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Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.

I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true?

user75908
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4 Answers4

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Let $H=D_8$, the group of symmetries of a square under flips and rotations. Let $F$ be the subgroup of flips about the vertical axis of symmetry. Let $G$ the symmetries you can find by combinations of such flips and $180$ degree rotations. You can show that $F$ is normal in $G$, and $G$ is normal in $H$.

Now let $h$ be a 90 degree clockwise turn and let $f$ be a flip. You can show that $hfh^{-1}$ is not a flip or the identity, so $F$ is not a normal subgroup of $H$.

Federica Maggioni
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$$\{\,(1)\,,\,(12)(34)\,\}\lhd\left\{\;(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\,\right\}\lhd A_4\;,\;\;\{\,(1)\,,\,(12)(34)\,\}\not\!\triangleleft A_4$$

DonAntonio
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  • Is there some sort of lemma that can be used to speed up the verification that $G={(1),(12)(34),(13)(24),(14)(23)}\triangleleft A_4$? I guess one should at least verify that if $\sigma$ is a $3$-cycle, then $\sigma G\sigma^{-1}=G$, is it necessary? – user75908 May 04 '13 at 10:35
  • Knowing the groups $,S_n,,,A_n,$ a little: the group $,G,$ is the union of two pretty simple conjugacy classes in $,S_4,$ (and also in $,S_4,$ , btw) , and is thus normal. Of course, you can verify that $,G,$ is invariang under conjugation by all kind of cycles possible in $,S_4,$ , but that will take a little longer...:) – DonAntonio May 04 '13 at 10:38
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    @user75908: $G$ contains every permutation with the cycle structure $(ab)(cd)$ and conjugation by a permutation preserves cycle structure. – Mikko Korhonen May 04 '13 at 10:39
  • Right, thanks for the precision, it is indeed a lot clearer with this result. – user75908 May 04 '13 at 10:46
  • Is it not true that $G=A_4$ or am I missing something here? – user113867 Jan 06 '14 at 23:03
  • @user113867 , if you want to refer to the question's original notation, then you're missing something: $;G; $ is the group with four elements, $;H=A_4;$ and $;F;$ is the group with two elements. – DonAntonio Jan 07 '14 at 04:43
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Take any finite solvable group $G$ which has a minimal normal subgroup $M$ which is not cyclic. Let $\langle x \rangle$ be any non-identity cyclic subgroup of $M.$ Then $\langle x \rangle \lhd M \lhd G,$ but $\langle x \rangle \not \lhd G$, since $M$ is minimal normal, but not cyclic. Now it's a matter of finding a solvable group with a non-cyclic minimal normal subgroup, which is not difficult.

Geoff Robinson
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Let $p$ be a prime, and let $G$ be a $p$-group of order $p^3$. Let $H \leq G$ be a non-normal subgroup of order $p$ (equivalently, $H$ is of order $p$ and not central). Then $H$ is contained in a subgroup $K \leq G$ of order $p^2$. In this case $H \trianglelefteq K \trianglelefteq G$, but $H$ is not normal in $G$.

For example, $G$ could be the Heisenberg group, which is the set

$$\left\{ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : a, b, c \in \mathbb{Z}_p \right\}$$

of matrices under multiplication. In the case $p = 2$ this group is isomorphic to $D_8$, which is the example given in another answer.

Mikko Korhonen
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