How to check which subgroups of $D_4$ are normal

Question

How do I check which subgroups of $D_4$ are normal? Trying all elements seems very cumbersome.

So far, I know only basic theorems like Lagrange's and the homomorphism as well as the isomorphism theorems.

Answer 1

Let a presentation for $D_{4}$ be given by $\langle r,s | r^{4}=s^{2}=1, rs=sr^{-1}\rangle$.

By Lagrange, any nontrivial subgroup will have either order $2$ or $4$. The cyclic subgroup generated by a rotation is a subgroup of order $4$. There are, as noted in the comments below, two other subgroups of order $4: \langle s, r^{2} \rangle = \{1, r^{2}, s, r^{2}s\}$ and $\langle rs, r^{2}\rangle = \{1, rs, r^{3}s, r^{2}\}$. It can be shown that any subgroup of index $2$ is normal; I will leave this to you as an exercise with the following suggestion. Consider the cosets of a subgroup of index $2$. What are they? Can you show the left and right cosets coincide?

Now, this leaves us with the subgroups of order $2$. You said you have identified these - great! Now, we will use a small "trick". Consider any subgroup $H= \{1, h\}$ of order $2$. If $H$ is a normal subgroup of $D_{4}$, then $gHg^{-1} = H$ for all $g \in D_{4}$. Since $g1g^{-1}=1$ for all $g\in D_{4}$, we see that if $H$ is normal, $ghg^{-1}=h$ for all $g\in D_{4}$, i.e. $h$ commutes with every element of $D_{4}$. Hence, to find the normal subgroups of order $2$ in $D_{4}$, it suffices to compute the center of $G$ and identify the elements of order $2$.

To compute the center of $D_{4}$ is straightforward, which I leave to you with some hints. One, use Lagrange's theorem and the fact that $Z(D_{4})$ is a subgroup of $D_{4}$ to make some conclusions about the possible order of $Z(D_{4})$. Two, leverage what you know about the relations on the generators of $D_{4}$ to speed up computations.