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}