I'm trying to find all of the subgroups of the dihedral group, $D_4$, of the square. I will exhibit the group as follows. Let $r$ be a counterclockwise reflection through $\frac{\pi}{2}$ radians, identified with the permutation $(1234)$, and $s$ a reflection through vertices $1$ and $3$, identified with the permutation $(24)$. Then we can write: $$D_4 = \{e, r, r^2, r^3, s, sr, sr^2, sr^3 \}.$$ By some quick calculation, the orders are: \begin{align*} |e| & = 1 \\ |r| & = 4 \\ |r^2| & = 2 \\ |r^3| & = 4 \\ |s| & = 2 \\ |sr| & = 2 \\ |sr^2| & = 2 \\ |sr^3| & = 2. \end{align*} Here is my attempt.
Since $|D_4| = 8$, the only possible subgroups have order $1$, $2$, $4$, or $8$. The only subgroup of order $1$ is the trivial subgroup. The only subgroup of order $8$ is $D_4$, the improper subgroup. Any element of order $2$ generates a cyclic subgroup of order $2$ isomorphic to $\mathbb{Z}_2$. We have five elements of order $2$, and hence five subgroups of order $2$: \begin{align*} \{e,r^2\}, \{e,s\}, \{e,sr\}, \{e,sr^2\}, \{e,sr^3\}. \end{align*} Finally, we consider subgroups of order $4$, which are either isomorphic to the cyclic group of order $4$, $\mathbb{Z}_4$, or the Klein-4 group, $\mathbb{Z}_2 \times \mathbb{Z}_2$. The cyclic subgroup of order $4$ contains an element of order $4$, so the only candidates are $r^3$ and $r$. Of course, composing $r^3$ with itself successively gives only rotations, so the only cyclic subgroup of order $4$ is the rotations: $$\{r, r^2, r^3, r^4\}.$$ Further, the three-non identity elements of the Klein four-group are all of order two, commute with one another, and in the case where we are given two of them, composition produces the third. Notice, furthermore, that such a subgroup necessarily contains at least two reflections, and composing two reflections gives a rotation, so if such a subgroup exists, it must contain $r^2$, the only rotation of order $2$. (It will never generate the identity rotation $e$, since every reflection is its own inverse.) Therefore, $e$ and $r^2$ are necessarily in the subgroup, and of the four remaining elements of order two, we must select two more, of which there are $\binom{4}{2} = 6$ possibilities. Notice that if $sr$ and $sr^2$ are in the group, we have that $(sr)(sr^2) = r^{-1} \left(s^2\right)r^2 = r^3 e r^2 = r$ is an element of the subgroup, but $r$ has order $4$, so $sr, sr^2$ cannot both be in such a subgroup. If $s$ and $sr^3$ are in the same subgroup, then $s(sr^3) = (ssr)r^3 = er^3 = r^3$ is in the subgroup, but $r^3$ has order $4$, so $s$ and $sr^3$ can likewise not be in such a subgroup. If $sr^2$ and $sr^3$ are in the same subgroup, we have that $(sr^2)(sr^3) = r^{-2} (ss)r^3 = r^2 e r^3 = r$ is in the subgroup, but $r$ has order $4$. Similarly, if $s$ and $sr$ are in the same group, then $s(sr) = (ss)r = er = r$ is in the subgroup, but, again, $r$ has order $4$. This leaves as the only candidates $s$ and $sr^2$ and $sr$ and $sr^3$. We notice that pairs, when composed in other direction, give $r^2$. Therefore, our two subgroups of order $4$ that are isomorphic to the Klein four-group are \begin{align*} \{e, r^2, s, sr^2\}, \; \{sr, sr^3\}. \end{align*} Therefore, we have one subgroup of order $1$, five subgroups of order $2$, three subgroups of order $4$, and one subgroup of order $8$. Therefore, there are $10$ total subgroups of $D_4$.
I am least certain on finding subgroups isomorphic to the Klein group. Is there a more methodical way to do it? Is this approach even correct?