I have troubles to solve this kind of exercises. For example:

Let $$G_1=\langle x,y |x^3=y^4=1\rangle,~~~G_2=\langle x,y |x^6=y^6=(xy)^3=1\rangle. $$ I want to check that $G_1$ is an infinite nonabelian group and in $G_2$ we have $xy^2x \neq 1$ .

For the first part, I have seen that it is useful to define a group homomorphism and then see that the image (that we know) is infinite and nonabelian. For the second part, similarly we can define a $\phi$ such that $\phi (xy^2x) \neq 1$.

How can I define this homomorphisms? There is any general procedure for this?


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    $\begingroup$ Tried reductio ad absurdum with your technique? $\endgroup$ – Zelos Malum Dec 14 '15 at 8:09
  • $\begingroup$ No, since I should define that $\phi$'s. $\endgroup$ – Peter Dec 14 '15 at 8:19
  • $\begingroup$ You can still use it by defining it to a finite group and show it gives a contradiction. $\endgroup$ – Zelos Malum Dec 14 '15 at 8:20

You can describe a finitely generated free nonabelian group $G$ as the collection of all finite strings of powers (positive and negative) of the generators.

A finite presentation generates a normal subgroup of that free group, such as in your example the collection of all (conjugates of) finite strings of powers of $x^3$ and $y^4,$ or in general the normal closure of the set of relations.

The quotient group is your group and it is the image of a natural homomorphism.

You need to verify that $xyx^{-1}y^{-1}$ is not in the normal closure of $\{x^3,y^4\}$ and that $xy^2x$ is not in the normal closure of $\{x^6,y^6,x^3y^3\},$ perhaps by providing an explicit description of the form of the strings in those normal closures.

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    $\begingroup$ Good luck proving this with elements! $\endgroup$ – Martin Brandenburg Dec 14 '15 at 8:39

Let $H = {\rm SL}(2,{\mathbb Z})$ and define $\phi:G_1 \to H$ by $$ \phi(x) = \left(\begin{array}{rr}0&1\\-1&-1\end{array}\right), \phi(y) = \left(\begin{array}{rr}0&1\\-1&0\end{array}\right).$$

Let $K=S_6$ and define $\psi:G_2 \to K$ by $$\psi(x)=(1,2,3,4,5,6),\,\psi(y)=(1, 2, 6, 5, 4, 3).$$


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