald Jul 13 '13 at 4:11
• @JonasMeyer Perhaps the confusion is notation: by $\langle a, b \rangle$, I believe he means the free group on two letters. – Eric Auld Jul 13 '13 at 4:12
• Thanks @Eric, Danial, for clearing up my confusion; I deleted my obsolete comments. – Jonas Meyer Jul 13 '13 at 4:23
• Okay, your new notation is clearer. I think your efforts here may be misguided: note that $\langle a,b\vert abab^{-1}\rangle\neq\langle a,b\rangle/\langle abab^{-1}\rangle$. Or did you have some other connection in mind? – Skatche Jul 13 '13 at 4:24

$H$ 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.

• So how to prove that if $N$ is the normal closure of $H$ then $\langle a,b \rangle /N \cong \langle a,b|abab^{-1}\rangle$? – Ronald Jul 13 '13 at 4:32
• but by definition the relation $abab^{-1}$ should be the generator of $N$ not for $H$!! or maybe I am confused – Ronald Jul 13 '13 at 4:47
• I shouldn't have claimed that's the definition; I'm not sure that's true. I deleted the comment. – Eric Auld Jul 13 '13 at 4:54

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.

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 $n$.