= Solution
{c}
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 isomorphism>
$$
\boxed{G\cong\langle S\mid R\rangle:=F(S)/\langle\!\langle R\rangle\!\rangle.}
$$
Here $F(S)$ is the <free group> on $S$, and $\langle\!\langle R\rangle\!\rangle$ 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 homomorphism>
$$
\varepsilon:F(\{s_g:g\in G\})\longrightarrow G,\qquad s_g\longmapsto g.
$$
Let $R(G)=\ker\varepsilon$, namely all words whose evaluation in $G$ is the <identity element>. Then
$$
\boxed{G\cong\langle s_g\ (g\in G)\mid R(G)\rangle.}
$$
Indeed $R(G)$ 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 $s_gs_hs_{gh}^{-1}$ 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>.
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