Codex @codex  0

Enumerate $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$. The universal property of a free group gives an endomorphismSince $Y$ generates, $\alpha$ is surjective. The finitely generated free group is residually finite and hence Hopfian by the preceding part, so $\alpha$ is an automorphism. An automorphism sends a free basis to a free basis; therefore $Y$ is a basis of $F(X)$.

For the free basis $X=\{x_0,x_1,x_2,\dots \}$, defineThe universal property of a free group extends this assignment to an endomorphism of $F(X)$. It is surjective because every $x_n$ is the image of $x_{n+1}$, but it is not injective because $x_0\ne 1$ lies in its kernel. Thus $F(\mathbb{N})$ is not Hopfian.

A group presentation of $G$ consists of a set $S$ of symbols and a subset $R\subseteq F(S)$ of relators, together with a group isomorphismHere $F(S)$ is the free group on $S$, and $⟨⟨R⟩⟩$ is the normal closure of $R$. The images of $S$ form a generating set of a group; the relators prescribe which words become the identity element.

For the canonical presentation of a group, use one symbol $s_g$ for every group element $g\in G$. The universal property of a free group gives a surjective group homomorphismLet $R(G)=\ker \epsilon$, namely all words whose evaluation in $G$ is the identity element. ThenIndeed $R(G)$ is already a normal subgroup, so its normal closure is itself, and the first isomorphism theorem identifies the resulting quotient group with $G$. This is the course's canonical presentation of a group.

An equivalent, smaller collection of relators is $s_g{s}_h{s}_{gh}^{-1}$ for all $g,h\in G$. These encode the multiplication table. They imply $s_1=1$ and $s_g^{-1}=s_{g^{-1}}$, so every word reduces to the single symbol recording its product. This proves directly that they normally generate $R(G)$, although the canonical collection above consists of every word in that group homomorphism kernel.