# How to solve this? We have the ring (ZZ_8,+,*)Demonstrate that the invertible elements of this ring are:hat1,hat3,hat5,hat7.

Apr 11, 2017

We have to prove that these elements are the only ones with an inverse

#### Explanation:

The elements of ${\mathbb{Z}}_{8}$ are $\hat{0}$, $\hat{1}$, $\hat{2}$, $\hat{3}$. $\hat{4}$, $\hat{5}$, $\hat{6}$ and $\hat{7}$. Now:

The inverse of $\hat{1}$ is the same $\hat{1}$ since $\hat{1}$ * $\hat{1}$ = $\hat{1}$. Similarly:

The inverse of $\hat{3}$ is the same $\hat{3}$ since $\hat{3}$ * $\hat{3}$ = $\hat{1}$

The inverse of $\hat{5}$ is the same $\hat{5}$ since $\hat{5}$ * $\hat{5}$ = $\hat{1}$

The inverse of $\hat{7}$ is the same $\hat{7}$ since $\hat{7}$ * $\hat{7}$ = $\hat{1}$. So these elements are invertible.

Now, let's consider $\hat{2}$. The possible products are:

$\hat{2}$ * $\hat{0}$ = $\hat{0}$

$\hat{2}$ * $\hat{1}$ = $\hat{2}$

$\hat{2}$ * $\hat{2}$ = $\hat{4}$

$\hat{2}$ * $\hat{3}$ = $\hat{6}$

$\hat{2}$ * $\hat{4}$ = $\hat{0}$

$\hat{2}$ * $\hat{5}$ = $\hat{2}$

$\hat{2}$ * $\hat{6}$ = $\hat{4}$

$\hat{2}$ * $\hat{7}$ = $\hat{6}$

We can see that no product gives us $\hat{1}$. The same way we can prove that $\hat{4}$ and $\hat{6}$ have no inverses. By definition $\hat{0}$ is not invertible.

Therefore, the only invertible elements are $\hat{1}$, $\hat{3}$, $\hat{5}$ and $\hat{7}$

In general terms, invertible elements in these rings are the ones coprime with the order (8 in this case) of the ring