# Normal subgroup of subgroups normal to D4

## Question:

Let $G=\left\{I,\mathbb{R},\mathbb{R}^2,\mathbb{R}^3,H,V,P,O \right\}=D_4$ (The symmetries of the square).

Exhibit two subgroups, $K$ and $L$ of $G$ s.t. $L \subset K \subset G$ (not proper subsets). Also, $L$ is normal in $K$, $K$ is normal in $G$, but $L$ is not normal in $G$.

## Attempted Solution:

So I have that $K$ must equal one of the following subgroups: $\left\{I,P,O,\mathbb{R}^2\right\}$, $\left\{I,H,V,\mathbb{R}^2\right\}$, $\left\{I,\mathbb{R},\mathbb{R}^2,\mathbb{R}^3\right\}$, or $\left\{I,\mathbb{R}^2\right\}$. I'm excluding $G$ and $\left\{I\right\}$ since $L\not=K$ and $K\not=G$.

I know that $L$ must be a normal subgroup to one of these four, but I am having difficulty finding these particular normal subgroups of $K$. I have a multiplication table drawn out as well to help me, but to no avail.

• The groups of order 4 are abelian, so all their subgroups are normal subgroups. You just have to find one that's not normal in $G$. – Gerry Myerson Nov 15 '15 at 6:37
• So would $L=\left\{I,H\right\}$ and $K=\left\{I,H,P, \mathbb{R}^2 \right\}$ satisfy the problem? – lollercide Nov 15 '15 at 6:43
• $\{\,1,H,P, R^2\,\}$ is not a subgroup. – Gerry Myerson Nov 15 '15 at 7:42
• You're right. I meant to type $K=\left\{I,H,V,\mathbb{R}^2\right\}$ – lollercide Nov 15 '15 at 7:57
• Then you have an answer. Write it up! Post it as an answer! – Gerry Myerson Nov 15 '15 at 7:59

From the fact that groups of order 4 are abelian and all subgroups of abelian groups are normal, I choose from the following: $\left\{I,P,O,\mathbb{R}^2\right\}$, $\left\{I,H,V,\mathbb{R}^2\right\}$, $\left\{I,\mathbb{R},\mathbb{R}^2,\mathbb{R}^3\right\}$.
Choosing $K=\left\{I,H,V,\mathbb{R}^2\right\}$, I then choose $L=\left<H\right>=\left\{I,H\right\}.$
Because $L$ is a normal subgroup in $K$ but not a normal subgroup in $G$, and becase $K$ is a normal subgroup to $G$, we have satisfied the conditions.