# 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 subgr

• 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. – Hagen von Eitzen Nov 3 '15 at 17:09
• An an example of this is the dihedral group $D_4$ of 8 elements. – vnd Nov 3 '15 at 17:10

## 3 Answers

$G=S_4$, $H = \langle (12)(34) \rangle$ 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$}\end{cases}$, $H=\{\,(a,b,1)\mid a,b\in\{\pm1\}\,\}$, $K=\{\,(a,1,1)\mid a=\pm1\,\}$.

This helpful post by John Cook also spells out an equivalent counterexample. https://www.johndcook.com/blog/2011/09/07/normal-subgroups/

Hiding in plain sight, the symmetries of a square has at least TWO nonisomorphic normal subgroups of order 4. But the cyclic subgroup of order 4 is NOT the one you want. A useful H is the group generated by 180 degree rotation and by spinning about the y axis. This is a Klein 4 group, (grab a rectangular note book, start rotating and flipping to see the 4 configurations ). Moreover H is normal in D_8, (try conjugation by a rotation).

Of course one can also check by brute force, we are in the world of the eight 2x2 matrices under multiplication, all entries are 0 1 or -1, and all determinants are 1 or -1. So, as John points out, if K is the 2 element subgroup which includes flip about the y axis. Conjugate the flip by a 90 degree rotation, first rotate, then flip, then unrotate. Now your note book is not only flipped, but also upside down, kicking us out of K