# Prove that if H is a normal subgroup of G and K is a normal subgroup of H, then K may not be a normal subgroup of G.

I was doing a course on algebra and had this question written in my notes.

Prove that if $$H$$ is a normal subgroup of $$G$$ and $$K$$ is a normal subgroup of $$H$$, then $$K$$ may not be a normal subgroup of $$G$$.

Now as I understand to prove a subgroup $$K$$ normal to $$G$$ I have to do $$g^{-1}kg$$ belongs to $$K$$. Clearly, This equation will be satisfied for all $$g$$ belonging to $$H$$ (as $$K$$ is a normal subgroup to H) but not necessarily for $$g$$ belonging to $$G-H$$.

Formally, I am at a loss how to show this that there may exist an element which will not satisfy this. Moreover I feel I am missing something as I have still not used $$H$$ is a subgroup.

• Theproblem with "prove that something may happen" is that you cannot work with general $G,H,K$. Instead, you could (and should) exhibit a specific counterexample. Nov 3, 2015 at 17:09
• An an example of this is the dihedral group $D_4$ of 8 elements.
– vnd
Nov 3, 2015 at 17:10

$G=S_4$, $H = \langle (12)(34) \rangle$ and $K=\{(12)(34),(13)(42),(23)(41),e \}$. $H$ is normal in $K$, $K$ is normal in $G$. But $H$ is not normal in $G$.
What could an example of could $K\lhd H\lhd G$ with $K\not\lhd G$ look like? One characterization of "$H\lhd G$" is that $H$ is fixed (though not pointwise fixed) under conjugation with elements of $G$. Likewise, $K$ is fixed under conjugation with elements of $H$, but we do not want it to be fixed under conjugation with elements of $G$ - at least not with all elements of $G$. Nevertheless, all $gKg^{-1}$ will still be subgroups of $H$; so we want $H$ to contain at least two distinct "copies" of $K$. So we might try with $K=C_2\lhd H=C_2\oplus C_2$ and find $G$ such that it sometimes switches the summands of $H$. This can be achieved by a semidirect product of $H$ with $C_2$ where $C_2$ acts on $H$ by switching the summands.
So all spelled out: Let $G=\{-1,1\}^3$ with $(a,b,c)*(d,e,f)=\begin{cases}(ad,be,cf)&\text{if$c=1$}\\(ae,bd,cf)&\text{if$c=-1$}\end{cases}$, $H=\{\,(a,b,1)\mid a,b\in\{\pm1\}\,\}$, $K=\{\,(a,1,1)\mid a=\pm1\,\}$.