Let $a\mathbf{Z}=\{ax:x\in\mathbf{Z}\}$ denote the set of all multiples of $a$.Prove that for any integers $a_1,\cdots,a_k$,
\begin{equation}
\bigcap_{i=1}^ka_i\mathbf{Z}=[a_1,\cdots,a_k]\mathbf{Z}
\end{equation}
Remark1:The proof is simple.I only want to point out that $$
\bigcap_{i=1}^ka_i\mathbf{Z} $$ is a subgroup of the cyclic group $\mathbf{Z}$,we find its generator $[a_1,\cdots,a_k]$.
Remark2:It not true that
\begin{equation}
\bigcup_{i=1}^ka_i\mathbf{Z}=(a_1,\cdots,a_k)\mathbf{Z}
\end{equation}