Let $R_2 = \{1, \rho_2 \} \subset D_4$. Show that $R_2$ is a normal subgroup of $D_4$.

Question

Let $R_2 = \{1, \rho_2 \} \subset D_4$. Show that $R_2$ is a normal subgroup of $D_4$.

I have the definition that a subgroup $H$ of $G$ is normal if $gH=Hg$ for all $g \in G$.

Now if $\rho’ \in D_4$, then $\rho’R_2= \{\rho’ \cdot 1, \rho’ \cdot \rho_2\}$.

Clearly $p’ \cdot 1 = 1 \cdot p’$, but I’m not sure about $\rho’ \cdot \rho_2 = \rho_2 \cdot \rho’ $. Looking at the multiplicative table it seems that $\rho_2 $ is commutative? Thus $\rho’R_2 =R_2 \rho’ $ which would make $R_2$ normal?

Answer 1

Instead of proving it directly, like you are trying to do, we can use the first isomorphism theorem, which links normal subgroups to homomorphisms.

Define the map

$$\begin{align} \varphi: D_4 &\to \Bbb Z_2^2,\\ \rho & \mapsto (1,0),\\ \sigma &\mapsto (0,1). \end{align}$$

Here $\rho$ is the rotation of the square by $\pi/2$ radians and $\sigma$ is the flip about the vertical axis of symmetry.

Verify that $\varphi$ defines a homomorphism. (It is defined on a generating set of $D_4$, sending each element to the elements of a generating set of $\Bbb Z_2\times\Bbb Z_2$. The way to think about this is to note that every element has the form either $\rho^n$ or $\rho^n\sigma$, so verify it for each of these elements.${}^\dagger$) Its kernel is $R_2$. Thus, by the first isomorphism theorem, $R_2\unlhd D_4$.

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$\dagger$: I thank @user1729 for this.

Answer 2

The relation $rs=sr^{-1}$ is part of the definition for the dihedral groups - it should be all you need.

In your notation you seem to have $\rho$ for rotations and $\rho'$ for reflections. So what I've said means $\rho\rho'=\rho'\rho^{-1}$, note $\rho^{-1}$ is $\rho^3$.