How do I prove that $\mathbb{Z}_n^*$ is a group?

Question

$\mathbb{Z}_{n}^{*}=\left \{ a\Big|\gcd\left (a,n \right )=1 \right \}$
Now, my problem is that:
If $a_1,a_2\in \mathbb{Z}_{n}^{*}$, how do I prove that $a_1\cdot a_2\in \mathbb{Z}_{n}^{*}$?

Can you help me and give me a hint how I need to prove it?

Thank you!

Answer 1

If $a_1\times a_2\notin\mathbb{Z}_n^*$, then $\gcd(a_1\times a_2,n)>1$. If $p$ is a prime factor of $\gcd(a_1\times a_2,n)$, then $p\mid a_1\times a_2$. What can you deduce from the fact that a prime number divides a product?

Answer 2

Hint: Since $\gcd\left (a_1,n \right )=1, \gcd\left (a_2,n \right )=1$, we have $a_1x_1+ny_1=1$ and $a_2x_2+ny_2=1$. Now $(a_1x_1)(a_2x_2)=(1-ny_1)(1-ny_2)=1+n (ny_1y_2-y_1-y_2). $ Rearrange the equation and conclude.

Answer 3

Alternative answer:

$\mathbb Z_n^*$ is the set of elements of $\mathbb Z_n$ that have multiplicative inverses.

If $a_1, a_2 \in \mathbb Z_n^*,$ then $\exists \,a_1^{-1}, a_2^{-1} \in \mathbb Z_n^*$ such that $a_1a_1^{-1}=1$ and $a_2a_2^{-1}=1$.

Note $a_1a_1^{-1}a_2a_2^{-1}=1=a_1a_2a_1^{-1}a_2^{-1}$ so $a_1a_2, a_1^{-1} a_2^{-1} = (a_1 a_2)^{-1} \in \mathbb Z_n^*.$