Question

Show by example that if M is a normal subgroup of N and if N is a normal subgroup of a group G , then M need not be a normal subgroup of G; in other words, normality isn’t transitive. [Hint: Consider $M=\left\{\upsilon ,r_o\right\}$and $N=\left\{h, \upsilon , r_{2,}, r_0\right\}$in $D_4\left..\right]$

Short answer

It is proved that if M is a normal subgroup of N and if N is a normal subgroup of G , then M need not be a normal subgroup of G.

Step-by-step solution

Required Theorem

Theorem 8.11: The following conditions on a subgroup N of a group G are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every$a\in G$ , Where$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every $a\in G$ , Where $\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every $a\in G$.
5. $aN{a}^{-1}\subseteq N$for every$a\in G$ .