Consider the homomorphism from A5 to Z_60. Show that the kernel is equal to A5.
I know that A5 is simple, and thus it has no non-trivial, proper subgroups. So the kernel must either be {e} or all of A5. But how do I show it's equal to A5/not equal to the trivial subgroup?
If the kernel were the trivial subgroup then we would have $A_5 \cong Z_{60}$ (by the first isomorphism theorem and the fact that they have the same cardinality).