- #1

cbarker1

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MHB

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$\newcommand{\Z}{\mathbb{Z}}$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.

**\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:

**,[c]\in{(\Z/n\Z)}^{\times}$ . We know that the $\Z$ is associative.**

$[a]\cdot($[a]\cdot(

**\cdot[c])=[a] \cdot [bc]=[abc]=[ab] \cdot [c]=([a]\cdot****) \cdot [c]$.**

$[1]\cdot[x]=[1\cdot x]= [x], \forall [x] \in {(\Z/n\Z)}^{\times}$

$[x] \cdot [1]=[x \cdot 1]=[x], \forall [x] \in {(\Z/n\Z)}^{\times}$

Thanks,

Cbarker1- Identity

$[1]\cdot[x]=[1\cdot x]= [x], \forall [x] \in {(\Z/n\Z)}^{\times}$

$[x] \cdot [1]=[x \cdot 1]=[x], \forall [x] \in {(\Z/n\Z)}^{\times}$

- inverse

Thanks,

Cbarker1