Why is $\left(\mathbb{Z}_{51}\right)^* \cong \mathbb{Z}_2 \times \mathbb{Z}_{16}$?

Question

I have to show that $\left(\mathbb{Z}_{51}\right)^* \cong \mathbb{Z}_2 \times \mathbb{Z}_{16}$.

I know that $\mathbb{Z}_{51}\cong\mathbb{Z}_3 \times \mathbb{Z}_{17}$ and that $(\mathbb{Z}_p)^*\cong \mathbb{Z}_{p-1}$, but does it hold that $(\mathbb{Z}_p \times \mathbb{Z}_{q})^*\cong \mathbb{Z}_p^* \times \mathbb{Z}_{q}^*$

Else how would i prove it?

Answer 1

If $A$ and $B$ are two different ring then $(A\times B)^*$ is indeed isomorphic to $A^*\times B^*$, via the canonical identity map.

$i)$ If $(a,b)\in A\times B$ is invertible, there exists $(c,d)\in A\times B$ such that $(a,b)(c,d)=(1,1)$. This implies that $ac=1$ in $A$ and $bd=1$ in $B$. Hence, $(a,b)\in A^*\times B^*$.

$ii)$ If $(a,b)\in A^*\times B^*$, then $a\in A^*$ and $b\in B^*$ which implies the existence of $(c,d)\in A\times B$ with $(a,b)(c,d)=(1,1)$.