Doubt in tranformations of perpendicular directions coordinates to the motion of frame in Lorentz transformation Consider a frame $S$ which is at rest. The frams $S'$ at $t=0$ coincides with $S$ and then start moving with velocity $v$ in the $+x$ direction.
In Galilean transformation we can easily see that $y'=y$
But while deriving the transformation formulas for Lorentz transformation, we take $y'=y$
If we consider $y'=y'(x,y,z,t)$ then definitely it is linear with respect to all its arguments so that Newton's law holds.
So, we can write $y'=a_1x+a_2y+a_3z+a_4t\tag{1}$
I am trying to prove $y'=y$ from $y'=y(x,y,z,t)$
If the particle is on $x-axis$ in $S$ then it remains on $x-axis$ on $S'$
So, $y'=y=0$ and $z'=z=0$
From (1),
$0=a_1x+a_4t$
As $x$ and $t$ are independent of each other.
We can conclude that $a_1=a_4=0$
Thus (1) becomes $y'=a_2y+a_3z \tag{2}$
If the particle is in $x-z$ plane in $S$ at $y=0$ the particle remains in $x-z$ plane at $y'=0$ otherwise we get a contradiction that $S'$ moves in $y$ direction also.
So, (2) becomes, $0=a_3z,\;\;\forall z$
$\implies a_3=0$
Thus (2) becomes, $y'=a_2y\tag{3}$
But I have trouble in proving $a_2=1$.
Some authors use the concept of symmetry, but I am not able to understand their argument completely.
Please help in understanding how $a_2=1$?
I am very confused.
 A: 
I think we must use the symmetry concept.
So, for notation convenience consider that your frames $\,\rm S,S'\,$ are $\,\rm S_1,S_2\,$ respectively as in above Figure-01(1). Your equation (3) is then(2)
\begin{equation}
y_2\boldsymbol{=}a_2\,y_1
\tag{A-01}\label{A-01}
\end{equation}
Below this configuration consider two frames $\,\rm S_3,S_4\,$ as follows :
The system $\,\rm S_3\,$ is at rest with respect to $\,\rm S_2\,$ with reverse the $\,x-,y-\,$ axes so
\begin{align}
x_3\boldsymbol{=-}x_2
\tag{A-02a}\label{A-02a}\\
y_3\boldsymbol{=+}y_2
\tag{A-02b}\label{A-02b}\\
z_3\boldsymbol{=-}z_2
\tag{A-02c}\label{A-02c}
\end{align}
The system $\,\rm S_4\,$ is at rest with respect to $\,\rm S_1\,$ with reverse the $\,x-,y-\,$ axes so
\begin{align}
x_4\boldsymbol{=-}x_1
\tag{A-03a}\label{A-03a}\\
y_4\boldsymbol{=+}y_1
\tag{A-03b}\label{A-03b}\\
z_4\boldsymbol{=-}z_1
\tag{A-03c}\label{A-03c}
\end{align}
The configuration of frames $\,\rm S_3,S_4\,$ is exactly that of $\,\rm S_1,S_2$ :  that is the second frame of a pair is moving along the common $\,x-$axis with velocity $\,\boldsymbol{\upsilon}\,$ with respect to the first frame of the pair. So, corresponding to equation \eqref{A-01} we have
\begin{equation}
y_4\boldsymbol{=}a_2\,y_3
\tag{A-04}\label{A-04}
\end{equation}
which by equations \eqref{A-03b}, \eqref{A-02b} yields
\begin{equation}
y_1\boldsymbol{=}a_2\,y_2
\tag{A-05}\label{A-05}
\end{equation}
Combining equations \eqref{A-01}, \eqref{A-05} we have
\begin{equation}
y_1\boldsymbol{=}a^2_2\,y_1
\tag{A-06}\label{A-06}
\end{equation}
so
\begin{equation}
a^2_2\boldsymbol{=}1 \quad \boldsymbol{\Longrightarrow} \quad a_2\boldsymbol{=\pm}1
\tag{A-07}\label{A-07}
\end{equation}
We choose $a_2\boldsymbol{=+}1$ in order to exclude space inversion.
The Figure-02 below is a back view of Figure-01.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
(1)
This Figure is extracted from my answers here Schutz's geometrical proof that spacetime interval is invariant and here Special Relativity - Reference Frames S and S′ relative velocity.

(2)
To be precise the coefficient $\,a_2\,$ must be considered as dependent on the velocity $\,\boldsymbol{\upsilon}\,$ and we must write $\,a_2(\boldsymbol{\upsilon})$. This fact doesn't modify the final conclusion.

A: the lines element are
$$s^2=-c^2 t^2+x^2+y^2+z^2\\
{s'}^2=-{c'}^2 {t'}^2 + {x'}^2 + {y'}^2 + {z'}^2\tag 1$$
with this Ansatz:
$$c=c'\\
x'=A\,(x-v\,t)\\
t'=B\,x+C\,t\\
y'=a_2\,y\\
z'=z$$
and ${s'}^2-s^2=0$ you obtain:
$$\underbrace{\left(1-A^2 + c^2 B^2\right)}_{eq_1=0} x^2 + \underbrace{\left(2\,A^2 v + 2\,c^2 CB\right)}_{eq_2=0} t\,x + \\ \underbrace{\left(-c^2 - A^2 v^2 + c^2 C^2
\right)}_{eq_3=0} t^2 +\underbrace{(1-a_2^2)}_{eq_4=0}\,y^2= 0.$$
You have now four  equations for the four  unknowns $A~,B~,C~,a_2$.
The solution
$$A = \gamma\\B = -\frac{\beta}{c}\,\gamma\\ C = \gamma\\a_2=1$$
where
$$\gamma=\frac{1}{\sqrt{1-\beta^2}}\quad \beta = \frac{v}{c}$$
$\Rightarrow$
\begin{align*}
 \begin{bmatrix}
   t' \\
   x' \\
   y' \\
   z' \\
 \end{bmatrix}
 = &\left[ \begin {array}{cccc} {\gamma}&-{\frac {{\gamma}\,v}{{c}^{2}}}&0
&0\\  -{\gamma}\,v&{\gamma}&0&0\\  0
&0&1&0\\  0&0&0&1\end {array} \right]\,
\begin{bmatrix}
   t \\
   x \\
   y \\
   z \\
 \end{bmatrix}
\end{align*}
