For the group of order 117 you can take a subgroup H of order 3 and a subgroup K of order 13. The one with order 13 happens to be normal by showing the number of Sylow 13-subgroups is one. Since |H|,|K| are relatively prime it means |HK| = |H||K| = 39, furthermore HK is a subgroup since K is a normal subgroup. Thus, HK is a subgroup of index p=3, where p is the smallest prime dividing |G|, thus HK is a normal subgroup. Finally, HK is isomorphic to H x K which is a cyclic group since H and K are cyclic groups of relative prime orders.

For the group of order 118 you can take a subgroup H of order 59, it happens to be normal again by using Sylow theorems. It is cyclic since H it is prime order. And finally it is an index 2 subgroup.