Peskin and Schroeder mention a few times that the anomalous dimension of a gauge interaction operator is zero. The justification for this is that the charge operator shouldn't get modified under anomalous dimensions. I'm trying to show explicitly for the simplest case, QED, but I'm not getting the right answer so I hoping to get some help. I detail my calculation below, though I suspect my problem is conceptual as opposed to a silly mistake.
We need to consider the diagrams,
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We use $\overline{MS}$ with massless fermions and only keep the divergent pieces. The first diagram is, \begin{align} \require{cancel} i G & = \int d^4 \ell \frac{ \gamma _\mu \cancel{ \ell } \gamma _\nu \cancel{ \ell } \gamma ^\mu }{ \ell ^6 } ( i ) ^2 ( - i ) ^2 ( - i e ) ^3 \\ & = -\frac{ i e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \end{align} The counterterms for the fermion and photon propagators are \begin{equation} - i \frac{ 2e ^2 }{ 16 \pi ^2 \epsilon } \quad \mbox{and} \quad - i \left( \frac{ 8 }{ 3} \frac{ e ^2 }{ 16 \pi ^2 \epsilon } \right) \left( q _\mu q _\nu - q ^2 g _{ \mu \nu } \right) \end{equation} where $q$ is the momentum of the photon. This gives for the last three diagrams, \begin{equation} - i\frac{ e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \quad , \quad - i\frac{ e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \quad , \quad - i \frac{ 8 }{ 3} \frac{ e ^3 }{ 16 \pi ^2 \epsilon } \left( \frac{q _\nu q _\mu}{q^2} - g _{ \nu \mu } \right) \gamma ^\mu \end{equation} , respectively.
Now if I understand correctly, for these not to give the 3-point interaction an anomalous dimension they must cancel, but I don't see how they would so I feel like my understanding of anomalous dimensions is off. Any ideas about what's going on here?