# Generators and relation in group and *non-triviality*

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?

• 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. – Tobias Kildetoft Oct 21 '18 at 7:35

## 1 Answer

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 $$A$$.