Give an example showing that normality is not transitive: i.e. \(H\trianglelefteq K \trianglelefteq G\) with \(H\) not normal in \(G\).
Solution
Take the dihedral group \[ G=D_4=\langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle \] of order \(8\). Set \[ K=\langle r^2,s\rangle=\{1,r^2,s,sr^2\}, \qquad H=\langle s\rangle=\{1,s\}. \]
Since \([G:K]=2\), we have \(K\trianglelefteq G\). Since \([K:H]=2\), we have \(H\trianglelefteq K\).
But \[ rsr^{-1}=sr^2\notin H, \] so \(rHr^{-1}\ne H\). Hence \[ H\trianglelefteq K\trianglelefteq G \quad\text{but}\quad H\not\trianglelefteq G. \] Thus normality is not transitive.