# Are normal subgroups transitive?

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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.

• $K$ characteristic in $H$ and $H$ normal in $G$ then $K$ is normal in $G$ – jim Dec 10 '12 at 7:48
• Take a look at $D_8$, the dihedral group with 8 elements. – Hans Giebenrath Dec 10 '12 at 7:51
• @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 $D_4$. – Mario Carneiro Dec 10 '12 at 8:02
• @PatrickDaSilva: They do, $\langle r^2,s\rangle$ is normal in $D_8$ and contains $\langle s \rangle$, which is not normal in $D_8$. – Hans Giebenrath Dec 10 '12 at 9:08
• @Hans : I guess I am tired for saying false things. Sorry to have doubted you. – Patrick Da Silva Dec 10 '12 at 9:16

## 3 Answers

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 $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.

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

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 $S_4$.

• 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). – Patrick Da Silva Dec 10 '12 at 7:59
• @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. – Mario Carneiro Dec 10 '12 at 9:09
• @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 $[B:A]=2$ – jim Dec 10 '12 at 10:56

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 $$N \lhd G$$.