The GNS null space is invariant under left translation

For fixed $g\in G$, the support of $L_{g}f$ is $g\mathrm{supp}(f)$, so $L_g$ preserves finite support and is complex-linear. The group law gives $L_e=I$, $L_g{L}_h=L_{gh}$, and $L_g^{-1}=L_{g^{-1}}$.

For finitely supported $f,h$, [A2] gives $B_{\phi }(L_{g}f,L_{g}h)={\sum }_{a,b\in G}f(g^{-1}a)\overline{h(g^{-1}b)}\phi (b^{-1}a)\mathrm{.}$ Only finitely many terms are nonzero. Substitute $a=gx$ and $b=gy$; then $b^{-1}a=(gy{)}^{-1}(gx)=y^{-1}x$, so the sum becomes ${\sum }_{x,y\in G}f(x)\overline{h(y)}\phi (y^{-1}x)=B_{\phi }(f,h)$. Thus left translation preserves the whole form.

If $f,h\in N_{\phi }$, [A1] makes all four terms in $B_{\phi }(f+h,f+h)$ vanish, and sesquilinearity gives $B_{\phi }(\lambda f,\lambda f)=\mathrm{\mid }\lambda {\mathrm{\mid }}^2{B}_{\phi }(f,f)=0$ for every $\lambda \in \mathbb{C}$. Hence $N_{\phi }$ is a complex linear subspace. Step 2.1 implies $L_g{N}_{\phi }\subseteq N_{\phi }$; applying it to $g^{-1}$ and using step 1.1 gives $L_g{N}_{\phi }=N_{\phi }$. Therefore if $f-h\in N_{\phi }$, then $L_{g}f-L_{g}h=L_g(f-h)\in N_{\phi }$, so $[f]\mapsto [L_{g}f]$ is well-defined on the quotient. Its inverse is induced by $L_{g^{-1}}$, and the identities in step 1.1 descend to the quotient.