Normality Relation is not Transitive/Proof 1

From ProofWiki
Jump to navigation Jump to search

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 $A_4$ 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:

$T \not \lhd A_4$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Normality_Relation_is_not_Transitive/Proof_1&oldid=520520"