Prove elements generate a free group

Solution 1:

This proof uses covering spaces, but I'm posting it at the request of orlandpm (the OP).

Consider the $2$-oriented graph, $G$:

This space is the wedge sum of two circles. Therefore its fundamental group is the free group on two generators $\langle x, y\rangle$.

This space has the following covering space, $\widetilde G$:

Since the map $p_* : \pi_1(\widetilde G) \to \pi_1(G)$ induced by the covering space $p : \widetilde G \to G$ is injective, it follows that the fundamental group of $\widetilde G$ is isomorphic to $\langle x^2, xy, y^2\rangle$.

At the same time, $\widetilde G$ is homotopy equivalent to the wedge sum of three circles. Therefore its fundamental group is the free group on three generators, $F_3$.

We conclude that $$ \langle x^2, xy, y^2\rangle \cong F_3. $$

For the theory behind this, check out section $1.3$ of Hatcher's Algebraic Topology book (available freely online). (Images courtesy of the book.)

Solution 2:

Although you can do this by bare hands, it is easier to do it after learning a bit of general theory about subgroups of free groups. These three elements generate the subgroup of the free group $F$ on $\{x,y\}$ consisting of words of even length, which clearly has index $2$ in $F$ with transversal $\{1,x\}$. The Schreier generators of the subgroup with respect to this transversal are $x^2,xy,yx^{-1}$, which are known to generate it freely. We can then replace $yx^{-1}$ by $(yx^{-1})(xy)=y^2$ to deduce that the given three elements are free generators.

You can also use the Nielsen theory to transform any set of elements to a set that freely generates the subgroup it generates. In this case, $x^2,xy,y^2$ is not Nielsen reduce, because $l(y^2(xy)^{-1}x^2) = l(yx)=2 \le l(y^2)+l(x^2)-l(xy)$, but again replacing $y^2$ by $y^{-1}x$ results in a Nielsen reduced set.

Solution 3:

Here is a “bare-hands” approach.

Let $X$ be your group generated by $x^2,y^2$ and $xy$, $A$ be the free group on three generators $a_1,a_2,a_3$. There is a unique group homomorphism $f: A \to X$, sending $a_1$ to $x^2$, $a_2$ to $y^2$, $a_3$ to $xy$. That homomorphism is obviously surjective. So all you need to show is that $f$ is injective, i.e. $f(a)\neq e$ whenever $a\neq e$.

In both $X$ and $A$, every element has a unique reduced writing (i.e. expression not containing terms of the form $tt^{-1}$ or $t^{-1}t$). This allows us to identify each element with a unique word, called its “normal form”, and saying that an element ends with a certain word.

Let $a\in A,a\neq e$. Then $a$ must end with something, there are six cases, and I claim that

(1) If $a$ ends with $a_1$ (in $A$), then $f(a)$ ends with one of $x^2,yx,y^{-1}x,y^2$ (in $X$).

(2) If $a$ ends with $a_2$ (in $A$), then $f(a)$ ends with $y^2$ (in $X$).

(3) If $a$ ends with $a_3$ (in $A$), then $f(a)$ ends with one of $xy,x^{-1}y$ (in $X$).

(4) If $a$ ends with $a_1^{-1}$ (in $A$), then $f(a)$ ends with $(x^{-1})^2$ (in $X$).

(5) If $a$ ends with $a_2^{-1}$ (in $A$), then $f(a)$ ends with one of $xy^{-1},x^{-1}y^{-1},(y^{-1})^2$ (in $X$).

(6) If $a$ ends with $a_3^{-1}$ (in $A$), then $f(a)$ ends with one of $yx^{-1},y^{-1}x^{-1}$ (in $X$).

Once this property is stated, its verification by induction on the length of $a$ and by case disjunction is purely mechanical. I can supply further details if you need them.

So we have by this disjunction in six cases, that $f(a)\neq e$ : $f$ is injective, which concludes the proof.

Solution 4:

Instead of the choice in the OP, we could select differently in order to get wisdom of the problem with respect to "orderings of words".

Among the 12 possibilities of length two reduced words in $F_2$ we can prove that the the first three in the ordering: $$x^2<xy<xy^{-1}<x^{-2}<x^{-1}y<x^{-1}y^{-1}<yx<yx^{-1}<y^2<y^{-1}x<y^{-1}x^{-1}<y^{-2}$$ generate the other nine.

Let $S=\{ x^2\ ,\ xy\ ,\ xy^{-1}\}$ be ours alternative choice, then we get $S^{-1}=\{ x^{-2}\ ,\ y^{-1}x^{-1}\ ,\ yx^{-1}\}$.

Now, there are six remaining cases:

• $x^{-1}y=x^{-2}(yx^{-1})^{-1}y^2$ then $y^{-1}x=y^{-2}(yx^{-1})x^2$

• $x^{-1}y^{-1}=x^{-2}(xy)y^{-2}$ then $yx=y^2(xy)^{-1}x^2$

• $y^2=(yx^{-1})(xy)$ then $y^{-2}=(xy)^{-1}(yx^{-1})^{-1}$,

that is, the product of the first three reduced words of length two and their inverses generate all the twelve reduced words of length two.

So, any other word of even length in $F_2$ can be generated by $S\cup S^{-1}$.

Now, is a matter of following the Schreier's algorithm to warrant that the subgroup $U=\langle\{ x^2\ ,\ xy\ ,\ xy^{-1}\}\rangle$ is indeed free.

For, we must be sure that $F=U\cup Ux$ is a partition. This is the case because if one admits that $U$ carries all the even length words, then $Ux$ carries the odd length words. Also the presence of $xy^{-1}\in U$ implies $Ux=Uy$.

So a set of transversals is $\Sigma=\{1,x\}$, which they are minimal in their coset and which together with $S$ they are going to give us the set $$\Sigma S=\{\ x^2\ ,\ xy\ ,\ xy^{-1}\ ,\ x^3\ ,\ x^2y\ ,\ x^2y^{-1}\ \},$$ hence one receives $\overline{\Sigma S}=\{1,x\}$.

According to Schreier's method the set $\{gs\overline{gs}^{-1}\ |\ g\in\Sigma,s\in S\}$ carries the free generator for $U$. In our case one computes: \begin{eqnarray*} x^2\overline{x^2}^{-1}&=&x^2,\\ xy\overline{xy}^{-1}&=&xy,\\ xy^{-1}\overline{xy^{-1}}^{-1}&=&xy^{-1},\\ x^3\overline{x^3}^{-1}&=&x^2,\\ x^2y\overline{x^2y}^{-1}&=&x^2yx^{-1},\\ x^2y^{-1}\overline{x^2y^{-1}}^{-1}&=&x^2y^{-1}x^{-1} \end{eqnarray*} Which they all are in $U$. Hence $\{\ x^2\ ,\ xy\ ,\ xy^{-1}\ \}$ is a set of free generator for $U$.