Question

In Exercises 15 and 16, find an element of infinite order and an element of finite order in the given quotient group. There are many correct answers. Remember that $\mathbf{\mathbb{Z}}$is an additive group.$\mathbf{(}\mathbf{\mathbb{Z}}\mathbf{\times }\mathbf{\mathbb{Z}}\mathbf{)}\mathbf{/}\mathbf{\left(}\mathbf{⟨}\mathbf{6}\mathbf{,}\mathbf{9}\mathbf{⟩}\mathbf{\right)}$

Short answer

The element of finite order is $\left(2,3\right)$

The element of infinite order is $\left(1,0\right)$

Step-by-step solution

Finding a finite element in (ℤ×ℤ)/(⟨6,9⟩)

An element $(a,b)\in (\mathbb{Z}\times \mathbb{Z})/⟨6,9⟩$ can only be finite if, for some $(\mathrm{a},\mathrm{b})\in (\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})$, $\mathrm{n}(\mathrm{a},\mathrm{b})\in \left(⟨6,9⟩\right)$.

For example, consider an element $(2,3)$; we get:

$3(2,3)\in \left(⟨6,9⟩\right)$

Therefore,$(2,3)$ is a finite element of $(\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})/⟨6,9⟩$.

Finding an infinite element in (ℤ×ℤ)/(⟨6,9⟩)

As we know, an element $(\mathrm{a},\mathrm{b})\in (\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})/⟨6,9⟩$ can only be finite if $(a,b)\in (\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})$, $n(a,b)\in \left(⟨6,9⟩\right)$.

Since $n(1,0)\notin ⟨6,9⟩$for any value of n, so, it is an infinite element.

Therefore, the finite element is $(2,3)$ and the infinite element is $(1,0)$.