combinatorial-group-theoryfree-groupsgroup-presentationgroup-theory
Let $G=\langle a,b \mid baba^{-1}=1\rangle$. Show that the subgroup generated by $a$ is infinite.
My attempt
Suppose $\langle a\rangle$ is finite so $a^k = 1$ for some $k \in \mathbb{Z}$. So I would like to prove that the relation $a^k=1$ is not a consequence of the original relation $baba^{-1}=1$. I also have been able to prove that every element $g$ in $G$ can be written as $g= a^n b^m$ for $n,m \in \mathbb{Z} $
Any hint will be appreciated.
The abelianization of a free group is a free abelian group: in particular, it is torsion-free. For your presentation, the relation $$2(x-y)=0$$ holds in the abelianization. Therefore, $G$ is free iff it is infinite cyclic. However, there clearly exists a morphism from $G$ onto $\mathbb{Z}_2 \times \mathbb{Z}_2$, so $G$ cannot be cyclic.
Added: In fact, your group $G$ is the fundamental group of the connected sum of two projective planes, that is of the Klein bottle $K$.
But we know that there exists a two-sheeted covering $\mathbb{T}^2 \to K$ from the torus $\mathbb{T}^2$. Therefore, $G$ has a subgroup of finite index two isomorphic to $\mathbb{Z}^2$. In particular, $G$ cannot be free.
Algebraically, it can be verified that $\langle x^2,xy^{-1} \rangle$ is a subgroup isomorphic to $\mathbb{Z}^2$ of index two in $\langle x,y \mid x^2=y^2 \rangle$.
You need to put more effort into solving the problem yourself, so I am just going to provide some hints to get you started. First show that $(xyx)(x^2y) = (x^2y)(xyx)$. That proves that the subgroup in the hint is abelian.
To show it's normal, show that $x(xyx)x^{-1} = x^2y$ and $x(x^2y)x^{-1} = (xyx)^{-1}(x^2y)^{-1}$. So now we can see that we need to put $a=xyx$, $b=x^2y$, and $t=x^{-1}$
Now, putting $A = \langle a,b \rangle$, it is easy to see that $\langle A,x \rangle = G$, so we have shown that $A \unlhd G$ and clearly $|G:A| \le 3$.
To complete the proof, prove that the semidirect product ${\mathbb Z}^2 \rtimes \langle t \rangle$ with the specified action satisfies the relations of the presentation of $G$.
Best Answer
Hint: consider the homomorphism $f$ from the free group $F = \langle x, y \rangle$ to the free group $H = \langle z \rangle$ that maps $x$ to $z$ and $y$ to $1$. The element $yxyx^{-1}$ lies in the kernel of $f$, so $f$ factors as $h \circ g$, where $g : F \to G$ and $h : G \to H$ are homomorphisms with $g(x) = a$, $g(y) = b$, $h(a) = z$ and $h(b) = 1$. The elements $f(x^n) = z^n$ for $n \in \Bbb{Z}$ are all distinct, hence the elements $g(x^n) = g(x)^n = a^n$ are also all distinct (if $g(x^n) = g(x^m)$, we have have $z^n = f(x^n) = h(g(x^n)) = h(g(x^m)) = f(x^m) = z^m$).