Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $[H,H]$?

share|improve this question
    
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
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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 $[H,H][H,K]$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.