Show $\mathbb{Z}_n^∗$ is an abelian group where $\mathbb{Z}_n^*=\{1 \le a \le (n-1)\mid \gcd(a,n) = 1\}$.

Question

I need to show that $\mathbb{Z}_n^∗$ is an abelian group where $\mathbb{Z}_n^*=\{1 \le a \le (n-1)\mid \gcd(a,n) = 1\}$.

I have tried specific examples for this such as $\mathbb{Z}_8$ and closure, identity, inverse, and associativity hold. But I do not know how to show this for any $n$.

I know that the Identity for any $n$ will be $1$ but I do not know how to show closure, inverse, and associativity.

Answer 1

If $R$ is a commutative ring with $1$ then the group of units $R^{\times}$ is always abelian. Here we have $R=\Bbb Z/n\Bbb Z$ and $$R^{\times}=U(n)=(\Bbb Z/n\Bbb Z)^{\times}=\Bbb Z_n^{\times}.$$ One can check that the units form a group and that one obtaines your group as the group of units from the commutative ring $\Bbb Z/n\Bbb Z$.

The converse of the above need not be true. There are non-commutative rings with an abelian group of units: Noncommutative rings with abelian group of units, e.g., the polynomial ring $K[X,Y]$ over a field $K$.

Answer 2

Hints:

• First define the group law in this set (starting point: multiplication in $\mathbf Z$).
• For the inverse, use a Bézout's relation $\;ua+vn=1$ and translate it modulo $n$.
• Associativity should be deduced from associativity of multiplication in $\mathbf Z$.

Answer 3

Well-defined: Suppose $a\equiv b\pmod{n}$. Then $n\mid a-b$, i.e., $a-b=kn$ for some $k\in\Bbb Z$. So $a=kn+b$. Hence $\gcd(a, n)=\gcd(b, n)$.

Closure: Let $[g]_n, [h]_n\in\Bbb Z_n^*$. Then $\gcd(g, n)=1=\gcd(h,n)$. Thus $\gcd(gh, n)=1$. (Hint: Use Bézout's Identity.)

Associativity: This follows from that of $\Bbb Z$ under multiplication.

Identity: The identity is $[1]_n$. Since multiplication is well-defined, it suffices to show that $[1\times g]_n=[g]_n$, which is quite clear to see.

Inverse: Suppose $[g]_n\in\Bbb Z_n^*$. Then, by Bézout's Identity, there exist $x, y\in \Bbb Z$ such that $xg+yn=1$; that is, $xg=1-yn$; so $[x]_n[g]_n=[1]_n$. Thus $[g]_n^{-1}=[x]_n$.

Commutativity: This follows from commutativity of $\Bbb Z$ under multiplication.

Hence $\Bbb Z_n^*$ is an abelian group.

Answer 4

You mean eqivalence classes modulo $n$, not integers themselves. The identity element is $1+Z_n$Abelianness and associativity are inherited from the integers.Note that if $a \equiv b \mod n$ then $$\gcd(a,n)=1 $$ iff $$\gcd(b,n)=1$$ so it is legitimate to write $$\gcd((a+Z_n),n)=1$$. If $\gcd(a,n)=1$ and $gcd(\alpha,n)=1$, then, by the uniqueness of prime decomposition,$ gcd(a\alpha,n)=1 $, which shows that the set of equivalence classes $\{a+Z_n|\gcd(a,n)=1 \} $ is stable under multiplication.Suppose that $\gcd(a,n)=1, 1 \le a \le n-1$. Then there exist integers $u$ and $v$ such that $$ua+vn=1$$ Then $\gcd(u,n)=1$. Let $$U \equiv u \mod n, 1 \le U \le n-1$$ Thus $$u+Z_n=U+Z_n$$ is the multiplicative inverse of $a+Z_n$