Normality Relation is not Transitive/Proof 2

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Theorem

Let $G$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $K$ be a normal subgroup of $N$.


Then it is not necessarily the case that $K$ is a normal subgroup of $G$.


Proof

Proof by Counterexample:

Let $D_4$ denote the dihedral group $D_4$.


Let $D_4$ be presented in matrix representation:

Let $\mathbf I, \mathbf A, \mathbf B, \mathbf C$ denote the following four elements of the matrix space $\map {\MM_\Z} 2$:

$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

\qquad \mathbf A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad \mathbf B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \qquad \mathbf C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$


The set:

$D_4 = \set {\mathbf I, -\mathbf I, \mathbf A, -\mathbf A, \mathbf B, -\mathbf B, \mathbf C, -\mathbf C}$

under the operation of conventional matrix multiplication, forms the dihedral group $D_4$.


Its Cayley table is given as:

$\begin{array}{r|rrrrrrrr}
          &  \mathbf I &  \mathbf A &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C \\

\hline 

\mathbf I &  \mathbf I &  \mathbf A &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C \\
\mathbf A &  \mathbf A &  \mathbf I & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I &  \mathbf C &  \mathbf B \\
\mathbf B &  \mathbf B &  \mathbf C & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C &  \mathbf I &  \mathbf A \\
\mathbf C &  \mathbf C &  \mathbf B &  \mathbf A &  \mathbf I & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I \\

-\mathbf I & -\mathbf I & -\mathbf A & -\mathbf B & -\mathbf C & \mathbf I & \mathbf A & \mathbf B & \mathbf C \\ -\mathbf A & -\mathbf A & -\mathbf I & \mathbf C & \mathbf B & \mathbf A & \mathbf I & -\mathbf C & -\mathbf B \\ -\mathbf B & -\mathbf B & -\mathbf C & \mathbf I & \mathbf A & \mathbf B & -\mathbf C & -\mathbf I & -\mathbf A \\ -\mathbf C & -\mathbf C & -\mathbf B & -\mathbf A & -\mathbf I & \mathbf C & \mathbf B & \mathbf A & \mathbf I \end{array}$


Consider the subgroup $H$ whose underlying set is:

$H = \set {\mathbf I, \mathbf A, -\mathbf I, -\mathbf A}$

From Subgroup of Index 2 is Normal, $H$ is normal in $D_4$.

Consider the subgroup $H$ whose underlying set is:

$K = \set {\mathbf I, \mathbf A} = \gen {\mathbf A}$

From Subgroup of Index 2 is Normal, $K$ is normal in $H$.


It remains to be demonstrated that $K$ is not normal in $D_4$.

From the Cayley table:

$\mathbf C \mathbf A \mathbf C^{-1} = \mathbf B \mathbf C = -\mathbf A \notin K$

Hence $K$ is not normal in $D_4$.

$\blacksquare$


Sources

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