Lagrangian: electron in magnetic field (what are the conserved quantities?) Given an electron subjected to a magnetic field $\vec{B} = (0,0,B_0)$
I have chosen the Lagrangian to be $L = T-V = \frac{1}{2}\,m\,\Vert\mathbf{\dot r}\Vert^2+ e\,\mathbf{\dot r}\boldsymbol \cdot\mathbf{A}$.

In between question: $V = -e\,\mathbf{\dot r}\boldsymbol \cdot\mathbf{A}$ is a generalised potential usually given by $V = e\,(\phi-\mathbf{\dot r}\boldsymbol \cdot\mathbf{A})$. Whatsoever I only get the right Hamiltonian same as on the sheet if I use the shortened potential. Why is that? $\phi= 0?$

With the vector potential set on $\mathbf{A} = (-B_0/2\,y, B_0/2\,x,0)$ the total Lagrangian becomes:
$L = \frac{1}{2}\,m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2 \right) -e\,\dot{x}\,B_0/2\,y+e\,\dot{y}\,B_0/2\,x$.
In order to get more information on conserved quantities I use Euler-Lagrange-Equations: $\dfrac{\mathrm{d}}{\mathrm{dt}}\partial_{\textstyle \dot{r}}L-\partial_{\textstyle {r}} L= 0$:
$m\,\ddot{x}- e\,B_0/2\,\dot{y}-e\,\dot{y}\,B_0/2 = 0$
$m\,\ddot{y}+ e\,B_0/2\,\dot{x}+e\,\dot{x}\,B_0/2 = 0$
$m\,\ddot{z}  \qquad \qquad \qquad \qquad =0$

Final question: what are the conserved quantities now? Based on the last line I get $m\,\dfrac{\mathrm{d}}{\mathrm{dt}}\dot{z} = 0$. Hence momentum in $z$-direction is conserved. But what else? There must be another quantity.

 A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
\newcommand{\e}{\bl=}
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\newcommand{\gr}{\bl>}
\newcommand{\les}{\bl<}
\newcommand{\greq}{\bl\ge}
\newcommand{\leseq}{\bl\le}
\newcommand{\plr}[1]{\left(#1\right)}
\newcommand{\blr}[1]{\left[#1\right]}
\newcommand{\lara}[1]{\langle#1\rangle}
\newcommand{\lav}[1]{\langle#1|}
\newcommand{\vra}[1]{|#1\rangle}
\newcommand{\lavra}[2]{\langle#1|#2\rangle}
\newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle}
\newcommand{\vp}{\vphantom{\dfrac{a}{b}}}
\newcommand{\hp}[1]{\hphantom{#1}} 
\newcommand{\x}{\bl\times}
\newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad}
\newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad}
\newcommand{\tl}[1]{\tag{#1}\label{#1}}
\newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$
Hint :
First note that
\begin{equation}
\mathbf E\e\m\bl\nabla\phi\m\dfrac{\partial \mathbf A}{\partial t}
\tl{a}
\end{equation}
Since in our case
\begin{equation}
\mathbf E\e\bl 0 \quad \texttt{and} \quad \dfrac{\partial \mathbf A}{\partial t}\e\bl 0
\tl{b}
\end{equation}
we have
\begin{equation}
\bl\nabla\phi\e\bl 0 \quad \bl\implies \quad \phi\e \texttt{constant}
\tl{c}
\end{equation}
Adding any constant in the Lagrangian $\,L\,$ doesn't affect the equations of motion so without loss of generality we set $\,\phi\e 0$.
I think that using a 3-dimensional version of the Beltrami identity
a constant of the motion would be
\begin{equation}
\mathbf{\dot r}\bl\cdot\dfrac{\partial L}{\partial \mathbf{\dot r}}\m L\e \texttt{constant}
\tl{1}
\end{equation}
Also, looking in the first two Euler-Lagrange equations what are our thoughts about the complex function
\begin{equation}
\mathrm w \e x\p i\,y 
\tl{2}
\end{equation}
Could we translate these equations to a simple 2nd order linear differential equation with respect to $\,\mathrm w\,$? (try to relate this with the axial symmetry of @LSS's answer).
$\hebl$
Responds to OP comments
First, for the $\,\texttt{constant}\,$ of equation \eqref{1} we have
\begin{equation}
\begin{split}
&\mathbf{\dot r}\bl\cdot\dfrac{\partial L}{\partial \mathbf{\dot r}}\m L  \e \texttt{constant}\quad \bl\implies\\
&\mathbf{\dot r}\bl\cdot\underbrace{\plr{m\,\mathbf{\dot r}\p e\mathbf A }}_{\partial L/\partial \mathbf{\dot r}}\m\underbrace{\plr{1/2\,m\,\Vert\mathbf{\dot r}\Vert^2\p e\,\mathbf{\dot r}\bl\cdot\mathbf{A}}}_{L}\e \texttt{constant}\quad \bl\implies
\end{split}
\nonumber
\end{equation}
\begin{equation}
\Vert\mathbf{\dot r}\Vert\e \texttt{constant}
\tl{R-1}
\end{equation}
the well-known result that the magnetic force as perpetually normal to the velocity of the particle doesn't change the magnitude of this velocity and doesn't produce work keeping unchanged its non-relativistic kinetic energy.
Second, for the complex function $\,\mathrm w\e x\p i\,y \,$ of equation \eqref{2} we have from the first two Euler-Lagrange equations
\begin{align}
\ddot{x} & \e  \p\omega\,\dot{y}
\tl{R-2a}\\
\ddot{y} & \e  \m\omega\,\dot{x}
\tl{R-2b}\\
\texttt{where}\quad \omega & \e\dfrac{e\,B_0}{2m}
\tl{R-2c}
\end{align}
So,
\begin{equation}
\ddot{x}\p i\,\ddot{y}\e\m i\,\omega\plr{\dot{x}\p i\,\dot{y}}
\tl{R-3}
\end{equation}
or
\begin{equation}
\ddot{\mathrm w}\e\m i\,\omega\,\dot{\mathrm w}
\tl{R-4}
\end{equation}
etc...
$\hebl$
ADDENDUM A
Motivated by the Euler-Lagrange-Equations of the post and @hft's answer I realize the conservation of the linear momentum \begin{equation}
\mathbf p_{\mathbf q}\e m\,\mathbf{\dot r}\m e\,\mathbf r\x\mathbf B \e m\,\mathbf{\dot r}\m e\,\mathbf r\x\plr{\bl\nabla\x\mathbf A}
\tl{A-1} 
\end{equation}
Now I try to find if this is the generalized momentum
conjugate to a cyclic coordinate $\,\mathbf q\,$ the latter satisfying :
\begin{equation}
\dfrac{\partial L}{\partial \mathbf{\dot q}}\e\mathbf p_{\mathbf q}\e m\,\mathbf{\dot r}\m e\,\mathbf r\x\mathbf B\qquad \texttt{and} \qquad \dfrac{\partial L}{\partial \mathbf q}\e 0
\tl{A-2}
\end{equation}
$\hebl$
ADDENDUM B
We will express the Lagrangian in cylindrical coordinates as suggested by @LSS's answer.
\begin{equation}
\mathbf r\e 
\begin{bmatrix}
\:x\:\vp\\
\:y\:\vp\\
\:z\:\vp
\end{bmatrix}
\e
\begin{bmatrix}
\:\rho\cos\phi\:\vp\\
\:\rho\sin\phi\:\vp\\
z\vp
\end{bmatrix}\quad\bl\implies\quad
\mathbf{\dot r}\e
\begin{bmatrix}
\:\:\:\dot{\!\!\rho}\cos\phi\m\rho\:\:\dot{\!\!\phi}\sin\phi\:\vp\\
\:\:\:\dot{\!\!\rho}\sin\phi\p\rho\:\:\dot{\!\!\phi}\cos\phi\:\vp\\
\dot z\vp
\end{bmatrix}
\tl{B-01}
\end{equation}
So
\begin{equation}
\Vert\mathbf{\dot r}\Vert^2\e\:\:\dot{\!\!\rho}^{\:2}\p\rho^2\:\:\dot{\!\!\phi}{}^{\;2}\p\dot{z}^2 
\tl{B-02}
\end{equation}
Also
\begin{equation}
\mathbf{\dot r}\bl\cdot\mathbf A\e 
\dfrac{B}{2}
\begin{bmatrix}
\:\:\:\dot{\!\!\rho}\cos\phi\m\rho\:\:\dot{\!\!\phi}\sin\phi\:\vp\\
\:\:\:\dot{\!\!\rho}\sin\phi\p\rho\:\:\dot{\!\!\phi}\cos\phi\:\vp\\
z\vp
\end{bmatrix}^{\bl\top}
\begin{bmatrix}
\:\m\rho\sin\phi\:\vp\\
\:\hphantom{\m}\rho\cos\phi\:\vp\\
0\vp
\end{bmatrix}\e 
\dfrac{B}{2}\rho^2\:\:\dot{\!\!\phi}
\tl{B-03} 
\end{equation}
From equations \eqref{B-02},\eqref{B-03} we have the following expression of the Lagrangian in cylindrical coordinates
\begin{equation}
L\plr{\rho,\phi,z,\:\:\dot{\!\!\rho},\:\:\dot{\!\!\phi},\dot z}\e\frac12 m\plr{\:\:\dot{\!\!\rho}^{\:2}\p\rho^2\:\:\dot{\!\!\phi}{}^{\;2}\p\dot{z}^2}\p\dfrac{e\,B}{2}\rho^2\:\:\dot{\!\!\phi}
\tl{B-04} 
\end{equation}
In above Lagrangian  the coordinates $\:\phi,z\:$ are cyclic, that is the Lagrangian is independent of them, so the conjugate momenta are conserved as shown also in the Euler-Lagrange equations of motion below
\begin{equation}
\left.
\begin{cases}
\dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}}\m\dfrac{\partial L}{\partial \rho}\e0\\ 
\dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\phi}}}\m\dfrac{\partial L}{\partial \phi}\e0\\ 
\dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \dot{z}}}\m\dfrac{\partial L}{\partial z}\e0
\end{cases}
\right\}
\quad\bl\implies\quad
\left.
\begin{cases}
\dfrac{\mathrm d}{\mathrm dt}\plr{m\:\:\dot{\!\!\rho}} \e m\rho\:\:\dot{\!\!\phi}{}^{\;2}\p e\,B\rho\:\:\dot{\!\!\phi}\vphantom{\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}}}\\  
\dfrac{\mathrm d}{\mathrm dt}\plr{m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2 }\e0\\ 
\dfrac{\mathrm d}{\mathrm dt}\plr{m\dot{z}}\e0
\end{cases}
\right\}
\tl{B-05}
\end{equation}
The conjugate momenta are
\begin{align}
p_\rho&\e\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}\e m\:\:\dot{\!\!\rho} 
\tl{B-06.1}\\
p_\phi&\e\dfrac{\partial L}{\partial \:\:\dot{\!\!\phi}}\e m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2
\tl{B-06.2}\\
p_z&\e\dfrac{\partial L}{\partial \dot z}\e m\dot z 
\tl{B-06.3}
\end{align}
The conserved momenta conjugate to the cyclic coordinates, angle $\,\phi\,$ and cartesian $\,z$, are the angular momentum $\,p_\phi\,$ and the linear momentum $\,p_z\,$ respectively
\begin{equation}
\dot{p}_\phi\e0\,,\qquad \dot{p}_z\e0
\tl{B-07} 
\end{equation}
Note that the angular momentum $\,p_\phi\,$ if expressed by the initial cartesian coordinates yields
\begin{equation}
p_\phi\e m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2\e m\plr{x\dot y\m y\dot x}\p\dfrac{e\,B}{2}\plr{x^2\p y^2} 
\tl{B-08} 
\end{equation}
in agreement with @Thomas Fritsch's answer, the item
\begin{equation}
m\rho^2\:\:\dot{\!\!\phi}\e m\plr{x\dot y\m y\dot x}
\tl{B-09} 
\end{equation}
being the angular momentum of the particle with respect to the $\,z\m$axis.
A: The system has axial symmetry, that is, if you rotate around the z axis the system remains the same.
I would recommend you to go to the cylindrical frame. In fact, if you do that, you will realize that the expression is independent of $\phi$.
A: 
$m\,\ddot{x}- e\,B_0/2\,\dot{y}-e\,\dot{y}\,B_0/2 = 0$


$m\,\ddot{y}+ e\,B_0/2\,\dot{x}+e\,\dot{x}\,B_0/2 = 0$


$m\,\ddot{z}  \qquad \qquad \qquad \qquad =0$


Final question: what are the conserved quantities now? Based on the last line I get $m\,\dfrac{\mathrm{d}}{\mathrm{dt}}\dot{z} = 0$. Hence momentum in $z$-direction is conserved. But what else? There must be another quantity.

There are lots of conserved quantities. For example, any quantity that is a function of only the z momentum is conserved.
In addition, based on the first two equation that you wrote, we can also see that:
$\frac{d}{dt}\left( m\,\dot{x}- e\,B_0/2\,y-e\,y\,B_0/2 \right)= 0$
and
$\frac{d}{dt}\left(m\,\dot{y}+ e\,B_0/2\,x+e\,x\,B_0/2 \right)= 0\;.$
Note: I'm assuming, for example, that your notation $B_0/2\,y$ means $y$ times $B_0$ divided by 2, not $B_0$ divided by two and divided by $y$. Your notation is a little confusing here.
So... I'll just combine a couple terms for you and define a couple more constant quantities:
$$
\alpha = \left( m\,\dot{x}- e\,B_0\,y \right)
$$
and
$$
\beta = \left(m\,\dot{y}+ e\,B_0\,x \right)
$$
A: So you have got these Euler-Lagrange equations:
$$\begin{align}
m\ddot{x}&=eB_0\dot{y} \\
m\ddot{y}&=-eB_0\dot{x} \\
m\ddot{z}&=0
\end{align}$$

*

*The magnetic force is known to do no work on the particle.
So we can guess, the kinetic energy
$$E_\text{kin}=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)$$
should be conserved. This can indeed be verified by using the
Euler-Lagrange equations:
$$\begin{align}
\frac{d}{dt}E_\text{kin}
&=\frac{d}{dt}\left(\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)\right) \\
&=\ ... \text{  (I leave the calculation to you)} \\
&=0
\end{align}$$
and hence
$$E_\text{kin}=\text{const}$$


*The system has a rotational symmetry around the $z$-axis. And
hence there should be a conserved quantity related to this symmetry.
Without having the magnetic field (i.e. with $B_0=0$) this
conserved quantity would be the $z$-component of angular momentum:
$$L_z=m(x\dot{y}-y\dot{x}).$$
But with the magnetic field this is not true anymore,
as can be checked with the Euler-Langrange equations:
$$\begin{align}
\frac{d}{dt}L_z
&=\frac{d}{dt}(m(x\dot{y}-y\dot{x})) \\
&=\ ... \text{  (I leave the calculation to you)} \\
&=-eB_0(x\dot{x}+y\dot{y})
\end{align}$$
This is obviously not zero, and hence $L_z$ is not conserved.
But luckily we can rewrite the right side as a total time derivative
$$\frac{d}{dt}L_z=-\frac{d}{dt}\left(\frac{1}{2}eB_0(x^2+y^2)\right)$$
and then bring everything to the left side
$$\frac{d}{dt}\left(L_z+\frac{1}{2}eB_0(x^2+y^2)\right)=0$$
So we have found the conserved quantity
$$L_z+\frac{1}{2}eB_0(x^2+y^2)=\text{const}.$$


*You can repeat the procedure above with the $x$- and $y$-components
of momentum ($p_x=m\dot{x}$ and $p_y=m\dot{y}$) to find
two more conserved quantities (the same two as given in @hft's answer):
$$p_x-eB_0y=\text{const}$$
$$p_y+eB_0x=\text{const}$$
