Say we have a presentation of a finite group. Does adding additional relations to the presentation always decrease the size of the group?

Question

It seems like it should, but I'm not sure how to prove it.

EDIT: I'm talking about nontrivial new relations here, i. e. ones that do not follow directly from the old ones.

Answer 1

Yes, it does (understanding decreasing as not strict):

Adding extra relations means taking a quotient of your original group $G$. That means that there exist a surjective map from your original group $G$ to the other one with extra relations, say $H$. Being $G$ and $H$ finite necessarily $|G| \geq |H|$.

Answer 2

Not necessarily. In case the "additional relations" are a consequence of the already existing ones in the presentation, the group remains the same.

Answer 3

The answer to the title question is that additional relations do not always decrease the size of the group. Take for example the finite group $$ \langle a,b\ |\ aba^{-1}b^{-2}=a^{-2}b^{-1}ab=1 \rangle. $$ Now adding any new relations in $a$ and $b$ will not decrease the size of the group. Of course, the relations will follow from the "old" ones, because the group in question here is trivial, and one cannot decrease its size.

Answer 4

There's some not quite trivial mathematical logic that seems relevant here.

Say you have a presentation with a set of relations $S$, that give a group $G$. Say you add a new relation $R$. If the new set of relations defines the same group $G$ then it follows that every group satisfying $S$ also satisfies $R$, and now the Completeness Theorem from logic shows that yes, there is a proof of $R$ from $S$ and the axioms of group theory.

Edit: I've been asked to expand on this. Omitting details about the language containing a constant symbol for each generator in the presentation, etc.: Consider the relations in $S$ as sentences of the form "[some word in the generators]=$e$". Suppose there is no proof of $R$ from $A\cup S$, where $A$ is the set of axioms for group theory. Then the Completeness Theorem shows that there is a model $H$ of $A\cup S$ in which $R$ does not hold. In particular $H$ is a group, being a model of $A$. Let $H'$ be the subgroup of $H$ generated by (the interpretations of) the generators in the original presentation. Then $H'$ is a group in which $S$ holds but $R$ does not. And being generated by the original generators, $H'$ is a quotient of $G$. Hence $R$ does not hold in $G$.