If $H$ is a normal subgroup of $K$ and $K$ is a normal subgroup of $G$ can we say that $H$ is a normal subgroup of $G$.I could not prove it and cannot find a suitable counter example

Will the results holds for $G$ abelian ?If else what will be counter example ?

  • $\begingroup$ Concerning the second question: in abelian groups all subgroups are normal, so this will hold. $\endgroup$ – lisyarus Mar 11 '15 at 17:42
  • $\begingroup$ okey can $S_{3}$ is a counter example for first part...? $\endgroup$ – Madhu Mar 11 '15 at 17:45
  • $\begingroup$ You have three groups in the question - $H \triangleleft K \triangleleft G$. To provide a counter example, you should specify three groups. $\endgroup$ – lisyarus Mar 11 '15 at 17:46

It is not true that $H \lhd K \lhd G$ implies $H \lhd G$ (although the stronger condition that $ H \text{ char } K\ \lhd G$ will force $H \lhd G$).

A good tactic would be to choose a group $G$ with an abelian normal subgroup $K$. Any subgroup of $K$ must be abelian, hence normal in $K$. Your job is to find a subgroup of $K$ that's not normal in $G$.

The alternating group on $4$ letters, $A_4$ is a good choice for $G$. There, you can find such a chain $H \lhd K \lhd G$ with $H \not\lhd G$.


To find a counter-example, we need a non-abelian group (every subgroup of an abelian group is normal). Unfortunately, $S_3$ (the smallest non-abelian group) doesn't work, since it has only one non-trivial proper normal subgroup, $A_3$, and $A_3$ is simple.

The next smallest non-abelian group is $D_4$, the dihedral group of order $8$. Here, we have two promising subgroups, $V = \{e,r^2,s,r^2s\}$ (where $r$ is a rotation of order $4$ and $s$ is some reflection), and $\langle r\rangle = \{e,r,r^2,r^3\}$, both of these are of index $2$, so are automatically normal. So these look good for $K$.

Clearly, we want a subgroup of order $2$ for $H$. Now $\{e,r^2\}$ is a subgroup of both possible $K$'s, but this isn't so good, since $r^2$ is central in $D_4$ (it commutes with everything). Since that's the ONLY subgroup of $\langle r\rangle$ of order $2$, we focus instead on $K = V$. It has another subgroup of order $2$, $\{e,s\}$ (We could also use $\{e,r^2s\}$, try it yourself and see!).

Now: $rsr^{-1} = rsr^3 = r(sr)r^2 = r(r^3s)r^2 = sr^2 = $

$(sr)r = (r^3s)r = r^3(sr) = r^3(r^3s) = r^2s \not \in \{e,s\}$

so that $rHr^{-1} \neq H$ for $H =\{e,s\}$, that is, $H$ is not normal in $D_4$, even though $H$ is normal in $V$, and $V$ is normal in $D_4$.

Unless I am mistaken, this is the minimal counter-example (in other word, look at small groups first, before trying large and difficult ones).


If $G$ is abelian then every subgroup its subgroup is normal: $gK = Kg$. Because $H$ also is subgroup of $G$ the answer is yes and proof is straightforward.

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    $\begingroup$ Try $G=S_4$, $K=\{(12)(34),(13)(42),(23)(41),e \}$ and $H = \langle (12)(34) \rangle$. You can show that $K$ is normal in $G$ (it might be seen easier by realizing that $K$ is isomorphic to $V_4$) and you can see that $H$ is normal in $K$. But is $K$ normal in $G$? $\endgroup$ – graydad Mar 11 '15 at 18:47

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