proving a commutator identity

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.

• These are all normal subgroups, yeah? Because you talk about $XH$ like it's a subgroup, and say that $[H,K]$ is a subgroup of both $H$ and $K$. Can you clarify in the question? Normality makes this easy! Jan 21 at 6:09
• In general $[H,K]$ is a normal subgroup of $\langle H,K \rangle$, but it is contained in $H$ if and only if $K$ normalizes $H$. Jan 21 at 9:03
• how do you define $\langle H,K\rangle$ ? The smallest group containing $H\cup K$? And how about $K$ normalizes $H$, is it $H\triangleleft K$ ? This suppose $H$ to be a subgroup of $K$. I am missing a lot of language here, do you have a reference where I can read about this? Better than the wiki page? Jan 21 at 17:23
• $XH$ is a subgroup if and only if $XH=HX$, that is, it suffices to have either $X$ or $H$ to be normal, but not necessarily, $XH$ can be a subgroup with both $X$ and $H$ not normal. Jan 21 at 17:27
• $\langle H,K \rangle$ is the subgroup of $G$ generated by $H$ and $K$, which is indeed the smallest subgroup containing $H \cup K$. $K$ normalizes $H$ means $K \le N_G(H) := \{ g \in G \mid g^{-1}Hg = K\}$. At the moment your question contains false statements and you need to edit it. I would guess that you might want to assume that $H$ and $K$ are normal subgroups of $G$, but really it is your responsibility to ask a question that makes sense. Jan 21 at 17:28

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.
• Since $X \le HX$ and $H \le HX$, we have $[K,X] \le [K,HX]$ and $[K,H] \le [K,HX]$, so $[K,X][K,H] \le [K,HX]$. Jan 22 at 15:23