Normality is not transitive 
Let $G=S_3\times S_3$ where $S_3$ is the symmetric group. Let $p=
        \begin{pmatrix}
        1 & 2 & 3 \\
        2 & 3 & 1 \\
        \end{pmatrix}
$, let $L=(p)$, $K=L\times L$ and $H=\{(I_3,I_3),(p,p),(p^2,p^2)\}$. Show that $K\triangleleft G$, $H\triangleleft K$ but $H$ no is a normal subgroup of $G$.

I wonder if there a quick way to do this exercise, without having to develop each of the products.
 A: $K \trianglelefteq G$ is easy to be seen because each $L \trianglelefteq S_3$ as $L$ has index 2 in $S_3$. For $H \trianglelefteq K$, we can use the fact:
If $H$ has a prime index $p$ in $G$ and there is no prime divisor of $|G|$ less than $p$, then $H \trianglelefteq G$. 
$|K:H| = 3$ and there is no smaller prime dividing $|K|=9$, so we see that $H \trianglelefteq K$. 
Now $H$ not being normal in $G$ should be easy done by one computation $((1,2),e)*(p,p)*((1,2),e)  = ((1,3,2),e)\notin H$
A: Using generators, there won't be much computation needed. 
$K$ is generated by $(1,p)$ and $(p,1)$. If $H$ is invariant under conjugation by the generators of $K$, then $H$ is normal in $K$.
Lagrange's theorem proves that $S_3$ is generated by $p$ and any transposition, e.g. $t=
        \begin{pmatrix}
        1 & 2 & 3 \\
        2 & 1 & 3 \\
        \end{pmatrix}$. So $G$ is generated by $(1,p),(1,t),(p,1),(t,1)$.
Conjugating these subgroups by the '$p$' generators will be easy, you just need to worry about $tpt^{-1}$.
A: Observe that $|L| = 3$ and $|S_3| = 6$, so the index of $L$ in $S_3$ is $2$. Therefore, $L \lhd S_3$. Consequently, $K = L \times L \lhd S_3 \times S_3 = G$.
Then, observe that $K$ is abelian, because it is a direct product of abelian groups, so all of its subgroups are normal. Therefore $H \lhd K$.
To see that $H$ is not normal in $G$, define
$$\tau = \begin{pmatrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{pmatrix} \quad \text{and} \quad
\mu = \begin{pmatrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{pmatrix}$$
You can confirm that $\tau p \tau^{-1} = p^2$ and $\mu p \mu^{-1} = p$, and therefore
$$(\tau, \mu)(p, p)(\tau^{-1}, \mu^{-1}) = (p^2, p) \not\in H$$
