Can a Free product with amalgamation of $\mathbb{Z}*\mathbb{Z}$ be isomorphic to $\mathbb{Z}\times \mathbb{Z}$?

Question

I was trying to compute the fundamental group of torus by forcing myself of using the Seifert Van Kampen theorem. One knows that the answer is the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $.

But from the way I tried to solve it something goes wrong and I can't find out what it is.

Here's what I do:

First I remove the Circle $AB$ from the torus which gives me an open set $X_1$.

For the second one I remove the circle $BC$ from the torus instead of the other one which gives me another open set $X_2$.

Now $X_0:=X_1 \cap X_2$ is the torus without those two circles, which is a simply connected subset of torus.

Now if applying the Seifert Van Kampen theorem to these path connected subsets of the torus you'll get

$\pi_1(\mathbb{T}) = \mathbb{Z}*\mathbb{Z}$

while writing this I realized something that solves the fundamental group problem I asked. However I'm still interested to know the answer of the question below:

Is there a free product with amalgamation of $\Bbb Z*\Bbb Z$ isomorphic to the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $?

Answer 1

As mentioned by OP in comments, the question asks: Is there a group $G$ which is an amalgam of the form $$ A\star_{C} B, $$ with the factors $A\cong B\cong {\mathbb Z}$, such that $G$ is isomorphic to ${\mathbb Z}^2$?

This question has a negative answer.

Proof. Suppose, to the contrary, that there exists an isomorphism $$ \phi: A\star_{C} B\to {\mathbb Z}^2. $$

There are two cases to consider.

a. $C$ is a nontrivial group. Then the images $\phi(A), \phi(B)$ are infinite cyclic subgroups of ${\mathbb Z}^2$ which have nontrivial intersection $\phi(C)$. But any two cyclic subgroups with nontrivial intersection in ${\mathbb Z}^n$ generate a cyclic subgroup. (I leave it as an exercise.)

This would imply that ${\mathbb Z}^2$ is cyclic (since $\phi$ is surjective). The latter is obviously false. (Again, a linear algebra exercise.)

b. $C$ is the trivial subgroup. Then $G\cong A\star B\cong F_2$, the free group on two generators. But a free group on two generators (say, $a, b$) is nonabelian, for instance, since the word $aba^{-1}b^{-1}$ is reduced and nonempty, hence, represents a nontrivial element of $F_2$ (by one of the equivalent definitions of $F_2$). qed

Remark. Incidentally, an amalgam of the form $A\star_{C} B$ with $A, B, C$ infinite cyclic, can contain ${\mathbb Z}^2$ as an index two subgroup. I leave this as an exercise in algebraic topology of surfaces.

Answer 2

Actually, while writing my question I realized where am I making a mistake!

I wanted to delete my post but since I thought it could be instructive, I decided to post the question.

So If you're interested think on the question asked and then read the answer by surfing the mouse in the box below!

What $X_1$ and $X_2$ do cover is not the torus but the torus minus one point! Try to visualize that this will in fact give you the $8$ symbol whose fundamental group is $\mathbb{Z}*\mathbb{Z}$.

Answer 3

Is there a free product with amalgamation of $\Bbb Z*\Bbb Z$ isomorphic to the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $?

No.

One way to look at $\Bbb Z\ast \Bbb Z$ is as the group given by the presentation

$$\langle a,b\mid \varnothing \rangle.\tag{1}$$ Similarly, $\Bbb Z\times\Bbb Z$ is given by

$$\langle a,b\mid ab=ba\rangle.\tag{2}$$

In order to go from $(1)$ to $(2)$ via a free product with amalgamation, we must be able to write

$$\Bbb Z\ast_{H=K}\Bbb Z\cong\langle a\mid \varnothing \rangle \ast_{w(a)=\widetilde{w}(b)}\langle b\mid \varnothing \rangle$$

in the form of $(2)$,

where $w(a)$ is some word over the alphabet $\{a\}$ (so, a power of $a$) that generates some subgroup $H\le \Bbb Z\cong\langle a\mid\varnothing \rangle $; similarly for $\widetilde{w}(b)$, $\{b\}$, and $K$.${}^\dagger$

But subgroups of cyclic groups are themselves cyclic.

Also, notice that $ab=ba$ in $(2)$ cannot be written as $a^h=b^k$ for any integers $h,k$.

For more on the topic, see Magnus et al.'s, "Combinatorial Group Theory: [. . .]".

I hope this helps :)

---

$\dagger$: As pointed out in the discussion in the Group Theory chat room, if I have understood the OP's question sufficiently, the case when the subgroup $H=K$ one amalgamates with is trivial needs covering. In that case, $$\Bbb Z\ast_{\{e\}}\Bbb Z=\Bbb Z\ast \Bbb Z.$$