Every finite group has a finite presentation from its multiplication table

Let $X=\{x_g:g\in G\}$ and $R=\{x_g{x}_h{x}_{gh}^{-1}:(g,h)\in G\times G\}$. The map $g\mapsto x_g$ is a bijection, so [F2] makes $X$ finite. By [L2], $G\times G$ is finite, so by [F2] fix a bijection $c:G\times G\to n$ for some $n\in \mathbb{N}$, and let $q$ send $(g,h)$ to $x_g{x}_h{x}_{gh}^{-1}$, so that $R$ is the image of $q$. Sending each $r\in R$ to the least element of the nonempty set $\{k<n:q(c^{-1}(k))=r\}$, which exists by [F4], is an injection of $R$ into $n$; it is a bijection onto its image, that image is finite by [F3], and [F2] transports finiteness back, so $R$ is finite.

The assignment $x_g\mapsto g$ sends each relator $x_g{x}_h{x}_{gh}^{-1}$ to $gh(gh{)}^{-1}=e_G$, so [L1] gives a homomorphism $\pi :P:=⟨X\mid R⟩\to G$.

By [F5] every relator of $R$ is the identity in $P$, so $[x_g][x_h][x_{gh}{]}^{-1}=e$ and hence $[x_g][x_h]=[x_{gh}]$; therefore $\sigma :G\to P$, $\sigma (g)=[x_g]$, is a homomorphism.

The composite $\pi \circ \sigma$ fixes every $g\in G$; the composite $\sigma \circ \pi$ fixes every generator class $[x_g]$, and uniqueness in [L1] makes it the identity on $P$. Thus $\pi$ and $\sigma$ are inverse isomorphisms.

Both $X$ and $R$ are finite and $P\cong G$, so [F1] shows that $G$ has the displayed finite presentation, including when $G$ is the one-element group.