under Z*32, consider H={1,17} and K={1,15}

consider Z*32/H and Z*32/K

Z*32/H = {H, 3H, ...}

Note that |Z*32| = phi(32) = 16 and |H| = 2 

thus |Z*32/H| = 8

there are abelian groups of order 8 as Z8 or Z4xZ2 or Z2xZ2xZ2

|3H| = 4 since 3H 9H 27H 81H = 17H

thus Z*32/H is not Z2xZ2xZ2

|5H| = 4 since 5H 25H=-7H -35H=-3H -15H=17H 

Note that if Z*32/H is cyclic then there is a unique cyclic subgroup of order of 4.

but <3H> != <5H> 

thus Z*32/H = Z4xZ2

and |Z*32/K| = 8

similarly,

|3K| = 4 since 3K 9K 27K 81K=17K

|5K| = 8 since 

5K 

25K 

125K=-3K 

-15K 

-75K=21K

105K=9K

45K=13K

65K=K

thus Z*32/K = Z8