0
$\begingroup$

Consider the following basic problem: let $G$ be a group with following generators and relations $$G=\langle x: x^2=1, x^3=1\rangle.$$ It is easy to play with generators-relations to conclude that $x=1$ and group is trivial.

Then we consider the following problem: let $H$ be the group $$H=\langle x,y: x^3=1, y^2=1, yxy^{-1}=x^{-1}\rangle.$$ It is well-known (symmetric) group. I asked myself, with symbolic computations, can we show that $x\neq 1$ or that $H$ is non-trivial. But when I assumed $x=1$, I didn't find any contradiction. It seems that to prove non-triviality of this group, we should use universal property of free groups or groups defined by generators and relations.

Q. Is it true that to prove a certain (non-trivial) group, defined in terms of generators and relations, is non-trivial, we must use universal properties of free groups?

$\endgroup$
  • 5
    $\begingroup$ To some extend yes, because the trivial group always satisfies the relations. To show that it is not trivial, you want to find a non-trivial group satisfying all the relations. $\endgroup$ – Tobias Kildetoft Oct 21 '18 at 7:35
2
$\begingroup$

Think of any set of generators and relations (a relation being some word/combination of the generators and inverses is equal to the identity). Suppose this is a presentation of the group $A$.

Set all the generators equal to the identity, and all the relations will be satisfied.

Set one generator $x$ equal to the identity and you have the presentation of a group $B$ with one fewer generators. It may or may not collapse to the identity.

$B$ is a homomorphic image of $A$ with $x$ in the kernel of the homomorphism.

In your example, $x$ generates a normal subgroup and $B$ is non-trivial. If you had set $y=1$ you would find that $B$ would be trivial, because the subgroup generated by $y$ is not normal, and is not contained in a non-trivial normal subgroup.

Note that the generators and relations give a homomorphism from the free group on those generators to the maximal group which fits. Apply any homomorphism to $A$ and you will find that the image $C$ satisfies all the relations simply by applying the properties of the homomorphism.

Whether a presentation gives the trivial group or not is a hard problem. The way to show it doesn't is to exhibit a non-trivial group which satisfies the relations. This need not be the whole group $A$ but could be a homomorphic image of $A$.

$\endgroup$

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.