On $\mathbb{Z}_n^*$ group
- Tolaso J Kos
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On $\mathbb{Z}_n^*$ group
Prove that $(\mathbb{Z}_n^*, \otimes)$ is a group if and only if $n$ is prime.
Imagination is much more important than knowledge.
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Re: On $\mathbb{Z}_n^*$ group
Suppose that \(\displaystyle{\left(\mathbb{Z}_{n}^{\star},\cdot\right)}\) is group.
If \(\displaystyle{n}\) is not a prime number, then \(\displaystyle{n=a\,b\,,1<a\,,b<n}\) . Therefore,
\(\displaystyle{[n]=[a\,b]\implies [0]=[a]\,\implies [a]^{-1}\,[0]=[a]^{-1}\,[a]\,\implies [0]=}\)
so \(\displaystyle{n\mid b\implies n\leq b}\), a contradiction. So, \(\displaystyle{n}\) is a prime number.
Now, suppose that \(\displaystyle{n=p}\) is a prime number.
Let \(\displaystyle{[a]\,,\in\mathbb{Z}_{p}^{\star}=\mathbb{Z}_{p}-\left\{[0]\right\}}\) .
If \(\displaystyle{[a]\,=[0]\iff [a\,b]=[0]}\), then \(\displaystyle{p\mid a\,b}\) and since
\(\displaystyle{p}\) is a prime number, we get \(\displaystyle{p\mid a}\) or \(\displaystyle{p\mid b}\)
that is \(\displaystyle{[a]=[0]}\), or \(\displaystyle{=[0]}\) , a contradiction.
We conclude that the multiplication is a well defined map on \(\displaystyle{\mathbb{Z}_{p}^{\star}}\) .
The multiplication is associative and commutative and
\(\displaystyle{[a]\cdot [1]=[a]=[1]\cdot [a]\,,\forall\,[a]\in\mathbb{Z}_{p}^{\star}}\), which means that
the element \(\displaystyle{[1]}\) is an identity element.
Finally, if \(\displaystyle{[a]\in\mathbb{Z}_{p}^{\star}}\), then :
\(\displaystyle{\begin{aligned} (a,p)=1&\implies (\exists\,x\in\mathbb{Z})(\exists\,y\in\mathbb{Z}), a\,x+p\,y=1\\&\implies [a\,x+p\,y]=[1]\\&\implies [a]\,[x]=[1]=[x]\,[a]\end{aligned}}\)
so, each element \(\displaystyle{[a]\in\mathbb{Z}_{p}^{\star}}\) is invertible and the pair
\(\displaystyle{\left(\mathbb{Z}_{p}^{\star},\cdot\right)}\) is an abelian group of order \(\displaystyle{p-1}\) .
By this way, you have also that the ring \(\displaystyle{\left(\mathbb{Z}_{n},+,\cdot\right)}\)
is a field if, and only if, \(\displaystyle{n}\) is a prime natural number.
If \(\displaystyle{n}\) is not a prime number, then \(\displaystyle{n=a\,b\,,1<a\,,b<n}\) . Therefore,
\(\displaystyle{[n]=[a\,b]\implies [0]=[a]\,\implies [a]^{-1}\,[0]=[a]^{-1}\,[a]\,\implies [0]=}\)
so \(\displaystyle{n\mid b\implies n\leq b}\), a contradiction. So, \(\displaystyle{n}\) is a prime number.
Now, suppose that \(\displaystyle{n=p}\) is a prime number.
Let \(\displaystyle{[a]\,,\in\mathbb{Z}_{p}^{\star}=\mathbb{Z}_{p}-\left\{[0]\right\}}\) .
If \(\displaystyle{[a]\,=[0]\iff [a\,b]=[0]}\), then \(\displaystyle{p\mid a\,b}\) and since
\(\displaystyle{p}\) is a prime number, we get \(\displaystyle{p\mid a}\) or \(\displaystyle{p\mid b}\)
that is \(\displaystyle{[a]=[0]}\), or \(\displaystyle{=[0]}\) , a contradiction.
We conclude that the multiplication is a well defined map on \(\displaystyle{\mathbb{Z}_{p}^{\star}}\) .
The multiplication is associative and commutative and
\(\displaystyle{[a]\cdot [1]=[a]=[1]\cdot [a]\,,\forall\,[a]\in\mathbb{Z}_{p}^{\star}}\), which means that
the element \(\displaystyle{[1]}\) is an identity element.
Finally, if \(\displaystyle{[a]\in\mathbb{Z}_{p}^{\star}}\), then :
\(\displaystyle{\begin{aligned} (a,p)=1&\implies (\exists\,x\in\mathbb{Z})(\exists\,y\in\mathbb{Z}), a\,x+p\,y=1\\&\implies [a\,x+p\,y]=[1]\\&\implies [a]\,[x]=[1]=[x]\,[a]\end{aligned}}\)
so, each element \(\displaystyle{[a]\in\mathbb{Z}_{p}^{\star}}\) is invertible and the pair
\(\displaystyle{\left(\mathbb{Z}_{p}^{\star},\cdot\right)}\) is an abelian group of order \(\displaystyle{p-1}\) .
By this way, you have also that the ring \(\displaystyle{\left(\mathbb{Z}_{n},+,\cdot\right)}\)
is a field if, and only if, \(\displaystyle{n}\) is a prime natural number.
- Grigorios Kostakos
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Re: On $\mathbb{Z}_n^*$ group
Two remarks:
1) In order to be the finite set $\mathbb{Z}_{n}^{\ast}$ a group, it is sufficient to be proved that the set is closed under multiplication.
2) If we use Fermat's little theorem we have the exact inverse of an element $[{\rm{k}}]_{{\rm{n}}}\in\mathbb{Z}_{n}^{\ast}$ :
Because for a prime number ${\rm{n}}$ , $(\forall\,\rm{k}\in\{{1,2,\ldots,{\rm{n}}-1}\})$ holds $\rm{k}^{{\rm{n}}-1}\equiv 1 \,{\rm{mod}}\, {\rm{n}}$, we have that
$$[{\rm{k}}^{{\rm{n}}-1}]_{{\rm{n}}}=[{1}]_{{\rm{n}}}\quad\Rightarrow\quad[{\rm{k}}^{{\rm{n}}-2}]_{{\rm{n}}}\otimes[\,{\rm{k}}]_{{\rm{n}}}=[{\rm{k}}]_{{\rm{n}}}\otimes[{\rm{k}}^{{\rm{n}}-2}]_{{\rm{n}}}=[{1}]_{{\rm{n}}}$$
1) In order to be the finite set $\mathbb{Z}_{n}^{\ast}$ a group, it is sufficient to be proved that the set is closed under multiplication.
2) If we use Fermat's little theorem we have the exact inverse of an element $[{\rm{k}}]_{{\rm{n}}}\in\mathbb{Z}_{n}^{\ast}$ :
Because for a prime number ${\rm{n}}$ , $(\forall\,\rm{k}\in\{{1,2,\ldots,{\rm{n}}-1}\})$ holds $\rm{k}^{{\rm{n}}-1}\equiv 1 \,{\rm{mod}}\, {\rm{n}}$, we have that
$$[{\rm{k}}^{{\rm{n}}-1}]_{{\rm{n}}}=[{1}]_{{\rm{n}}}\quad\Rightarrow\quad[{\rm{k}}^{{\rm{n}}-2}]_{{\rm{n}}}\otimes[\,{\rm{k}}]_{{\rm{n}}}=[{\rm{k}}]_{{\rm{n}}}\otimes[{\rm{k}}^{{\rm{n}}-2}]_{{\rm{n}}}=[{1}]_{{\rm{n}}}$$
Grigorios Kostakos
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