How to find the kernel of $\sigma: \Bbb Z_{17}\to G $

Question

Let G be a group,$ e_G$ is the unit of G and $\sigma: \Bbb Z_{17}\to G $ is a homomorphism which is not injective. so find $Ker(\sigma)$

it is probably wrong but we know $Ker(\sigma)\neq\{\bar 0\}$ and $Ker(\sigma) \triangleleft \Bbb Z_{17}$. since 17 is prime $Z_{17} $ has only two subgroup which are $\{\bar 0\} $ and $\Bbb Z_{17}$ so $Ker(\sigma) $ should equal to $\Bbb Z_{17}$

Answer 1

you are right.the kernel should equal to $Z_{17}$