# 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, 2012 at 7:48
• Take a look at $D_8$, the dihedral group with 8 elements. Dec 10, 2012 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 $D_4$. Dec 10, 2012 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 $D_8$. Dec 10, 2012 at 9:08
• @Hans : I guess I am tired for saying false things. Sorry to have doubted you. Dec 10, 2012 at 9:16

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 $$f[(12)(34)]f^{-1}$$ 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. Dec 10, 2012 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). Dec 10, 2012 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. Dec 10, 2012 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, 2012 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 $$N \lhd G$$.