The Abelianization of $\langle x, a \mid a^2x=xa\rangle$ I wish to verify the following statement (which comes from Fox, "A Quick Trip Through Knot Theory", although that is probably not important).
"$\Gamma=\pi_1 (M)=\langle x, a \mid a^2x=xa\rangle$ so the homology of $M$ is infinite cyclic."
So, I need to find the Abelianization of the fundamental group. Using the relations I get
$$y_1:=[x,a]=x^{-1}ax,\qquad y_2:=[a,x]=x^{-1}a^{-1}x$$
generate $[\Gamma,\Gamma]$. Now thinking of $\Gamma/[\Gamma,\Gamma]$ as left cosets I find
$$xy_1=ax,\qquad ay_1=ax^{-1}ax$$
$$xy_2=a^{-1}x,\qquad ay_2=ax^{-1}a^{-1}x$$
but I don't see how this is infinite cyclic. Using various combinations of the relations don't seem to get me to a single coset. Is there some trick I am missing related to $[\Gamma,\Gamma]$ apparently being the conjugates of $a$? Or did I just screw up something else?
 A: To go from a group presentation to a presentation of the abelianisation, you have to add the commutator relation for each pair of generators. In this case, it says $ax = xa$. From your relation you get $axa = xa$, and so $a = 1$. Thus the abelianisation is just $\mathbb{Z}$ (as it should be for a knot complement).
A: Abelianizing you get the relation $a^2x=xa=ax$ from which $a=1$, thus only the $x$ generator remains and you have a copy of $\mathbb{Z}$.
A: The homomorphism $\Gamma\rightarrow\mathbb{Z}$ that maps $x$ to $1$ and $a$ to $0$ is well-defined and surjective. Since $\mathbb{Z}$ is abelian it factors through the abelianisation, so we have a surjective homomorphism $\Gamma_{ab}\rightarrow\mathbb{Z}$ that maps $\overline{x}$ to $1$. On the other hand $\overline{a}^2\overline{x}=\overline{x}\overline{a}$ in $\Gamma_{ab}$ implies $\overline{a}=\overline{1}$ and so $\Gamma_{ab}$ is generated by $\overline{x}$ which implies that the above homomorphism is an isomorphism.
PS: $\overline{y}$ is the projection of $y\in\Gamma$ to $\Gamma_{ab}$.
