Why is $\langle S|-\rangle$ not isomorphic to $\langle T|R\rangle$

Question

Let $S,T$ be sets with $|S|>|T|$ and $R$ some relations on $T$.
Why is then $\langle S|-\rangle$ not isomorphic to $\langle T|R\rangle$

This came up when I wanted to solve a different problem, which I also asked on this site. Unfortunately, the answers provided used a completely different strategy and I still wonder about how to prove this.

Intuitively its clear, but I look for a clear proof.

Answer 1

The group is a quotient group. The images of the elements of T generate this quotient group. If $<T|R>$ is free then any set of free generators has less than or equal to |T| generators. Thus it can't be isomorphic to $<S|->$.

Answer 2

For completeness, the following includes a proof (admittedly, using the axiom of choice) of the closely related fact that free groups are isomorphic only if they are over sets of the sae cardinality.

Assume $\phi\colon\langle T\mid R\rangle\to \langle S\rangle $ is an isomorphism (or just an epimorphism). Together with the canonical projection $\pi\colon \langle T\rangle\to \langle T\mid R\rangle$, we obtain an onto homomorphism $\phi\circ\pi\colon \langle T\rangle\to \langle S\rangle$. Let $V$ be an $\Bbb F_2$ vector space of dimension $|S|$. A bijection from $S$ to a basis of $V$ gives rise to an onto homomorphism $\psi\colon \langle S\rangle \to V$. Then $\psi(\phi(\pi(T)))$ is a generating system of $V$, hence contains a basis of $V$. We conclude that $$ |T|\ge |\psi(\phi(\pi(T)))|\ge \dim V=|S|,$$ contradicting the assumption that $|S|>|T|$.

Answer 3

I feel a little like cheating here, because that is basically an adaptation of the idea provided from an answer to the linked question... but again they say that mathematics is the art of finding good analogies.

Before I start allow me to introduce the commutator subgroup. For every group $G$ by $G'$ I denote the subgroup of $G$ generated by the elemets of the form $xyx^{-1}y^{-1}$ where $x,y \in G$. Such $G'$ is a normal subgroup and $G/G'$ is the smallest commutative quotient group of $G$, for more information take a look at this.

Assume that you have an isomorphism $$f \colon \langle S | - \rangle \longrightarrow \langle T | R \rangle$$ the this isomorphism should preserve the commutator subgroups, that is $$f(\langle S| - \rangle')=\langle T|R\rangle'\ .$$

Because of this we get an isomorphism $$\bar f \colon \langle S| - \rangle / \langle S| - \rangle' \longrightarrow \langle T|R\rangle/\langle T|R\rangle'$$ of abelian groups.

With a little effort one can easily prove that the first quotient is isomorphic to the free abelian group with a basis of cardinality $|S|$.

But the second group is generated by $|T|$ elements where $|T| < |S|$. This is an absurd because the cardinality of a basis of a free abelian group is the minimal number of generators for the group, so $\bar f$ cannot be an isomorphism and neither $f$ can.