How to determine $\Gamma_{15}$?

Question

For context,yesterday I asked How to determine $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5$?.

Now I learned of new theorem that made me curious:

Theorem:

Let $m,n \in \mathbb{Z}$ such that $gcd(m,n)=1$. Then the map $\Omega: \Gamma_n \times \Gamma_m \rightarrow \Gamma_{nm}, (\chi,\chi') \mapsto \chi \chi'$ is a (group)- isomorphism.

In How to determine $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5$? we determined $\Gamma_3$ and $\Gamma_5$.

Now given $\Gamma_3$ and $\Gamma_5$ the Theorem mentioned should make it possible to calculate $\Gamma_{15}$ in an easy way.

I did calculate $\Gamma_3$ and $\Gamma_5$ explicitly and here are the values:

For $\Gamma_3$ we get the following table:

$\begin{matrix} \Gamma_3 & k=1 & k=2 \\ \chi_0(k) &1 & 1\\ \chi_1(k) & 1 & -1 \end{matrix}$

and for $\Gamma_5$ we get this table ( I denote the elements of $\Gamma_5$ as $\Psi$ for better readability):

$\begin{matrix} \Gamma_5 & k=1 & k=2 & k=3 & k=4\\ \Psi _0(k) & 1& 1 & 1 & 1\\ \Psi _1(k)& 1 & -1 &-1 &1 \\ \Psi _2(k)& 1& i& i &-1 \\ \Psi _3(k)& 1 & -i &i & -1 \end{matrix}$.

I will denote the elements of $\Gamma_{15}$ as $\Phi$. $(\mathbb{Z}/15\mathbb{Z})^{\times}=\{1,2,4,7,8,11,13,14\}$.

Now since $gcd(3,5)=1$, the map $\Omega: (\chi,\Psi) \mapsto \chi \cdot \Psi=\Phi$ is an Isomorphism. And we can calculate the table

$\begin{matrix} \Gamma_3 \times \Gamma_5 & k=(1,1) & k=(1,2) &k=(1,3) &k=(1,4) &k=(2,1) & k=(2,2) & k=(2,3) & k=(2,4)\\ \chi_0 \cdot \Psi_0& 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \chi_0 \cdot \Psi_1& 1 & -1 & -1 & 1 & 1 & -1 & -1 &1 \\ \chi_0 \cdot \Psi_2& 1 & i &i & -1 & 1 & i & i & -1\\ \chi_0 \cdot \Psi_3& 1 & -i & i & -1 & 1 & -i & i & -1 \\ \chi_1 \cdot \Psi_0& 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1\\ \chi_1 \cdot \Psi_1& 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1\\ \chi_1 \cdot \Psi_2 & 1 & i & i & -1 & -1 & -i & -i & 1 \\ \chi_1 \cdot \Psi_3& 1 & -i & i & -1 & -1 & i & -i & 1 \end{matrix}$

My only problem now is that in this table we have the values for $k=(1,1),(1,2),(1,3)$,.... and so on. Instead I would like to have the proper values for $k$ regarding $\Gamma_{15}$ i.e. $k=1,2,4,7,8,11,13,14$.

The Question now is, how can I assign those tupels $(k_1,k_2)$ to it's corresponding elements (in $(\mathbb{Z}/15\mathbb{Z})^{\times}$)?

Answer 1

The Chinese remainder theorem states that the ring homomorphism $\mathbb{Z}/mn\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ that maps $x + mn\mathbb{Z}$ to $(x + m\mathbb{Z},x + n\mathbb{Z})$ is an isomorphism: Bézout's identity gives us $rm + sn = 1$; then the map taking $(x + m\mathbb{Z},y + n\mathbb{Z})$ to $xsn + yrm + mn\mathbb{Z}$ is a well-defined inverse. (This gives an isomorphism between the groups of units $(\mathbb{Z}/mn\mathbb{Z})^\times \cong (\mathbb{Z}/m\mathbb{Z})^\times \times (\mathbb{Z}/n\mathbb{Z})^\times$.) In other words: $xsn + yrm \equiv x \pmod{m}$ and $xsn + yrm \equiv y \pmod{n}$.

For the case of $(m,n) = (3,5)$, we have for instance that $2 \cdot 3 + (-1) \cdot 5 = 1$, so given $k_1$ and $k_2$ we have $-5k_1 + 6k_2$ (which you may want to reduce mod 15) is congruent to both $k_1$ mod $3$ and $k_2$ mod $5$. For example, the pair $(k_1,k_2)=(2,3)$ (mod 3 and 5) maps to $2 \cdot (-5) + 3 \cdot 6 = 8$.

Now, if you are just trying to build a table, you could take your list of elements of $(\mathbb{Z}/15\mathbb{Z})^\times$, reduce them mod $3$ and mod $5$, apply the corresponding characters and multiply. (e.g. $(\chi_1 \cdot \Psi_2)(8) = \chi_1(2) \Psi_2(3) = -1 \cdot i$)