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!
Property of commutators
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0Correct. Well done. The other direction is actually more difficult IMHO. – 2011-07-07
2 Answers
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.
In this same spirit, and Arturo's answer and remark on commutator identities, try to prove the Three Subgroups Lemma: if $N$ is a normal subgroup of $G$, and $K, L, M$ are subgroups of $G$ such that $[K,L,M] \subset N$, and $[L,M,K] \subset N$, then $[M,K,L] \subset N$. Here $[*,*,*] = [[*,*],*]$.