I need to understand why the following identity is true:
$$[XK, XH]= [X, X][X, K][X, H][K, H]$$ where $H,K$ are normal subgroups of some given group, and $X$ a subgroup (not necessarily normal). And $[H,K]= \langle h^{-1}k^{-1}hk\ ,\ h\in H, k\in K \rangle $ is a subgroup of $\langle H,K \rangle$.
I tried using the identities in the wiki page with no success, I started from the RHS and tried to obtain the LHS but it seems it is not a straight forward calculation or that I lack imagination in term of 'throwing in' the identity element $hh^{-1}$ appropriately.
We are assuming now that $H$ and $K$ are normal subgroups and $X$ is an arbitrary subgroup of a group $G$. I will use the definition $[x,y] := x^{-1}y^{-1}xy$ for the commutator.
We first show that $[KX,HX]= [K,HX][X,HX]$. Then, for $x \in X$, $k \in K$ and $y \in HX$, we have $$[kx,y] = [k,y]^x[x,y] = [k^x,y^x][x,y] \le [K,XH][X,XH]$$ so $[KX,HX] \le [K,HX][X,HX]$, and the reverse inclusion is clear.
Next we show $[X,HX] = [X,X][X,H]$. For $k \in K$, $x,y \in X$, we have $$ [y,hx] = [y,x][y,h]^x = [y,x][y^x,h^x] \in [X,X][X,H]$$ and again the reverse inclusion is clear.
Finally we show $[K,HX] = [K,X][K,H]$. For $x \in X$, $h \in H$, $k \in K$, we have $$[k,hx] = [k,x][k,h]^x = [k,x][k^x,h^x] \in [K,X][K,H]$$ and the reverse inclusion is clear.