The cyclic group Z8(integer modulo 8) has two nontrivial subgroups.

Let G be a cyclic group Z8 under group addition, G={0,2,4,6,1,3,5,7}. Then the subgroups of G is can be formed by the Cayley table:



For 0
(0,0)=0+0=0∈Z8
(0,2)=0+2=2∈Z8
(0,4)=0+4=4∈Z8
(0,6)=0+6=6∈Z8
(0,1)=0+1=1∈Z8
(0,3)=0+3=3∈Z8
(0,5)=0+5=5∈Z8
(0,7)=0+7=7∈Z8
For 2
(2,0)=2+0=2∈Z8
(2,2)=2+2=4∈Z8
(2,4)=2+4=6∈Z8
(2,6)=2+6=0∈Z8
(2,1)=2+1=3∈Z8
(2,3)=2+3=5∈Z8
(2,5)=2+5=7∈Z8
(2,7)=2+7=1∈Z8
For 4
(4,0)=4+0=4∈Z8
(4,2)=4+2=6∈Z8
(4,4)=4+4=0∈Z8
(4,6)=4+6=2∈Z8
(4,1)=4+1=5∈Z8
(4,3)=4+3=7∈Z8
(4,5)=4+5=1∈Z8
(4,7)=4+7=3∈Z8
For 6
(6,0)=6+0=6∈Z8
(6,2)=6+2=0∈Z8
(6,4)=6+4=2∈Z8
(6,6)=6+6=4∈Z8
(6,1)=6+1=7∈Z8
(6,3)=6+3=1∈Z8
(6,5)=6+5=3∈Z8
(6,7)=6+7=5∈Z8
For 1
(1,0)=1+0=1∈Z8
(1,2)=1+2=3∈Z8
(1,4)=1+4=5∈Z8
(1,6)=1+6=7∈Z8
(1,1)=1+1=2∈Z8
(1,3)=1+3=4∈Z8
(1,5)=1+5=6∈Z8
(1,7)=1+7=0∈Z8
For 3
(3,0)=3+0=3∈Z8
(3,2)=3+2=5∈Z8
(3,4)=3+4=7∈Z8
(3,6)=3+6=1∈Z8
(3,1)=3+1=4∈Z8
(3,3)=3+3=6∈Z8
(3,5)=3+5=0∈Z8
(3,7)=3+7=2∈Z8
For 5
(5,0)=5+0=5∈Z8
(5,2)=5+2=7∈Z8
(5,4)=5+4=1∈Z8
(5,6)=5+6=3∈Z8
(5,1)=5+1=6∈Z8
(5,3)=5+3=0∈Z8
(5,5)=5+5=2∈Z8
(5,7)=5+7=5∈Z8
For 7
(7,0)=7+0=7∈Z8
(7,2)=7+2=1∈Z8
(7,4)=7+4=3∈Z8
(7,6)=7+6=5∈Z8
(7,1)=7+1=0∈Z8
(7,3)=7+3=2∈Z8
(7,5)=7+5=4∈Z8
(7,7)=7+7=6∈Z8
Hence we can see that group has two nontrivial subgroups which are 
  1.  J={0,4}, 
  2.  H={0,2,4,6},
 where J is also a subgroup of H.
The Cayley table for H is the top-left quadrant of the Cayley table for G. The group G is cyclic, and so are its subgroups. In general, subgroups of cyclic groups are also cyclic.