Find the kernel of a homomorphism 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 $[H,H]$?
 A: 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. 
A: 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 $[H,H][H,K]$.
