Finding subgroups of $D_8$ $$D_8=\{(),(1234),(13)(24),(1432),(13),(24),(14)(23),(12)(34) \}$$
Am I right to say that to find the subgroups, we have to make sure that the identity can be generated or is in the group and the inverse of each element can be generated or is in the group?
Is there an easy way to do this though because it seems really time consuming.
 A: Think of $D_8$ geometrically - it's the symmetries of a square.
Now, for something to be a "subgroup" of a group, it should be closed under group operations. Now, what does this mean? It means that the subset of $D_8$ that you choose of the group must also form a group.
Now, we can make use of a couple of theorems to figure out the subgroup structure of $D_8$


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*Lagrange's theorem - the order of a subgroup must divide the order of the entire group. Since $D_8$ has 8 elements, we know that the subgroups can be of order $1$, $2$, $4$ or $8$.


Clearly, the subgroup of order $1$ is the trivial group $\{e\}$, and the subgroup of order $8$ is the entire group $D_8$.
Hence, the subgroups we need to check for are those of order $2$ and $4$. We have a complete classification of the groups of order $2$ and $4$.
We know that the only group of order $2$ is $\mathbb{Z} / 2\mathbb{Z}$.
The groups of order $4$ are the cyclic group $\mathbb{Z} / 4\mathbb{Z}$ and the Klein-$4$ group.
Since we know the different "types" of subgroups we can have, we can now hunt for the subgroups in the dihedral group.
To actually look for the subgroups, use the geometric information of $D_8$. We know that the elements are rotations and a reflection, along with the fact that the rotations form a cyclic subgroup.
Hence, by using both group classification and the "multiplication table" / "geometric information" of $D_8$, we can build up the entire picture.
A: Yes, you need to make sure the identity can be generated and the inverse is in the group.
