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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!

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2 Answers 2

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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.

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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

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

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  • $\begingroup$ 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
    Commented Apr 24, 2017 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
    Commented Apr 24, 2017 at 17:48

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