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If a,b is in (Z/nZ)*, then ab is in (Z/nZ)*

If a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}, then a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}.

Proof. By definition, if a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}, then 
\begin{align*}
ac &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } c \in \mathbb{Z}/n\mathbb{Z} \\
bd &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } d \in \mathbb{Z}/n\mathbb{Z}.
\end{align*}
Obviously, we will get 
\begin{align*}
1 = 1 \cdot 1 = ac \cdot bd = ab \cdot cd, cd \in \mathbb{Z}/n\mathbb{Z},
\end{align*}
which implies that ab \in (\mathbb{Z}/n\mathbb{Z})^{\times}
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