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?

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    $\begingroup$ 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

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 $A$.


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