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With the convention in the question,acts asAlthough $V(x+r)=V(x)$ for $r\in \Lambda$, the vector potential changes byConsequentlyfor a nonzero magnetic field in general.

Now let $K=p+eA$. Sinceand the derivative in the $r$ direction annihilates
$r\cdot (B\times x)$, the two terms in its exponential commute. The magnetic translation operator therefore acts as

Part (a) gives $[K_i,p_j-e{A}_j]=0$, so ${\mathcal{T}}_r$ commutes with the kinetic term. Its phase commutes with $V$, while its translation sends $V(x)$ to $V(x+r)=V(x)$. Hence