Question:
In the final result of the calculation, how do we get $$\pi_1(T)\cong (\mathbb{Z}*\mathbb{Z})/N \cong N \ $$ where $ \ N = \langle a, b : aba^{-1}b^{-1}=e \rangle $?
In particular, why is $ (\mathbb{Z}*\mathbb{Z})/N \cong N \ $?
Using the square representation of the torus $T$ (with identified opposite edges), one can use the cover of $T$ consisting of an open square, $U$ say, and a closed square with a closed disc removed from it, $V$ say.
This gives the free group
$$\pi_1(U)*\pi_1(V)= \{e\}*\pi_1(V)\cong\{e\}*(\mathbb{Z}*\mathbb{Z}) \cong \mathbb{Z}*\mathbb{Z}$$
where the associated homomorphism
$$ \ \phi : \{e\}* (\mathbb{Z}*\mathbb{Z}) \longrightarrow \pi_1(T)$$
has as kernel all those words $w \in \mathbb{Z}*\mathbb{Z} $ generated by the word $aba^{-1}b^{-1} \in \mathbb{Z}*\mathbb{Z}$, and this kernel is a normal subgroup denoted by $N = \langle a, b : aba^{-1}b^{-1}=e \rangle$. (I'm not certain about this.)
So, if $e$ is the identity in $ \pi_1(T)$, then $\ e=\phi(w)=aba^{-1}b^{-1} $ (since $\phi$ is just an embedding?).
Now, if my understanding is correct, then van Kampen says
$$\pi_1(T)\cong (\mathbb{Z}*\mathbb{Z})/N$$
where $(\mathbb{Z}*\mathbb{Z})/N$ is a quotient group?
However, from several different sources I get that in fact
$$\pi_1(T)\cong N = \langle a, b : aba^{-1}b^{-1}=e \rangle . $$
Thanks for any help making sense of this.
(Also, when we say "$N$ is generated by the word $aba^{-1}b^{-1}$", we include the conjugation operation – what does conjugation mean exactly in this case?)
Best Answer
Your understanding is correct, however, the kernel of $\phi$ is not $\langle a,b : aba^{-1}b^{-1}\rangle$. Rather, the kernel is all products of conjugates of $aba^{-1}b^{-1}$, or the "normal closure" of the subgroup generated by $\{aba^{-1}b^{-1}\}$. This is what
means. Since this may be the point of confusion, I'll expand on it more:
You can think of taking the normal closure of a subgroup as a process: start with the subgroup, and add the necessary elements so that it becomes normal. If $aba^{-1}b^{-1}$ is in the subgroup, and $g$ is some other element of that group, the normal closure must also include $gaba^{-1}b^{-1}g^{-1}$ (ie $aba^{-1}b^{-1}$ conjugated by $g$) if we want it to be normal. The normal closure is basically what you get when you keep on doing this.
Hopefully that clears up what is meant by "including the conjugation operation."
With this view, we can see that $N$ is just all products of conjugates of $aba^{-1}b^{-1}$. (It may take a moment to digest that statement)
Then, when we take the quotient, we can see that $(\mathbb{Z} \ast \mathbb{Z})/N$ is the free group of rank 2 with the added relation $aba^{-1}b^{-1}$. That is,
$$\pi_1(T) \cong (\mathbb{Z} \ast \mathbb{Z})/N \cong \langle a,b : aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2$$
Edit: in response to your comments: it is not true that $(\mathbb{Z} \ast \mathbb{Z})/N \cong N$.
Second edit: As for actually computing the quotient...
First, think about the coset $aba^{-1}b^{-1}N = aNbNa^{-1}Nb^{-1}N$. This must be equal to $N$, as $aba^{-1}b^{-1} \in N$. Then, $aNbNa^{-1}Nb^{-1}N = N$, so $aNbN = bNaN$, ie the quotient is abelian. So, we add the relation $aNbN = bNaN$. Now, are there any other relations? no:
If $wN = N$, then $w \in N$, so $w$ is a product of conjugates of $aba^{-1}b^{-1}$. Then, $wN$ is a product of conjugates of $aNbNa^{-1}Nb^{-1}N$, which is just the identity (in light of the previous paragraph). Then $aNbN = bNaN$ is the only relation, and we get the presentation $$ \langle a,b : aba^{-1}b^{-1}\rangle$$