# Does every presentation correspond to a group?

I am a graduate student. I do not know much about group presentations. I have only seen those things while studying Dihedral groups and Quaternion groups.

For example, $$D_n=\langle r,f\mid r^n=1,f^2=1,fr=rf^{-1}\rangle$$ and $$Q_8=\langle a,b\mid a^4=1,ab=ba^{-1},a^2=b^2\rangle$$. I know that groups can be represented in this way.

Is there actually exist a group satisfying those relations? If yes,then what is the justification? For example, can I get a group $$G=\langle a,b,c\mid a^2=b^3=c^4=1,ab=bc=ca\rangle$$?

• Any presentation defines some group. However, it may happen that two seemingly different presentations define the same group. See here for some examples.
– Pedro
Commented Nov 14, 2022 at 6:45

Yes. The group is the free group on the generators modulo the normal closure of (smallest normal subgroup generated by) the subgroup generated by the relators.

So in your example, say, $$G=F_3/\langle a^2,b^3,c^4,b^{-1}abc^{-1},c^{-1}aba,c^{-1}bca\rangle$$.

The group $$N$$ that we quotient by consists in the smallest normal subgroup containing all the reduced words that can be formed out of the relators.

Note that the same group can have multiple presentations. The group can turn out to be trivial or free, for instance.

The group $$G$$ that you defined as an example is in fact the trivial group.

The relations $$ab=bc=ca$$ tell you that the three generators $$a,b,c$$ are all conjugate in $$G$$, and so they must have the same order.

But then the first two relations $$a^2=b^3=1$$ tell you that this order must be $$1$$, so $$a=b=c=1$$ and $$G$$ is trivial.