The Abelianization of $\langle x, a \mid a^2x=xa\rangle$

Question

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?

Did you type the relation correctly? At present it's equivalent to $a^2=1$, and the abelianization will not be infinite cyclic (it will have an element of order $2$). — mdp, Jun 10 '13 at 16:56
Also, $Gamma$ is not free, hence the question tilte is somewhat misleading. — Hagen von Eitzen, Jun 10 '13 at 16:58
This posting repeatedly used the code ;|;. I changed it to \mid. I think that is standard usage. — Michael Hardy, Jun 10 '13 at 18:30
@MichaelHardy thanks, i can't believe i've been doing it wrong for so long — citedcorpse, Jun 11 '13 at 14:12

score 3 · Accepted Answer · answered Jun 10 '13 at 18:20

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).

score 3 · Answer 2 · answered Jun 10 '13 at 19:22

3

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 $\mathbb{Z}$.

answered Jun 10 '13 at 19:22

Edoardo Lanari

3,694

score 2 · Answer 3 · answered Jun 10 '13 at 18:19

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 $\Gamma_{ab}$.

The Abelianization of $\langle x, a \mid a^2x=xa\rangle$

3 Answers3