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[parent] normality of subgroups is not transitive (Example)

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. Let $ G$ be the subgroup of orientation-preserving isometries 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

    $\displaystyle g^{-1}\circ t\circ g = r^{-1}\circ t'\circ r, $
    where $ t'\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. 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
    $\displaystyle 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.
    \includegraphics{c2,2_in_d4}

    The subgroup

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

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

    $\displaystyle f\circ rf \circ f = fr \notin K. $



"normality of subgroups is not transitive" is owned by yark. [ full author list (2) | owner history (1) ]
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Cross-references: order, identity, Klein 4-group, isomorphic, generated by, square, graph, automorphisms, dihedral group, finite, origin, abelian group, rotate, integer, translations, rotations, plane, orientation-preserving, normal subgroups, subgroup, group
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This is version 10 of normality of subgroups is not transitive, born on 2002-06-30, modified 2006-12-18.
Object id is 3147, canonical name is NormalityOfSubgroupsIsNotTransitive.
Accessed 3849 times total.

Classification:
AMS MSC: 20A05 (Group theory and generalizations :: Foundations :: Axiomatics and elementary properties)
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