Question

In Exercise 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.$\mathbf{(}\mathbf{\mathbb{Z}}\mathbf{\times }\mathbf{\mathbb{Z}}\mathbf{)}\mathbf{/}\mathbf{\left(}\mathbf{⟨}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{⟩}\mathbf{\right)}$

Short answer

The element of finite order is $(1,1)$

The element of infinite order is $(1,\text{\hspace{0.17em}0})$

Step-by-step solution

Finding a finite element in (ℤ×ℤ)/(⟨5,5⟩)

An element $(a,b)\in (\mathbb{Z}\times \mathbb{Z})/⟨5,5⟩$can only be finite if for some $(a,b)\in (\mathbb{Z}\times \mathbb{Z})$, $n(a,b)\in \left(⟨5,5⟩\right)$.

For example, consider an element $(1,1)$. We get:

$5(1,1)\in \left(⟨5,5⟩\right)$

Therefore, $(1,1)$is a finite element of $(\mathbb{Z}\times \mathbb{Z})/⟨5,5⟩$.

Finding an infinite element in (ℤ×ℤ)/(⟨5,5⟩)

As we know,

An element $(\mathrm{a},\mathrm{b})\in (\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})/⟨5,5⟩$can only be finite if $(\mathrm{a},\mathrm{b})\in (\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}})$, $n(a,b)\in \left(⟨5,5⟩\right)$

Since $n(1,0)\notin ⟨5,5⟩$ for any value of n, it is an infinite element.

Therefore, an example of a finite element is $(1,1)$ , and an example of an infinite element is $(1,0)$.