Non-cyclic normal subgroup Suppose $H$ is a cyclic normal subgroup of $G$, then it is true that every subgroup of $H$ is normal in $G$. However, it is not true if $H$ is not cyclic. Is there any counterexample?
 A: in $S_n$, the group of permutation of $n$ elements, $A_n$ is the only normal subgroup for $n>4$, so any subgroup of $A_n$ is not normal in $S_n$
A: Take $\;A_4\;$ , then
$$\{(1)\,,\,(12)(34)\}\lhd \{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}\lhd A_4\;,\;\;\text{but}$$
$$\{(1)\,,\,(12)(34)\}\;\rlap{\;\;/}\lhd A_4$$
A: Consider the group $G=(C_3\times C_3)\rtimes C_2$ with the presentation $\{a,b\mid a^2=e,b=(b_1,b_2)\in C_3\times C_3,a(b_1,b_2)a=(b_2,b_1)\}.$ Then $G$ has order $18,$ and $N:=\left<(1,0)\right>$ is a normal subgroup of $H:=C_3\times C_3,$ (as $C_3\times C_3$ is abelian) which, by the definition of the semi-direct product, is normal in $G:$ i.e. $N\triangleleft H\triangleleft G.$
However, $a(1,0)a=(0,1)\not\in\left<(1,0)\right>,$ thus $N\not\triangleleft G.$
P.S. That $H\triangleleft G$ can also be seen by the fact that $[G:H]=2.$
Please point out any error that occurs; thanks in advance, and hope this helps.
Edit
To be more clear and to be coherent with the linked notation, $G$ is in fact the semi-direct product $G=H\rtimes_\phi C_2,$ where $\phi:C_2\rightarrow \operatorname{Aut}(C_3\times C_3)$ sends the generator of $C_2$ to $\begin{pmatrix}0&1\\1&0\end{pmatrix}\in \operatorname{GL}(2,3)\cong\operatorname{Aut}(C_3\times C_3).$
