# normality of subgroups is not transitive

Let $G$ be a group. A subgroup $K$ of a subgroup $H$ of $G$ is obviously a subgroup of $G$. It seems plausible that a similar situation would also hold for normal subgroups, but in fact it does not: even when $K\trianglelefteq H$ and $H\trianglelefteq G$, it is possible that $K\ntrianglelefteq G$. Here are two examples:

1. 1.

Let $G$ be the subgroup of orientation-preserving isometries (http://planetmath.org/Isometry) of the plane $\mathbb{R}^{2}$ ($G$ is just all rotations and translations), let $H$ be the subgroup of $G$ of translations, and let $K$ be the subgroup of $H$ of integer translations $\tau_{i,j}(x,y)=(x+i,y+j)$, where $i,j\in\mathbb{Z}$.

Any element $g\in G$ may be represented as $g=r_{1}\circ t_{1}=t_{2}\circ r_{2}$, where $r_{1,2}$ are rotations and $t_{1,2}$ are translations. So for any translation $t\in H$ we may write

 $g^{-1}\circ t\circ g=r^{-1}\circ t^{\prime}\circ r,$

where $t^{\prime}\in H$ is some other translation and $r$ is some rotation. But this is an orientation-preserving isometry of the plane that does not rotate, so it too must be a translation. Thus $G^{-1}HG=H$, and $H\trianglelefteq G$.

$H$ is an abelian group, so all its subgroups, $K$ included, are normal.

We claim that $K\ntrianglelefteq G$. Indeed, if $\rho\in G$ is rotation by $45^{\circ}$ about the origin, then $\rho^{-1}\circ\tau_{1,0}\circ\rho$ is not an integer translation.

2. 2.

A related example uses finite subgroups. Let $G=D_{4}$ be the dihedral group with eight elements (the group of automorphisms of the graph of the square). Then

 $D_{4}=\left\langle r,f\mid f^{2}=1,r^{4}=1,fr=r^{-1}f\right\rangle$

is generated by $r$, rotation, and $f$, flipping.

The subgroup

 $H=\langle rf,fr\rangle=\left\{1,rf,r^{2},fr\right\}\cong C_{2}\times C_{2}$

is isomorphic to the Klein 4-group – an identity and 3 elements of order 2. $H\trianglelefteq G$ since $[G:H]=2$. Finally, take

 $K=\langle rf\rangle=\left\{1,rf\right\}\trianglelefteq H.$

We claim that $K\ntrianglelefteq G$. And indeed,

 $f\circ rf\circ f=fr\notin K.$
Title normality of subgroups is not transitive NormalityOfSubgroupsIsNotTransitive 2013-03-22 12:49:27 2013-03-22 12:49:27 yark (2760) yark (2760) 13 yark (2760) Example msc 20A05 NormalIsNotTransitive