The commutator subgroup of a quotient in terms of the commutator subgroup and the kernel Let $H$ be the normal subgroup of $G$. 
Is it true that $(H[G,G])/H$ is isomorphic to $[G/H,G/H]$?
If so, I want to make a surjective homomorphism $\phi\colon H[G,G]\to [G/H,G/H]$ with Kernel $H$ to prove it.
But, if I define $\phi(h[g_1,g_2])=[g_1/H,g_2/H]$ then I have some trouble. 
 A: An easy argument certainly follows user8268's comment.
A few things should be noticed (as they might not seem totally obvious). At first since $H$ is normal in $G$,
$$ [g_1H,g_2H] = [g_1,g_2]H \in G/H.$$
Now restricting the canonical epimorphism $G\to G/H$ to $H[G,G]$, we may call it $\phi$, gives us a mapping, that satisfies
$$ \phi(h[g_1,g_2])= [g_1,g_2]H = [g_1H,g_2H] \in [G/H,G/H]$$
for $h\in H$ and $g_1,g_2\in G$. Since $H[G,G]$ is generated by the elements $h[g_1,g_2]$, $\phi(H[G,G])$ is certainly generated by the elements $[g_1H,g_2H]$, that generate $ [G/H,G/H]$. Hence $\phi$ is surjective and its kernel is still $H$ as it is included in $H[G,G]$. 
A: The commutator subgroup is a verbal subgroup. That means that for any group homomorphism $\varphi\colon G\to K$, you have that $\varphi([G,G])\subseteq [K,K]$. If $\varphi$ is onto, then the induced map on commutator subgroups is also onto.
Applying this to the quotient map $G/H$, you have that $\pi([G,G])$ maps $[G,G]$ onto $[G/H, G/H]$. The kernel of this map is $[G,G]\cap H$, so by the isomorphism theorems we have
$$\left[\frac{G}{H},\frac{G}{H}\right] \cong \frac{[G,G]}{[G,G]\cap H} \cong \frac{H[G,G]}{H}.$$
Follow through the definitions given by the isomorphism theorems, and you'll see that the isomorphism from $\frac{H[G,G]}{H}$ to $[G/H,G/H]$ is indeed given by mapping $huH$ to $\pi(u)$ (or by mapping from $H[G,G]$ onto $[G/H,G/H]$ via $hu\mapsto \pi(u)$). This is the map you give (though you only describe it for a generating set of $H[G,G]$). 
The same result holds for any verbal subgroup $\mathfrak{V}(G)$: if $N$ is normal, then 
$$\mathfrak{V}\left(\frac{G}{N}\right) \cong \frac{\mathfrak{V}(G)}{\mathfrak{V}(G)\cap N} \cong \frac{N\mathfrak{V}(G)}{N}.$$
E.g., for the subgroup generated by the $n$th powers, $G^n$, we have
$$\left(\frac{G}{N}\right)^n \cong \frac{NG^n}{N}.$$ 
