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Let the homomorphism $f:H \rtimes K \rightarrow K$ be defined by $f(hk)=k$.

Now, I will construct the homomorphism $f: [H \rtimes K ,H \rtimes K ] \rightarrow [K,K]$. How to find the kernel of $f$?. Is the kernel isomorphic with $[H,H]$?

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The kernel certainly contains $[H,H]$; but it is in general larger. For example, if $h\in H$ and $k\in K$ do not commute, then $[h,k]$ maps to $[1,k]=1$, but $[h,k]\notin[H,H]$. More generally, a generator of $[H\rtimes K,H\rtimes K]$ of the form $[hk,h'k']$ will lie in the kernel if and only if $[k,k'] = 1$. – Arturo Magidin Apr 13 '12 at 19:45

The kernel of the original map is $H$. Of course, the kernel of the restriction will be $H\cap[H\rtimes K,H\rtimes K]$; but this, in general, contains more than just $[H,H]$. Since $H\triangleleft H\rtimes K$, it also contains $[H,H\rtimes K]=\langle [h,h'k]\mid h,h'\in H,k\in K\rangle$, which is contained in $H$ by the normality of $H$, and certainly contained in $[H\rtimes K,H\rtimes K]$; this is nontrivial and strictly larger than $[H,H]$ unless the semidirect product is actually a direct product.

In fact, the kernel is exactly $[H,H\rtimes K]$. Indeed, it is easy to see that every generator of this group is contained in the kernel.

Conversely, suppose that $$[h_1k_1,h_2k_2]\cdots [h_{2m-1}k_{2m-1},h_{2m}k_{2m}]$$ is an element of the kernel. This means that $$[k_1,k_2]\cdots[k_{2m-1},k_{2m}]=1.$$

Using the identity $$[xy,zt] = [x,y]^y[y,t][x,z]^{yt}[y,z]^t$$ (note: my commutators are defined by $[a,b]=a^{-1}b^{-1}ab$) we can rewrite each $[h_{2i-1}k_{2i-1},h_{2i}k_{2i}]$ as $$[h_{2i-1},k_{2i}]^{k_{2i-1}}[k_{2i-1},k_{2i}][h_{2i-1},h_{2i}]^{k_{2i-1}k_{2i}}[k_{2i-1},h_{2i}]^{k_{2i}}.$$ Now, all terms except perhaps for $[k_{2i-1},k_{2i}]$ lie in $[H,H\rtimes K]$, and since $[H,H\rtimes K]$ is normal, we can rewrite $$[k_{2i-1},k_{2i}][h_{2i-1},h_{2i}]^{k_{2i-1}k_{2i}}[k_{2i-1},h_{2i}]^{k_{2i}}$$ as $$\alpha [k_{2i-1},k_{2i}]\text{ for some }\alpha\in [H,H\rtimes K].$$ Repeating this, starting from the rightmost factor and working towards the left, we can rewrite $$[h_1k_1,h_2k_2]\cdots [h_{2m-1}k_{2m-1},h_{2m}k_{2m}]$$ as $$x[k_1,k_2][k_3,k_4]\cdots[k_{2m-1},k_{2m}]$$ for some $x\in[H,H\rtimes K]$. And since $[k_1,k_2]\cdots[k_{2m-1},k_{2m}]=1$ by assumption, this proves that $$[h_1k_1,h_2k_2]\cdots [h_{2m-1}k_{2m-1},h_{2m}k_{2m}]\in [H,H\rtimes K]$$ as desired.

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But we do not have $[H,H] \le [H,K]$ in general. For example, in a direct product $[H,K]$ is trivial, but the kernel still contains $[H,H]$. It is also possible to have $[H,H]=[H,K]$ without it being a direct product. – Derek Holt Apr 14 '12 at 8:22
@Derek: I think I messed up when I wrote $[H,K]$; I keep thinking $K$ is the whole group, whereas it should be $[H,H\rtimes K]$. Thank you. – Arturo Magidin Apr 14 '12 at 19:34

The kernel of $f$ is $H$. Since $[xy,z]=[x,z]^y[y,z]$, and $G=HK$, we have $$ [G,G]=[HK,HK]=[H,HK][K,HK]=[H,H][H,K][K,K].$$

So the kernel of your restriction is obviously $H\cap [H,H][H,K][K,K]$. Dedekind's Lemma implies this is the same as $[H,H](H\cap [H,K][K,K])$, and applying Dedekind's Lemma once again reduces this to $[H,H][H,K](H\cap [K,K])$. But $H\cap K=\lbrace1\rbrace$, so the kernel you're after is just $[H,H][H,K]$.

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