System of generators for $\Bbb Z_9\times \Bbb Z_{18}$.

Question

Is $\{(4,3),(3,5)\}$ a system of generators for $\Bbb Z_9\times \Bbb Z_{18}$?

I tried to generate the elements using the straightforward method $(4,3), (8,6), (3, 9)$ ... but it takes too long.

Can I use a faster method?

Answer 1

Let $v_1=(4,3)$ and $v_2=(3,5)$. Then $$ \begin{pmatrix}v_1\\v_2\end{pmatrix} = \begin{pmatrix}4&3\\3&5\end{pmatrix} \begin{pmatrix}e_1\\e_2\end{pmatrix} $$ where $e_1=(1,0)$ and $e_2=(0,1)$. The matrix has determinant $11$, which is invertible in the ring $\Bbb Z_9\times \Bbb Z_{18}$. Therefore, $v_1$ and $v_2$ generate the same additive group as $e_1$ and $e_2$.

Explicitly, $$ \begin{pmatrix}e_1\\e_2\end{pmatrix} = \begin{pmatrix}7&3\\3&2\end{pmatrix} \begin{pmatrix}v_1\\v_2\end{pmatrix} $$

Answer 2

Hint: $$\gcd(4,9)=\gcd(5,18)=1.$$

Answer 3

The Smith normal form can be used here: $\begin{pmatrix}4&3\\3&5\\9&0\\0&18\end{pmatrix}\to\begin{pmatrix}4&3\\11&11\\9&0\\0&18\end{pmatrix}\to\begin{pmatrix}1&0\\0&1\\0&0\\0&0\end{pmatrix}$.

Thus the elements do generate.