# The Trouble with Normal Subgroups

1. May 9, 2008

### Tom Mattson

Staff Emeritus
[SOLVED] The Trouble with Normal Subgroups

1. The problem statement, all variables and given/known data
Find an example of a group $G$ and subgroups $H$ and $K$ such that $H$ is normal in $K$, $K$ is normal in $G$, but $H$ is not normal in $G$.

2. Relevant equations
None.

3. The attempt at a solution
I have many attempts. :grumpy:

In the beginning I was fixated on groups of orders 8 and 16, because I figured it would be easy to find nested subgroups with index 2 (that is, $[G:K] =2$ and $[K]=2$). Finding such groups is fairly easy, but finding them with $H$ not normal in $G$ has proved elusive.

Here are my false starts.

Attempt 1:
$G=D_8$, the dihedral group of order 8
$K=<b>$, the subgroup of order 4 generated by $b$
$H=<b^2>$, the subgroup of order 2 generated by $b^2$

Result: No good, $H$ is normal in $G$

Attempt 2:
$G=Q_8$, the quaternionic group of order 8
$K=\left<b\right>$
$H=\left<b^2\right>$

I didn't even bother finishing this one. It became obvious that this attempt would fail for the same reason that the last one failed.

Attempt 3:
Here's where I tried to be more clever.

$G=GL_2\left(\mathbb{Z}_3\right)$
$K=SL_2\left(\mathbb{Z}_3\right)$
$H=$ the subgroup generated by the elements of order 4 in $K$ (this is a subgroup of order 8).

Once again, $H$ is normal in both $K$ and $G$.

I also did some goofing around with $S_n$ and $A_n$, but to no avail.

Little help?

2. May 9, 2008

### zhentil

Try the dihedral group with order 8.

3. May 13, 2008

### Tom Mattson

Staff Emeritus
That was my first attempt. It didn't work.

4. May 13, 2008

### Tom Mattson

Staff Emeritus
Oh brother! I've got it. Like a dummy I was fixated on cyclic subgroups. I just considered a subgroup with two generators, and I got it.

FYI, $\left<ab^2\right>$ is normal in $D_8$, $\left<a,b^2\right>$ is normal in $\left<ab^2\right>$, but $\left<ab^2\right>$ is not normal in $D_8$, as $b^{-1}ab^2b=abb^3=ab^4=a$, which is not in $\left<ab^2\right>$.

So it was $D_8$ all along...