Prove $\left \langle n_{1}, n_{2},...,n_{k} \right \rangle =\left\langle (n_{1},n_{2},...,n_{k})\right \rangle$ in $\mathbb{Z}$ Prove that in $\mathbb{Z}$, $\left \langle n_{1}, n_{2},...,n_{k} \right \rangle =\left\langle (n_{1},n_{2},...,n_{k})\right \rangle$.
If somebody could explain to me what exactly the problem is asking. Is it talking about the cyclic subgroup of $\mathbb{Z}$? 

Probably $\langle n_1,\dots,n_k\rangle$ denotes the subgroup of $\mathbb{Z}$ generated by $n_1,\dots,n_k$, or in other words the set of all sums $x_1 n_1 + \cdots + x_k n_k$, where $x_1,\dots,x_k\in\mathbb{Z}$, while $(n_1,\dots,n_k)$ denotes the greatest common divisor of $n_1,\dots,n_k$, often denoted $\gcd(n_1,\dots,n_k)$ (or $\text{hcf}(n_1,\dots,n_k)$) precisely in order to avoid confusion of this sort, so $\langle(n_1,\dots,n_k)\rangle$ is the subgroup of $\mathbb{Z}$ generated by the single element $(n_1,\dots,n_k)$, or the set of all multiplies of $(n_1,\dots,n_k)$.
