Induced representation in Zee's Group Theory I am trying to understand the topic of Induced representation of the euclidean Group E(2) in A. Zee's Group theory in a Nutshell in Chapter IV.i3.
The Lie algebra of E(2) has three elements $P_1, P_2, J$ with commutation relations
$[P_1, P_2] = 0$ and $[J, P_{1,2}] = \pm i P _{2,1}$
Zee constructs an infinite dimensional representation by identifying the maximally commuting subalgebra $P_1, P_2$ and labeling states by $|\vec{p}\rangle = |p, \phi\rangle$, where $P_{1,2}|\vec{p}\rangle = \vec{p}_{1,2}  |\vec{p}\rangle$.
He then states that "evidently under Rotations $R(\theta)|p, \phi\rangle = |p, \phi + \theta \rangle$". While this seems sensible, I have trouble reconciling this with the Lie algebra commutation relations since when considering infinitesimal rotations (and setting $P = (P_1, P_2)^{T}$ )
$$[P, R(\delta \theta) ] |\vec{p}\rangle = P |p, \phi + \delta \theta\rangle 
- \vec{p}|p, \phi + \delta \theta\rangle \\
= 
( \begin{pmatrix} 
\text{cos}(\delta \theta) & -\text{sin}(\delta \theta) \\
\text{sin}(\delta \theta) & \text{cos}(\delta \theta) 
\end{pmatrix} - \mathbb{1} ) \ \vec{p} \ | p, \phi + \delta \theta\rangle   \\  \approx 
\quad
\begin{pmatrix} 
0 & -1 \\
1 & 0 
\end{pmatrix}
\quad  \delta  \theta \ \vec{p} \ | p, \phi + \delta \theta\rangle  .$$
This should equal $$[P, 1 + i \delta\theta J]|\vec{p}\rangle = i \delta\theta \ [P, J] |\vec{p}\rangle = \delta  \theta\quad
\begin{pmatrix} 
0 & -1 \\
1 & 0 
\end{pmatrix}
\quad    \vec{p} \ | p, \phi \rangle  ,$$
which it doesn't. So Zee's "evident" prescription doesn't seem to reproduce the correct commutation relations. 
My question is am I misunderstanding something basic or is there more going on, that has simply been swept under the rug?
 A: Well, in standard practice, several things are swept under rugs in QM... The basic relation you may check in standard tasteful texts such as Sakurai & Napolitano, etc, is the translation formula,
$$
\bbox [yellow]{|\phi + \delta \theta\rangle= e^{-i\delta \theta J} |\phi\rangle=(1-i\delta\theta J+...)|\phi\rangle  }~,
$$
so 
$$
\langle \chi|\phi+ \delta \theta\rangle= \delta (\chi -\phi -\delta \theta) \\ \approx  \delta (\chi -\phi)  -\delta \theta ~\partial_\chi \delta (\chi -\phi)
= \langle \chi|\phi \rangle -i\delta \theta\langle \chi|J|\phi \rangle .
$$
Yes, these are infinite-dimensional vectors and one may choose to be careful$^\dagger$;
but, by and large, dismissing $O(\delta\theta ^2)$ terms is generally consistent...
The length of the two-vector, p, is a canard, of course.
Your first and second expression agree to lowest order in $\delta \theta$.

$^\dagger$ As a reminder, $\hat p | x\rangle= i\hbar\partial_x |x\rangle$, so that $\exp(-i\hat p a/\hbar)| x\rangle= |x+a\rangle= \exp(  a \partial_x )| x\rangle$, that is, somewhat hyper-formally, $\langle y|x+a\rangle=\delta(x+a-y)=\sum_{n=0}^\infty \frac{(a\partial_x)^n}{n!}     ~\delta(x-y)$. Consequently, dotting by $\langle \psi|$, observe $\psi^*(x+a)=\sum_{n=0}^\infty \frac{(a\partial_x)^n}{n!}     ~ \psi^*(x)= \psi^*(x)+a \psi ~'^*(x)+...  $. 
