How many cyclic subgroups are in the Dihedral group? let $D_8$ be the group of symmetries of a square. Not counting the trivial subgroup, how many distinct cyclic subgroup s does $D_8$ contain? 
My result:
A group $D_8$, contains an element $a$ of order 4.
$<{a}> =(a^4:4\in Z)$
$(1.a,a^2,a^3,x,b,c,d)$ the reflections and symmetries in $D_8$.
$(1)(a)=a
\\
(a)(a)=a^2
\\
(a)(a^2)=a^3
\\
(a)(a^3)=a^4=1$  
$(1,a,a^2,a^3)$
Did I answer this correctly? If yes can someone clarify my writing or if it's wrong please guide me ...
 A: You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$$\{1, a^2\}$$
Four additional cyclic subgroups of order two are as follows:
$$\{1, x\}, \{1, b\}, \{1, c\}, \{1, d\}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: $\{1\}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
A: Your notation is not very clear, so I don't quite understand your answer. Note that your answer should be a natural number: the number of distinct cyclic subgroups of $D_8$ other than the trivial subgroup. 
To find out what that number is, you can just go over each and every element of $D_8$ and check what it generates. That will give you a list of all the cyclic subgroups. You then identify how many distinct ones there are, and that will be your answer. 
Note that $D_8$ only has $8$ elements so this is certainly easy enough to just do (also note that some texts call your $D_8$ by $D_4$).
A: No. of cyclic subgroups of $D_n $ ( order of $D_n$ is $2n$)  is 
$\tau{(n)} +$ $n$. Where $\tau(n)$ is number of positive divisor of $n$.
And total number of subgroups of $D_n$ is $\tau(n) + \sigma(n)$ where $\sigma(n)$ is sum of all positive divisor.
A: Below, we can see the subgroups concrete. It's done by GAP:
gap> LoadPackage("sonata");;
      f:=FreeGroup("a","b");;
      a:=f.1;;   b:=f.2;;
      s:=f/[a^2,b^4,(a*b)^2];
      e:=Subgroups(s);;
      for i in [1..8] do Print(e[i],"  ",IsCyclic(e[i]),"\n"); od;


   Group( <identity ...> )  true
   Group( [ b^2 ] )  true
   Group( [ a ] )  true
   Group( [ b^-1*a*b ] )  true
   Group( [ a*b ] )  true
   Group( [ b*a ] )  true
   Group( [ a, b^-2 ] )  false
   Group( [ b ] )  true

A: $D_{2n} = \{ s^i r^j / i \in \{0,1\}, j \in \{0,1,\cdots,n-1\}  \}$.
You can show that $|sr^j| = 2$ for each $j$, so $<sr^j>$ is a cyclic group.
On the other hand, $|r| = n$ then $r^j$ is a generator of the group $<r>$ if $(j,n)=1$, ie., $r^j$ generates another cyclic group if $j|n$. The number of divisors of $n$ is denoted by $\tau(n)$.
Therefore, there are $\tau(n)+n$ cyclic subgroups.
