Find subgroups of $D_4$.

Question

If r stands for counter-clockwise 90 degree rotation, s stands for horizontal flip. $D_4= \{1, r, r^2, r^3, s, rs, r^2s, r^3s\}$. What rule should I apply to find the subgroups of $D_4$? Should I just put elements with same order in the same subgroup?

Answer 1

The group $D_8$ is small enough that we can proceed naively, splitting cases according to whether each element occurs within a given subgroup.

Hint Let $H$ be a subgroup of $D_8$.

Case 1: $r \in H$. We have $H \supset \langle r \rangle \cong C_4$. If $H$ furthermore contains any of the four remaining elements $r^k s$, it contains at least $5$ elements and so is equal to all of $D_8$ (we invoke this reasoning later without comment). Thus, this exausts all of the subgroups that contain $r$. Since $r \in \langle r^3 \rangle$, this exhausts all subgroups that contain $r^3$, too.

Case 2: $r \not\in H, r^2 \in H$. Now $H \supset \langle r^2 \rangle \cong C_2$. If $H$ furthermore contains $s$, then $H \supset \{1, r^2, s, r^2 s\} \cong C_2 \times C_2$. Since $s \in \langle r^2 , r^2 s \rangle$, this exhausts all subgroups that contain $r^2$ and at least one of $s, r^2 s$. On the other hand, if $H$ does not contain $s$ but does contain $r s = s r^3$, then...

Case 3: $r, r^2, r^3 \not\in H$. ...

Can you finish this analysis?

Exercise We can view $D_8$ as the group of symmetries of a square. Which geometric objects are exactly preserved by each of the subgroups?

Answer 2

HINT

Pick elements one by one and see what happens to their generated subgroups (i.e. orbits under the operation $\cdot$). Then try to mix them with each other. E.g.

• $O(s): 1, s, s^2=1$
• $O(r^2): 1, r^2, (r^2)^2 = r^4 = 1$
• $O(r^2,s): 1, s, r^2, r^2 s, 1$

Can you find some other ones?

UPDATE

In more detail, to see what the orbit generated by $s$ and $r^2$ is, you apply the operation to all possible combinations of the base elements:

• $s$ generates just itself since $s^2=1$
• $r^2$ generates just itself since $(r^2)^2 = r^4 = 1$
• The last thing left is to multiply them, getting $r^2 s$ (and $sr^2$ if the operation was not commutative but here it is). Finally, $(sr^2) \cdot s = r^2$ and $(sr^2) \cdot r^2 = s$ so no new elements are generated here either.

Hence, the final orbit is $$O(r^2,s) = \{1,s,r^2, sr^2\}.$$