0
$\begingroup$

I am a bit new to presentations and free groups, so of course I am stuck on a very easy question.

I want to prove that $\langle x, y \mid xyx^{-1}y^{-1}\rangle \cong \mathbb{Z} \times \mathbb{Z}$. My proof ideas is the following:

We consider the map $f: \{x, y\} \longrightarrow \mathbb{Z} \times \mathbb{Z}$ with $f(x) = (1,0)$ and $f(y) = (0,1)$. Now by the universal mapping property, there exist a unique homomorphism $\phi: F(\{x, y\}) \longrightarrow \mathbb{Z} \times \mathbb{Z}$, such that $f = \phi \circ \iota$, where $\iota$ is the inclusion map.

Now we need to show, that $\ker \phi = \langle \langle xyx^{-1}y^{-1} \rangle \rangle$.

One direction, $\ker \phi \supseteq \langle \langle xyx^{-1}y^{-1} \rangle \rangle$, is of course fairly easy to verify, as every element in the RHS is of the form $\prod_{i=1}^{n} g_i [x, y]^{e_i} g_i^{-1}$, where $[x, y] = xyx^{-1}y^{-1}$, $g_i$ is any word in $F(\{x, y\})$, and $e_i = \pm 1$.

But how do I show, that every element in the kernel can be expressed in the designated form? Or is there a more convenient way of showing the other inclusion?

Or is my whole proof idea already doomed to fail?

By the way, I am fully aware that there is already a thread (Check in detail that $\langle x,y \mid xyx^{-1}y^{-1} \rangle$ is a presentation for $\mathbb{Z} \times \mathbb{Z}$.) with a highly similar question, but in all those answers, I do not see rigorously proven, why we can use the relation $\langle xyx^{-1}y^{-1}\rangle = \langle xy = yx \rangle$ to simply rearrange all our words in the form $x^ay^b$ and then map them to $\mathbb{Z} \times \mathbb{Z}$. Or put differently, why the kernel of our homomorphism, is PRECISELY $\langle \langle xyx^{-1}y^{-1}\rangle\rangle $.

Thank you!

$\endgroup$
1
$\begingroup$

We have to show that for all $w \in F(S)$, if $f(w)=0$ then $w \in \langle \langle xyx^{-1}y^{-1} \rangle \rangle$ ( the normal closure of $\{xyx^{-1}y^{-1} \}$. ) I leave you to fill in the details of the following claims. That is, we have to consider reduced words that start and end with one of $x,y$, and I have proven one example, but they are almost identical.


Lemma: If $w \in F(S)$, and $\sum a_i = \sum b_j = 0$ (these are the sum of exponents of $x$ and $y$ respectively in word $w$) then $w \in \langle \langle xyx^{-1}y^{-1} \rangle \rangle$.

Proof: Prove by induction. When $l(w)=4$: wlog, of form $x^ay^bx^{-a}y^{-b}$. Apply two smaller inductions to the following expressions, $$ x[x^{k}yx^{-k}y^{-1}]x^{-1} (xyx^{-1}y^{-1})= x^{k+1}yx^{-(k+1)}y^{-1}. $$ $$(x^kyx^{-k}y^{-1})y[x^ky^mx^{-k}y^{-m}]y^{-1} = x^{k}y^{m+1}x^{-k}y^{-(m+1)}.$$

For the inductive step, apply "conjugation" or "multiplication", for example $$ x^{a_1} y^{b_1} \cdots y^{b_s}x^{a_k} = x^{a_1}[y^{b_1} \cdots y^{b_{s}}x^{a_k+a_1}]x^{-a_1} . $$ $$ x^{a_1}y^{b_1} \cdots x^{a_k}y^{b_s} = [x^{a_1}y^{b_1} \cdots y^{b_{s-1}+b_s} x^{a_k} ] (x^{-a_k} y^{-b_s} x^{a_k} y^{b_s}). $$ inducting on the [elments in brackets].


The result then follows: Wlog, suppose $w=x^{a_1} y^{b_1} \cdots x^{a_k}y^{b_s}$ $$\phi(w)= \sum a_i \phi(x) + \sum b_j \phi(y) = 0 \Leftrightarrow \sum a_i = \sum b_j = 0$$ By lemma, $w \in \langle \langle xyx^{-1}y^{-1} \rangle \rangle$, as desired.

$\endgroup$
1
$\begingroup$

Too long for a comment:

Prove that an element in $\;F(x,y)=\;$ the free group in two variables $\;x,y\;$ belongs to the commutator subgroup $\;F'\;$ iff the power-sum of each and both $\;x,y\;$ in the reduced word representing the element is zero, where the power-sum of a letter in a reduced word is the sum of all the exponents of that letter, for example:

$$w=x^{-1}yx^3y^{-5}x^{-2}y^4\implies s_x(w)=s_y(w)=0\implies w\in F'$$

but

$$w=xyx^2y^{-3}xyx^{-4}\implies s_x(w)=0,\,s_y(w)=-1\implies w\notin F'$$

$\endgroup$
  • $\begingroup$ I guess this is precisely where I am struggling. One direction of the "iff" is straightforward, namely every element in the normal subgroup generated by xyx-1y-1 has zero powersum. But the other way is precisely what I'm not getting ... $\endgroup$ – Marcel S Apr 24 '17 at 17:39
  • $\begingroup$ @Marcel: try induction on the length of $\;w\;$ , and then you can use reduced cyclic words ... I guess you can also divide in cases: whereas the first and last letters are equal, etc. $\endgroup$ – DonAntonio Apr 24 '17 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.