As we know, the Dedekind eta function $\eta(\tau)$ acquires a phase $\exp(2\pi i/24)$ under the modular transformation: $\tau \rightarrow \tau+1$. Therefore $\eta(\tau)^2$ is invariant under $\tau \rightarrow \tau+12$. Here comes the question, is $\eta(\tau)^2$ invariant on $\Gamma(12)$, where $\Gamma(12)$ denotes the principal subgroup?



Is $\eta(\tau)^2$ a modular form of weight 1 on $\Gamma(12)$?
Yes, it is. There are standard criteria (stated for example in Ken Ono's book "Web of Modularity" - Theorems 1.64 and 1.65) that indicate when an eta quotient is modular on $\Gamma\_{0}(N)$, and these show that $\eta(12\tau)^{2}$ is a modular form of weight $1$ on $\Gamma\_{0}(144)$ with Nebentypus $\chi\_{-4}$ (with order of vanishing $1$ at every cusp, somewhat surprisingly). This function $\chi\_{-4}$ is the Dirichlet character where $\chi\_{-4}(n) = 0$ if $n$ is even, $\chi\_{-4}(n) = (-1)^{(n-1)/2}$ if $n$ is odd. 


For $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, define $g(\tau) | M = \det(M)^{1/2} (c\tau+d)^{-1} g\left(\frac{a\tau+b}{c\tau+d}\right)$ to be the weight $1$ "slash" operator. Saying that $g(\tau) = \eta(12\tau)^{2}$ has character $\chi\_{-4}$ means that if $M \in \Gamma\_{0}(144)$ we have that
$g | M = \chi\_{-4}(d) g$. 


If we let $f(\tau) = \eta(\tau)^{2}$, then $f | \begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix}$ transforms nicely under $\Gamma\_{0}(144)$, and hence $f(z)$ transforms nicely under $\begin{bmatrix} 12 & 0 \\ 0 & 1 \end{bmatrix} \Gamma\_{0}(144) \begin{bmatrix} 1/12 & 0 \\ 0 & 1 \end{bmatrix}$. In particular, if $M = \begin{bmatrix} a & b \\ 144c & d \end{bmatrix} \in \Gamma\_{0}(144)$, we get that $g | M = \chi\_{-4}(d) g$ and this gives that
$$
 f | \begin{bmatrix} a & 12b \\ 12c & d \end{bmatrix} = \chi\_{-4}(d) f.
$$
In particular, for any $N = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \in {\rm SL}\_{2}(\mathbb{Z})$ with $a' \equiv d' \equiv 1 \pmod{4}$ and $b' \equiv c' \equiv 0 \pmod{12}$, we have that $f | N = f$. The set of such $N$ is a group which has $\Gamma(12)$ as an index $4$ subgroup. 


