My question is as follows:

Let $G = \langle a,b \mid a^2=b^2=1 \rangle $ be a group generated elements $a, b$ and the equation $a^2=b^2=1$. Prove that $G$ is infinite and non-abelian.

I got the following hint:

Let $\mathbb{F}_2 = \langle x,y \rangle$ be the free group on two letters. Let $H = \langle \langle x^2,y^2 \rangle \rangle$ be the normal closure of $ \{ x^2,y^2 \}$ in $\mathbb{F}_2$, i.e. the smallest normal subgroup of $\mathbb{F}_2$ containing $\{ x^2 , y^2 \}$. Then we have $G \cong \mathbb{F}_2 /H$, induced by the map $\phi : \mathbb{F}_2 \to G$, given by $\phi(x) = a$ and $\phi (y) = b$. If we can show that no power of $xy$ lies in $H$, then that's equivalent to showing that $ab$ has infinite order in $G$. Following from this, it also implies that $(ab)^2$ is not trivial, i.e. that $a$ and $b$ do not commute, showing that $G$ is not Abelian.

I further have some questions:

(a) Why the map $\phi$ is a homomorphism?

(b) To obtain $G \cong \mathbb{F}_2 /H$ we actually need $H \subset \ker\phi$. Why it is the case?

(c) Why no power of $xy$ lies in $H$?

Any help would be appreciated.

  • 1
    $\begingroup$ In order to get a useful answer, you should say something about what you know and/or what you have tried in your efforts to answer this question yourself. Otherwise, the question is likely to be closed. $\endgroup$ – Lee Mosher Sep 7 '15 at 14:34

(a) The map $\phi$ is a homomorphism by property of the free group (it is actually its universal property, anytime you give the image of the generators you actually define a homomorphism).

(b) Actually the definition of a group by generators and relations (such as the one for $G$ here) is exactly that $G:=\mathbb{F}_{\text{generators}}/<<\text{relations}>>$.

In this setting, the homomorphism $\phi$ is actually defined as the canonical projection of $\mathbb{F}_{\text{generators}}$ on its quotient $G$. Hence $Ker(\phi)=<<\text{relations}>>=H$ by definition (or more precisely by the universal property of the quotient).

(c) To me, you have the good direction. Now what you need to show is that $(xy)^k\in H$ iff $k=0$. What I would show (and I think this easy) is the following assertion :

Any reduced form of any non-trivial word of $H$ contains at least one occurence of $x^{2l}$ or $y^{2l}$ for some $l\neq 0$.

Since $(xy)^k$ is never trivial for $k\neq 0$ and the absolute values of the powers of $x$ and $y$ in a reduced form of $(xy)^k$ are $1$ then we know that $(xy)^k\notin H$.

  • $\begingroup$ Thanks @ClémentGuérin. I understood parts (a) and (b). I couldn't figure out why your assertion is straightforward. Also, why $(xy)^k$ is never trivial for $k\ne0$? $\endgroup$ – user Sep 8 '15 at 14:48
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    $\begingroup$ For your second question, it is because $(xy)^k$ is considered as a "word" in $F_2$,in the free group, it is easy to see that $(xy)^k$ won't be trivial (for instance you could talk about the length of $(xy)^k$. For the first one, this is just that any word written as the product of conjugates of $x^2$ or $y^2$ will have even powers in its writing... $\endgroup$ – Clément Guérin Sep 8 '15 at 15:37
  • $\begingroup$ Thanks. I see. Your kindly answers hepled me a lot. $\endgroup$ – user Sep 8 '15 at 15:52

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