3

Prelude

Cconsider the finite abelian group $\mathcal G = \prod_{i=1}^A \mathbb Z_{a_i}$ and let $\mathbf s \in \mathcal G$. Let $\mathcal H = \operatorname{grp}({\mathbf s})$ be the subgroup of $\mathcal H$ generated by $\mathbf s$. Then, for some integer, $S$, there is an isomorphism $f:\mathbb Z_S \to \mathcal H$ described by $f(\bar t)= t\mathbf s$.

So there must be an isomorphism $\kappa:\mathcal H \to \mathbb Z_S$ which is the inverse mapping of $f$. I got curious about what $\kappa$ must look like. This is a theorem that I came up with. Please let me know if you find any faults with it. Especially let me know if someone has already done something like this before. I have posted the proof as an answer.

Setup

Let $\mathcal G = \prod_{i=1}^A \mathbb Z/a_i\mathbb Z$
Let $\mathbf s= (s_1, s_2, s_3, \dots, s_A)= (s_i)_{i=1}^A$.
Let $\mathcal H = \operatorname{grp}({\mathbf s})$ be the subgroup of $\mathcal H$ generated by $\mathbf s$.
Let $S = \operatorname{ord} \mathbf s$.

Create the following list of ordered integer $A$-tuples.

$\mathbf a = (a_i)_{i=1}^A$.

$\mathbf s = (s_i)_{i=1}^A$.

$\mathbf g = (\gcd(a_i,s_i))_{i=1}^A$.

$\mathbf m = \left(\dfrac{a_i}{g_i}\right)_{i=1}^A$.

$\mathbf p = (p_i)_{i=1}^A$ Where \begin{array}{ll} (1.) &p_i \mid m_i \; (1 \le i \le A)\\ (2.) &i \ne j \implies \gcd(p_i,p_j) = 1 \; (1 \le i,j \le A)\\ (3.) &S = \prod_{i=1}^A p_i = \operatorname{lcm}\{m_i\}_{i=1}^A\\ \end{array}

$\mathbf t = \left(\dfrac{a_i}{p_i}\right)_{i=1}^A$.

$\mathbf r = \left(\dfrac{a_i}{p_i\cdot g_i}\right)_{i=1}^A = \left(\dfrac{m_i}{p_i}\right)_{i=1}^A = \left(\dfrac{t_i}{g_i}\right)_{i=1}^A$.

$\mathbf{\alpha} = (\alpha_i)_{i=1}^A$.

\begin{array}{l} & \text{Note that } \gcd \left\{\dfrac{S}{p_i}\right\}_{i=1}^A = \dfrac{S}{\operatorname{lcm}\{p_i\}_{i=1}^A} = 1.\\ & \text{So there exists integers } \lambda_i \text{ such that } \sum_{i=1}^A \lambda_i \dfrac{S}{p_i} = 1. \\ & \text{We define }\alpha_i = \lambda_i \dfrac{S}{p_i} \pmod{S}.\\ \end{array}

$\mathbf {\omega} = (\omega_i)_{i=1}^A$.

$\qquad $ Note that $\; \gcd\left(\dfrac{s_i}{g_i}, m_i\right) =\gcd\left(\dfrac{s_i}{g_i}, \dfrac{a_i}{g_i}\right) = 1$

$\qquad$ We define $ \omega_i = \left(\dfrac{s_i}{g_i}\right)^{-1}\pmod{m_i}$

$\mathbf{\beta}_i \equiv \alpha_i \ \omega_i \pmod S \; (i=1..A)$.

Example

As an example, let $\mathcal G = \mathbb Z_{36} \times \mathbb Z_{40} \times \mathbb Z_{72}$. Let $\mathbf s= (\overline{16}, \overline{15}, \overline{30})$.

Then $\mathcal H = \operatorname{grp}({\mathbf s})$

$S = 72$. We find

 a:  36  40  72
 s:  16  15  30
 g:   4   5   6
 m:   9   8  12
 p:   9   8   1
 t:   4   5  72
 r:   1   1  12
 α:  -1   1   0
 ω:   7   3   5
 β:  -7   3   0

You may have noticed a preponderance of variables here. to avoid adding any more than is necessary, I will use the notation $<n>$ to mean the same thing as "$x$, where $x \in \{0, 1, 2, \dots, n-1\}$". So for example, $<5>$ represents a variable that can take on the value $0, 1, 2, 3, 4$.

Finally, here is my theorem.

T H E O R E M

For $\mathbf x \in \mathcal G$, define $\kappa = \sum_{i=1}^A \left \lfloor \dfrac{x_i}{g_i} \right \rfloor \beta_i \pmod S$.

Then every $\mathbf x \in \mathcal G$ can be expressed uniquely as

$$ \mathbf x = \kappa \mathbf s + ( p_i \, g_i <r_i> + <g_i>)_{i=1}^A$$

Continuing with our example, we see that, for $(x,y,z) \in \mathcal G$,

$\kappa = -7\lfloor x/4 \rfloor + 3\lfloor y/5 \rfloor \pmod{72}$.

Then every $(x,y,z) \in \mathcal G$ can be expressed uniquely as

$$(x, y, z) = \kappa(16, 15, 30) + (36<1> + <4>, 40<1> + <5>, 6<12> + <6>)$$

which simplifies to

$$(x, y, z) = \kappa(16, 15, 30) + (<4>, <5>, <72>)$$

Note that this implies that all triples of the form $(<4>, <5>, <72>)$ constitute a transversal of the cosets of $\mathcal H$ in $\mathcal G$.

Steven Alexis Gregory
  • 27,619

1 Answers1

3

Let $x = (\bar x_1, \bar x_2, \dotsc, \bar x_A) \in \mathcal G$. For each $1 \le i \le A$, there is an integer, $\gamma_i$, such that

$$ x_i = g_i \left\lfloor{\frac{x_i}{g_i}}\right\rfloor + \gamma_i \; \text{where} \; 0 \le \gamma_i < g_i \quad (1\le i\le A). \tag{1}\label{1} $$

Let $k$ be the unique integer for which \begin{equation*} 0 \le k < S \; \text{and} \; k\equiv \sum_{i=1}^A \beta_i \left\lfloor{\frac{x_i}{g_i}}\right\rfloor \pmod S. \end{equation*}

For all $i \ne j$, $\alpha_i \equiv 0 \pmod{p_j}$. Since $\sum_{i=1}^A \alpha_i \equiv 1 \pmod S$, we must have $\alpha_i \equiv 1 \pmod{p_i}$ for all $1 \le i \le A$. Hence $\beta_i \equiv \omega_i \pmod{p_i}$ for all $1 \le i \le A$. It follows that $$ k \equiv \omega_i \left \lfloor\frac{x_i}{g_i}\right\rfloor \pmod{p_i} \quad (1\le i\le A). \tag{2}\label{2} $$ Since $p_i|m_i$, then $\omega_i \equiv \left(\dfrac{s_i}{g_i}\right)^{-1} \pmod{p_i}$. So Equation $(2)$ can be rewritten as \begin{equation} \left\lfloor\frac{x_i}{g_i}\right\rfloor \equiv k\frac{s_i}{g_i} \pmod{p_i} \quad (1\le i\le A). \tag{3}\label{3} \end{equation} Then, for each $1 \le i \le A$, there must exists an integer $\lambda_i$ such that \begin{equation} \left\lfloor \frac{x_i}{g_i}\right\rfloor = k\frac{s_i}{g_i} + p_i \lambda_i \quad (1\le i\le A). \tag{4}\label{4} \end{equation} Substituting this into Equation $(1)$, we get \begin{equation} x_i = k s_i + g_i p_i \lambda_i + \gamma_i \text{ where }0 \le \gamma_i < g_i \quad (1\le i\le A). \tag{5}\label{5} \end{equation} Since $ r_i = \dfrac{a_i}{g_i p_i}$, then $a_i = g_i p_i r_i$. It follows that, for each $i$, there exist a unique integer, $\rho_i$, such that $1 \le \rho_i \le r_i$ and $g_i p_i \lambda_i \equiv g_i p_i \rho_i \pmod{a_i}$. Hence we can rewrite Equation $(5)$ as \begin{equation} x_i \equiv k s_i + g_i p_i \rho_i + \gamma_i \pmod{a_i} \quad (1\le i\le A) \tag{6}\label{6} \end{equation} where $0 \le k < S$, $0 \le \gamma_i < g_i$ and $0 \le \rho_i < r_i \quad (1 \le i \le A)$. We can rewrite Equation $(6)$ as $$ \mathbf x = \kappa \mathbf s + (g_i p_i <r_i> + <g_i>)_{i=1}^A \tag{7}\label{7} $$

We still need to prove that Equation $(7)$ is unique for each $\mathbf x \in \mathcal G$. So suppose that \begin{equation} x_i \equiv k s_i + g_i p_i \rho_i + \gamma_i \equiv k' s_i + g_i p_i \rho'_i + \gamma'_i \pmod{a_i} \quad (1\le i\le A) \tag{8}\label{8} \end{equation} for some $0 \le k, k' < S$, $0 \le \gamma_i, \gamma'_i < g_i$ and $0 \le \rho_i, \rho'_i < r_i$.

We start by noting that, for each $i$, $g_i|a_i$ and $g_i|s_i$. It follows that, modulo $g_i$, Equation $(8)$ reduces to $\gamma_i \equiv \gamma'_i \pmod{g_i}$. We conclude that $\gamma_i = \gamma'_i$ for all $1 \le i \le A$. Equation $(8)$ now simplifies to \begin{equation} k s_i + g_i p_i \rho_i \equiv k' s_i + g_i p_i \rho'_i \pmod{n_i} \quad (1\le i\le A) \tag{9}\label{9} \end{equation}

Since, for each $i$, $g_i|s_i$ and $\dfrac{n_i}{g_i} = m_i$, Equation $(9)$ can be reduced to

\begin{equation} k \frac{s_i}{g_i} + p_i \rho_i \equiv k' \frac{s_i}{g_i} + p_i \rho'_i \pmod{m_i} \quad (1\le i\le N) \tag{10}\label{10} \end{equation}

Rewriting Equation $(10)$ modulo $p_i$ and simplifying, we get $$ k \dfrac{s_i}{g_i} \equiv k' \dfrac{s_i}{g_i} \pmod{p_i} \quad (1\le i\le A) $$. Multiplying both sides by $h_i$, we get $$ k \equiv k' \pmod{p_i} \quad (1\le i\le N) $$. Since the $p_i$ are pairwise prime and $p_1 p_2 p_3 \dots p_N = S$, it follows that $k \equiv k' \pmod{S} \quad (1\le i\le A)$. Since $0 \le k, k' < S$, then $k = k'$. We can now simplify Equation $(10)$ to \begin{equation} p_i \rho_i \equiv p_i \rho'_i \pmod{m_i} \quad (1\le i\le A) \tag{11}\label{11} \end{equation} Because $\dfrac{m_i}{p_i} = r_i$, it follows that $\rho_i \equiv \rho'_i \pmod{r_i} \quad (1\le i\le A)$. Since $0 \le \rho_i, \rho'_i < r_i$, we conclude that $\rho_i = \rho'_i \quad (1 \le i \le N)$. Hence Equation $(7)$ is unique for each $\mathbf x \in \mathcal G$.

Steven Alexis Gregory
  • 27,619