Let $G$ a group and let $H,K,L$ normal subgroups of $G$. Then $[HK,L]=[H,L][K,L]$. I want to prove a property of commutators in a group $G$.
Let $G$ a group and let $H,K,L$ normal subgroups of $G$. Then $[HK,L]=[H,L][K,L]$.
I have already shown that $[HK,L]$ is contained in $[H,L][K,L]$. someone can help me for the other inclusion?
Thanks!

The key is that the commutator subgroup is not merely the collection of all commutators, but rather the subgroup generated by these products. That is,
$[HK,L]$ is not equal to the set $\{[hk,\ell]\mid h\in H, k\in K,\ell\in L\}$, but rather is the subgroup generated by such products. Because of that, you easily get that both $[H,L]$ and $[K,L]$ are contained in $[HK,L]$. This holds for any subgroups (even if $HK$ is not a subgroup).
To get that $[HK,L]$ is contained in $[H,L][K,L]$, you can use certain commutator identites; e.g.,
$$[xy,z] = [x,z]^y[y,z]$$
using the fact that the groups are normal. 
