MHB Proof the set with the multiplication is a group

[cbarker1]

Dear Everyone,

$\newcommand{\Z}{\mathbb{Z}}$Suppose the set is defined as:
$\begin{equation*}
{(\Z/n\Z)}^{\times}=\left\{\bar{a}\in \Z/n\Z|\ \text{there exists a}\ \bar{c}\in \Z/n\Z\ \text{with}\ \bar{a}\cdot\bar{c}=1\right\}
\end{equation*}$
for $n>1$
I am having some trouble
Proving that ${(\Z/n\Z)}^{\times}$ is an abelian group under multiplication on ${(\Z/n\Z)}$.

My Attempt:

• WTS: Multiplication on ${(\Z/n\Z)}^{\times}$ is closed.

By exercise 5 in homework 2, if $[a], \in {(\Z/n\Z)}^{\times}$, then $[a]\cdot \in {(\Z/n\Z)}^{\times}$. So we know that multiplication is closed under ${(\Z/n\Z)}^{\times}$.

• Associativity:

Let $[a],,[c]\in{(\Z/n\Z)}^{\times}$ . We know that the $\Z$ is associative.
$[a]\cdot(\cdot[c])=[a] \cdot [bc]=[abc]=[ab] \cdot [c]=([a]\cdot) \cdot [c]$.

• Identity

We know that $[1]$ is the identity for ${(\Z/n\Z)}^{\times}$.
$[1]\cdot[x]=[1\cdot x]= [x], \forall [x] \in {(\Z/n\Z)}^{\times}$
$[x] \cdot [1]=[x \cdot 1]=[x], \forall [x] \in {(\Z/n\Z)}^{\times}$

• inverse

We are given the right side inverse by the definition of the set. We need to show that the left side exist ( here is where I am having troubles).

Thanks,
Cbarker1

 

[I like Serena]

Hi Cbarker1,

Let's do commutativity first:
$$\forall [a],\in(\mathbb Z/n\mathbb Z)^\times:[a]\cdot=[ab]=[ba]=\cdot[a]$$

Now we can apply commutativity of multiplication to find the left side inverse.