Normality Relation is not Transitive

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 1

Proof by Counterexample:

Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:

$\begin{array}{c|cccc|cccc|cccc} \circ & e & t & u & v & a & b & c & d & p & q & r & s \\ \hline e & e & t & u & v & a & b & c & d & p & q & r & s \\ t & t & e & v & u & b & a & d & c & q & p & s & r \\ u & u & v & e & t & c & d & a & b & r & s & p & q \\ v & v & u & t & e & d & c & b & a & s & r & q & p \\ \hline a & a & c & d & b & p & r & s & q & e & u & v & t \\ b & b & d & c & a & q & s & r & p & t & v & u & e \\ c & c & a & b & d & r & p & q & s & u & e & t & v \\ d & d & b & a & c & s & q & p & r & v & t & e & u \\ \hline p & p & s & q & r & e & v & t & u & a & d & b & c \\ q & q & r & p & s & t & u & e & v & b & c & a & d \\ r & r & q & s & p & u & t & v & e & c & b & d & a \\ s & s & p & r & q & v & e & u & t & d & a & c & b \\ \end{array}$

From Normality of Subgroups of Alternating Group on 4 Letters:

$K := \set {e, t, u, v}$ is a normal subgroup of $A_4$

$T := \set {e, t}$ is not a normal subgroup of $A_4$.

But by Subgroup of Abelian Group is Normal:

$T$ is a normal subgroup of $K$.

Thus we have:

$T \lhd K$, $K \lhd A_4$

but:

$T \not \lhd A_4$

$\blacksquare$

Proof 2

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 $D_4$.

$\blacksquare$