Abelian group G is of order $240$, $sigma(g)=g^{16}in text{Aut}(G)$, then #{ker $sigma$} =?
0
$begingroup$
I'm told to determine if this is true: If $G$ is an abelian group of order $240$ . Then the kernel of the map $Gto G$ given by $sigma(g)=g^{16}$ has 16 elements. Since the group is not determined and there are $5$ abelian groups of order $240$ up to isomorphism. They are $mathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ , $mathbb{Z}_4bigotimesmathbb{Z}_2bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ , $mathbb{Z}_8bigotimesmathbb{Z}_2bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ , $mathbb{Z}_{16}bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ , $mathbb{Z}_4bigotimesmathbb{Z}_4bigotimesmathbb{Z}_3bigotimesmathbb{Z}_5$ . I'm not sure if I need to discuss each case.