I need to determine the subgroups of the dihedral group of order 8, $D_4$.

Solution 1:

By Lagrange's Theorem, the possible orders are $1, 2, 4,$ and $8$.

The only subgroup of order $1$ is $\{1\}$ and the only subgroup of order $8$ is $D_4$.

If $D_4$ has an order $2$ subgroup, it must be isomorphic to $\mathbb{Z}_2$ (this is the only group of order $2$ up to isomorphism). Such a group is cyclic, it is generated by an element of order $2$. Are there any such elements in $D_4$?

If $D_4$ has an order $4$ subgroup, it must be isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2\times\mathbb{Z}_2$ (these are the only groups of order $4$ up to isomorphism). In the former case, the group is cyclic, it is generated by an element of order $4$. Are there any such elements in $D_4$? In the latter case, the group is generated by two commuting elements of order $2$. Are there any such pairs of elements in $D_4$?

In summary, first find all the elements of order $2$ and all the elements of order $4$; each of them generates a cyclic subgroup. Then consider pairs of elements of order $2$ to find which of them generate subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.

Solution 2:

The notes by K. Conrad have a nice answer: the dihedral group $D_n$ is generated by a rotation $r$ and a reflection $s$ subject to the relations $r^n=s^2=1$ and $(rs)^2=1$.

Proposition: Every subgroup of $D_n$ is cyclic or dihedral. A complete listing of the subgroups (including $1$ and $D_n$) is as follows:

$(1)$ $\langle r^d \rangle$ for all divisors $d\mid n$.
$(2)$ $\langle r^d,r^is \rangle$, where $d\mid n$ and $0\le i\le d-1 $.

Very nice pictures of the subgroup diagram of $D_4$ can be found here.