Question about commutator of normal subgroups Let $\mathbf G$ be a group and $H, K \trianglelefteq \mathbf G$ ($H$ and $K$ are normal subgroups of $\mathbf G$).
It follows that $[H,K]$, the subgroup of $\mathbf G$ generated by elements of the form $h^{-1}k^{-1}hk$, with $h \in H$ and $k \in K$ is also a normal subgroup of $\mathbf G$.
Morevover, $[H,K] \subseteq H \cap K$.
Suppose $a,b,c,d \in G$ are such that:
$a^{-1}b \in [H,K]$ (whence $a^{-1}b \in H \cap K$),
$a^{-1}c, b^{-1}d \in H$,
$c^{-1}d \in K$.
Does it follow that $c^{-1}d \in [H,K]$?
It certainly follows that $c^{-1}d \in H \cap K$, since 
$$c^{-1}d = (c^{-1}a)(a^{-1}b)(b^{-1}d)$$
and each of the factors in parentheses is in $H$, and by hypothesis, $c^{-1}d \in K$.
Since each of $H,K$ and $[H,K]$ is normal it has an associated congruence, say
\begin{align}
\alpha &= \{ (x,y) \in G^2 : x^{-1}y \in H \},\\
\beta &= \{ (x,y) \in G^2 : x^{-1}y \in K \},\\
\gamma &= \{ (x,y) \in G^2 : x^{-1}y \in [H,K] \}.
\end{align}
The following diagram is a tentative of illustrating the problem, in this setting.
Of course it follows that $(a,b) \in \alpha$ (since $\gamma \leq \alpha$), whence $(c,d) \in \alpha$ (by transitivity), and therefore $(c,d) \in \alpha \wedge \beta$ (which is equivalent to the above observation that $c^{-1}d \in H \cap K$).

So the problem in the setting of congruences is: does it follow that $(c,d) \in \gamma$?

Let $G=H=K=\langle t \rangle$ where $t$ is of order $4$. 
Let $a=b=1$, $c=t^{-1}$, $d=t$. 
Then $a^{-1}b=1\in [H,K]$.
Also $a^{-1}c=t^{-1}\in H$, and $b^{-1}d=t\in H$.
Then $c^{-1}d=t^2\in K$. But $c^{-1}d\not=1$, so $c^{-1}d\not\in [H,K]$.
