39
$\begingroup$

Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?

Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.

If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.

$\endgroup$
  • $\begingroup$ $K$ characteristic in $H$ and $H$ normal in $G$ then $K$ is normal in $G$ $\endgroup$ – jim Dec 10 '12 at 7:48
  • 2
    $\begingroup$ Take a look at $D_8$, the dihedral group with 8 elements. $\endgroup$ – Hans Giebenrath Dec 10 '12 at 7:51
  • $\begingroup$ @HansGiebenrath I'm not seeing a counterexample in $D_4$. It has $C_4=\{1,r,r^2,r^3\}$ as a normal subgroup, but the only subgroup of $C_4$ is $C_2=\{1,r^2\}$, which is normal in $D_4$. $\endgroup$ – Mario Carneiro Dec 10 '12 at 8:02
  • 6
    $\begingroup$ @PatrickDaSilva: They do, $\langle r^2,s\rangle$ is normal in $D_8$ and contains $\langle s \rangle$, which is not normal in $D_8$. $\endgroup$ – Hans Giebenrath Dec 10 '12 at 9:08
  • 1
    $\begingroup$ @Hans : I guess I am tired for saying false things. Sorry to have doubted you. $\endgroup$ – Patrick Da Silva Dec 10 '12 at 9:16
33
$\begingroup$

Using some suggestions from the other commenters:

The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$

Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and $$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$ The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)

However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.

$\endgroup$
  • $\begingroup$ Great! You understand the theory verywell. +1! And sorry for the false comments. $\endgroup$ – Patrick Da Silva Dec 10 '12 at 9:17
15
$\begingroup$

Look at $S_4$ and its following subgroups $A = \langle (12)(34) \rangle$ and $B=\{(12)(34),(13)(42),(23)(41),e \}$. Try to show that $A$ is normal in $B$ and $B$ is normal in $S_4$ but $A$ is not normal in $S_4$.

$\endgroup$
  • $\begingroup$ Good eye! I like this example. How about you show it? The proof is not long. No calculations required. I challenge you (unless you were leaving the exercise to the OP). $\endgroup$ – Patrick Da Silva Dec 10 '12 at 7:59
  • $\begingroup$ @PatrickDaSilva I've fleshed out this argument in an answer below, but this leads to an interesting line of investigation: What is the smallest counterexample? As you point out, $|G|\geq 8$, but $D_8$ fails, and the quaternion group fails too since $\{1,-1\}\lhd Q_8$ is the only subgroup of order 2. $\endgroup$ – Mario Carneiro Dec 10 '12 at 9:09
  • 1
    $\begingroup$ @PatrickDaSilva $A_4$ is normal in $S_4$ and the sylow 2 subgroup of of $A_4$ is a unique group of order 4 hence the group of $B$ given above will be normal in $S_4$ and the normality of $A$ in $B$ is becuase of the fact $[B:A]=2$ $\endgroup$ – jim Dec 10 '12 at 10:56
0
$\begingroup$

We need a non-abelian group, since all subgroups of abelian groups are normal. One small candidate is $D_8$, the symmetries of a square, here in a little more detail about how we might go about finding examples:

Consider all the subgroups in $D_8$. It's useful to visualize the subgroups as a lattice:

(Picture of Dummit and Foote I found on the web)

Now we try to pick an $H$. For all the subgroups on the third row from the top, their only proper subgroup is the trivial subgroup, which is trivially normal to $G$, so it doesn't make sense to use any of the subgroups on the third row for $H$.

Our only options for $H$ now are the second row: $\langle s, r^2 \rangle$, $\langle r \rangle$, and $\langle rs, r^2 \rangle$. We observe that if $H = \langle r \rangle$, the proper subgroups $\langle r^2 \rangle$ and $1$ are both normal to $D_8$, so that case is excluded. Our candidates are $\langle s, r^2 \rangle$ and $\langle rs, r^2 \rangle$.

Take $H = \langle s, r^2 \rangle$ and $K = \langle s \rangle$. It's easy to verify that $\langle s \rangle$ is not normal to $D_8$. All that's left is to show $K \lhd H$ and $H \lhd G$.

This is not difficult if we remember that any element in $D_8$ can be written as $r^i s^j$ with $0 \le i \le 3$ and $j = 0, 1$. Also we have the identity $rs = sr^{-1}$ which can repeated as $r^k s = s r^{-k}$. So for $g \in D_8$, we want to prove or disprove $r^i s^j n s^j r^{-i} \in N$ to show $N \lhd G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.