This is a neat question [there is a fast solution]:
Classify, up to isomorphism, groups of order 77.
My attempted solution is as follows:
1.$\displaystyle Z_{77} = Z_{7} \times Z_{11}$.
Since 11 is not congruent to 1modulo7, it has one Sylow 7-subgroup and one Sylow 11-subgroup, which has an intersection at {e}. Thus, the group of order 77 is the direct product of a normal Sylow 7-subgroup and a normal Sylow 11-subgroup, which is cyclic.
I think there is only one group of order 77 up to isomorphism.