The cyclic group $Z_8$(integer modulo 8) has two nontrivial subgroups.

Let $G$ be a cyclic group $Z_8$ 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 $0$
$(0,0)=0+0=0\in{Z_8}$
$(0,2)=0+2=2\in{Z_8}$
$(0,4)=0+4=4\in{Z_8}$
$(0,6)=0+6=6\in{Z_8}$
$(0,1)=0+1=1\in{Z_8}$
$(0,3)=0+3=3\in{Z_8}$
$(0,5)=0+5=5\in{Z_8}$
$(0,7)=0+7=7\in{Z_8}$
For $2$
$(2,0)=2+0=2\in{Z_8}$
$(2,2)=2+2=4\in{Z_8}$
$(2,4)=2+4=6\in{Z_8}$
$(2,6)=2+6=0\in{Z_8}$
$(2,1)=2+1=3\in{Z_8}$
$(2,3)=2+3=5\in{Z_8}$
$(2,5)=2+5=7\in{Z_8}$
$(2,7)=2+7=1\in{Z_8}$
For $4$
$(4,0)=4+0=4\in{Z_8}$
$(4,2)=4+2=6\in{Z_8}$
$(4,4)=4+4=0\in{Z_8}$
$(4,6)=4+6=2\in{Z_8}$
$(4,1)=4+1=5\in{Z_8}$
$(4,3)=4+3=7\in{Z_8}$
$(4,5)=4+5=1\in{Z_8}$
$(4,7)=4+7=3\in{Z_8}$
For $6$
$(6,0)=6+0=6\in{Z_8}$
$(6,2)=6+2=0\in{Z_8}$
$(6,4)=6+4=2\in{Z_8}$
$(6,6)=6+6=4\in{Z_8}$
$(6,1)=6+1=7\in{Z_8}$
$(6,3)=6+3=1\in{Z_8}$
$(6,5)=6+5=3\in{Z_8}$
$(6,7)=6+7=5\in{Z_8}$
For $1$
$(1,0)=1+0=1\in{Z_8}$
$(1,2)=1+2=3\in{Z_8}$
$(1,4)=1+4=5\in{Z_8}$
$(1,6)=1+6=7\in{Z_8}$
$(1,1)=1+1=2\in{Z_8}$
$(1,3)=1+3=4\in{Z_8}$
$(1,5)=1+5=6\in{Z_8}$
$(1,7)=1+7=0\in{Z_8}$
For $3$
$(3,0)=3+0=3\in{Z_8}$
$(3,2)=3+2=5\in{Z_8}$
$(3,4)=3+4=7\in{Z_8}$
$(3,6)=3+6=1\in{Z_8}$
$(3,1)=3+1=4\in{Z_8}$
$(3,3)=3+3=6\in{Z_8}$
$(3,5)=3+5=0\in{Z_8}$
$(3,7)=3+7=2\in{Z_8}$
For $5$
$(5,0)=5+0=5\in{Z_8}$
$(5,2)=5+2=7\in{Z_8}$
$(5,4)=5+4=1\in{Z_8}$
$(5,6)=5+6=3\in{Z_8}$
$(5,1)=5+1=6\in{Z_8}$
$(5,3)=5+3=0\in{Z_8}$
$(5,5)=5+5=2\in{Z_8}$
$(5,7)=5+7=5\in{Z_8}$
For $7$
$(7,0)=7+0=7\in{Z_8}$
$(7,2)=7+2=1\in{Z_8}$
$(7,4)=7+4=3\in{Z_8}$
$(7,6)=7+6=5\in{Z_8}$
$(7,1)=7+1=0\in{Z_8}$
$(7,3)=7+3=2\in{Z_8}$
$(7,5)=7+5=4\in{Z_8}$
$(7,7)=7+7=6\in{Z_8}$
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.