permutation group Given: Suppose G= D6, the group of all symmetries of a regular hexagon. 
D6={e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b}
Question: Find and list all the distinct subgroups of D6. Explain how you know that you have found all of the subgroups, and how you know that all examples found are subgroups
Answer: I am asking for help to explaining how to find all of the subgroups....like this is so far what i have. This group has order 12; |D6| = 12, so the possible orders of these cyclic subgroups are    1,2,3,4,6,12. Now I am at point where i need to explain for each order, and i cant "describe" in words. like i know how to find it but putting into words hard
Could anyone help me with explaining??
 A: There are unique subgroups of orders $1$ and $12$, of course. 
For order $2$, such groups must be cyclic, so you can simply list all elements of order $2$.
Similarly for order $3$; each is cyclic, so list all elements of order $3$ and determine which ones generate the same subgroup of order $3$.
For order $4$, all groups of order $4$ are isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}/2\oplus\mathbb{Z}/2$. By inspection, there are no elements of order $4$, so you only need to consider the second case. Such a group must be generated by two elements of order $2$, and they have to commute. So which elements of order $2$ commute? From here, you can find a list.
For order $6$, it sounds like you know all of the subgroups of order $6$ and need to argue that there are no others. Note that a subgroup of order $6$ cannot itself have a subgroup of order $4$, so you can use your list of subgroups of order $4$ to indicate elements that a subgroup of order $6$ cannot contain. This lowers it down enough so that you can conclude there are no other subgroups of order $6$.
