Question

10. Show that$\left(1,0\right)$ and$\left(0,2\right)$ generate the additive group $\mathrm{\mathbb{Z}}\times {\mathrm{\mathbb{Z}}}_7$.

Short answer

The elements$\left(1,0\right)$ and$\left(0,2\right)$ generates the additive group $\mathrm{\mathbb{Z}}\times {\mathrm{\mathbb{Z}}}_7$.

Step-by-step solution

Cyclic Subgroup

Every element x in a group G, form a subgroup of all integer powers is called a cyclic subgroupof x.

Elements a,b in ℤ×ℤ7

Let $\left(a,b\right)\in \mathrm{\mathbb{Z}}\times {\mathrm{\mathbb{Z}}}_7$, if b is even then there exists a integer k such that$b=2k$ then $\left(a,b\right)=a\left(1,0\right)+k\left(0,b\right)$.

If b is odd then$b=b+7$ and the representative$b+7$ is even implies$b+7=2k$ so that$b=2k$ then $\left(a,b\right)=a\left(1,0\right)+k\left(0,b\right)$.

Since, 1 is the identity generates all the elements of$\mathrm{\mathbb{Z}}$ such that $\left(a,b\right)=a\left(1,0\right)+k\left(0,b\right)$.

Thus, all the elements of$\mathrm{\mathbb{Z}}\times {\mathrm{\mathbb{Z}}}_7$ are sum of integer multiples of$\left(1,0\right)$ and $\left(0,2\right)$.

Therefore, the elements$\left(1,0\right)$ and$\left(0,2\right)$ generates the whole additive group $\mathrm{\mathbb{Z}}\times {\mathrm{\mathbb{Z}}}_7$.