# What's the process for finding a group with specific conditions for it's normal subgroups

Find an example of a group $G$ with subgroups $K$ and $N$ such that $N$ is normal in $G$, $K$ is normal in $N$ but $K$ is not normal in $G$.

What is the process for answering a question such as this? Is there a way of determining it through some step-by-step procedure or am I supposed to just 'look it up' and write down and answer?

Here's as far as I've got

Let $g \in G, n \in N, k \in K$

Then we know that -

$g^{-1}ng \in N$

$n^{-1}kn \in K$

and we want to show $g^{-1}kg$ not $\in K$

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I'm not sure in general but a well-known counterexample starts with $G = A_4$ – Cocopuffs Nov 13 '12 at 18:48
@Cocopuffs: You did it right. :) – S. Snape Nov 13 '12 at 19:31

Here's some ideas. Good way to start with something like this is to first rule out what the groups $G$ and subgroups $K$ and $N$ cannot be like.

First of all, you want to find a group $G$ that actually has non-normal subgroups. So we can rule out $G$ abelian, there are definitely no examples there since every subgroup of an abelian group is normal.

Also, you want $K \neq N$ because if $K = N$ and $N$ is normal, then so is $K$. So $K$ has to be a proper subgroup of $N$ and of course $K \neq \{1\}$, so $N$ must have order $\geq 4$. Since $N$ also be a proper subgroup (else $N = G$ and $K$ is normal in $G$), the group $G$ must have order $\geq 8$.

There also one more thing you can notice. The group $N$ cannot be cyclic, because if $N$ is a normal cyclic subgroup of $G$, then every subgroup of $N$ is also normal in $G$.

So we start looking at non-abelian groups. The smallest one is $S_3$, but it's too small to have an example in it. The next ones are of order $8$, quaternion group $Q_8$ and dihedral group $D_8$. Every subgroup of $Q_8$ happens to be normal, but it turns out there is an example in $D_8$.

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Very nice answer, lots of concepts to take in in that post, cheers. – sonicboom Nov 13 '12 at 20:07
I can't see the example. If we let $G = D8$ then whether we take $N$ as the cyclic group of order 4 or the Klein 4 group, in either case we have take the subgroup $K$ as a cyclic groups of order 2 but this will be normal in $G$? – sonicboom Nov 14 '12 at 22:59
Ah...I get it now. – sonicboom Nov 14 '12 at 23:29

As @Cocopuffs notes nicely above, you could consider $A_4.\;\;$ In fact $$H=\{id,(1\ 2)(3\ 4),(1\ 3)(2\ 4),(1\ 4)(2\ 3)\}\trianglelefteq A_4$$ and $$K=\{id,(1\ 4)(2\ 3)\}\trianglelefteq H$$ but $K$ is not normal in $A_4.\;\;$ In fact $$(1\ 2\ 3)^{-1}(1\ 4)(2\ 3)(1\ 2\ 3)=(2\ 4)(3\ 1)\notin K$$

Let $G$ be a finite group such that $P\in \operatorname{Syl_p}(G),\;$ $$\text{ Ff}\;\;H\leq G\;\;\text{and}\;\;P\trianglelefteq H\trianglelefteq G,\;\;\text{then} \;\;P\trianglelefteq G.$$