commutator subgroup and semidirect product

Suppose $G$ is a solvable group such that $G = N \rtimes H$. Then I can show that $G' = M \rtimes H'$, where $G'$ is the commutator subgroup of $G$ and $M = N \cap G'$, $H' = H \cap G'$. I can also show that $H'$ is indeed the commutator subgroup of $H$. So $H$ and $H'$ are related without this relation explicitly depending on $G'$. Is this also true for $M$ and $N$, i.e. is there a relation between $M$ and $N$ that does not hinge on $G'$?

In the case of $N$ abelian of maximal order it might hold that $N \cong Z(G) \times M$, where $Z(G)$ is the center of $G$. At least it holds for the few groups I checked. I'm equally interested in the general and this special case.

• Also I have seen the notation [,] but I don't understand how I'm doing basic linear algebra. Would you mind to show me the connection – vuur Jul 16 '13 at 19:52
• Ok, I explained the comment sin a longer post. Deleting the comments, since stackexchange prefers answers to comments :-) – Jack Schmidt Jul 16 '13 at 20:46

Here is a fundamental example (if $G$ is finite solvable, then $G/\Phi(G)$ is of the form described here).

Let $H$ be a group of invertible $n \times n$ matrices over the field $\mathbb{Z}/p\mathbb{Z}$. Let $N$ be the elementary abelian group of order $p^n$, $N \cong C_p^n$. Then $H$ acts on $N$ by matrix multiplication. Let $G = N \rtimes H$.

Then $G' = [G,G] = [N,N][N,H][H,H] = [N,H] \rtimes H'$.

$[N,H] = \langle n^{-1} n^h : n \in N, h \in H \rangle$ in multiplicative notation, but in matrix notation, we just get $$[N,H] = \langle -n + n \cdot h : n \in N, h \in H \rangle = \langle n\cdot (h-1): n \in N, h \in H \rangle = \sum_{h \in H} \newcommand{\im}{\operatorname{im}}\im(h-1)$$

Coprime action

If $H$ has order coprime to $p$, then some fancy linear algebra shows that $\ker((h-1)^n) = \ker(h-1)$ and $\im(h-1)$ is a direct complement. In other words, by Fitting's lemma (applied to the semisimple operator $h-1$), we get $$N = \ker(h-1) \oplus \im(h-1) = C_{h}(N) \times [N,h]$$

Using some slightly fancier versions of these linear algebra ideas we even get $$N=\left( \bigcap_{h \in H} \ker(h-1) \right) \oplus \left(\sum_{h \in H} \im(h-1) \right) = C_H(N) \times [H,N]$$

Even if $N$ is not abelian similar ideas give: $N=C_H(N)[H,N]$, though the intersection may be non-identity.

Defining characteristic

If $H$ has order a power of $p$, then one gets sort of the opposite behavior. The minimum polynomial of $h$ divides $x^{p^n}-1 = (x-1)^{p^n}$, so every eigenvalue of $h-1$ is 0, and $h-1$ is nilpotent. Hence Fitting's lemma tells us that $N=\ker((h-1)^{p^n}) \times \im((h-1)^{p^n})$, but that is useless since $(h-1)^{p^n}=0$ and so the kernel is all of $N$ and the image is $1$.

If we try to apply this to $h-1$ directly without raising to the $p^n$th power, then things go very weird. Take $h=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Then $\im(h-1) = \{ (0,x) : x \in C_p \}$ but also $\ker(h-1) = \{ (0,x) :x \in C_p \}$. When $p=2$, this is the $D_8$ example.

If one wants larger $N$, then one can take $H=\langle h_1,h_2\rangle$ with $$h_1=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}, \qquad h_2=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$$ Then $\im(h_i-1)=\{ (0,0,x) :x \in C_p \}$ but $\ker(h_1-1) = \{ (0,y,z) : y,z \in C_p \}$ and $\ker(h_2-1) = \{ (x,0,z) : x,z \in C_p \}$ so $$\bigcap_{h \in H} \ker(h-1) = \{ (0,0,x) : x \in C_p \}$$ and $$\sum_{h \in H} \im(h-1) = \{ (0,0,x) : x \in C_p \}$$

When $p=2$, this is the $D_8 \operatorname{\sf Y} D_8$ example.

Notice how broken the decomposition is here.

References

Kurzweil–Stellmacher, Theory of Finite Groups, Chapter 8, is where this really started to make sense for me.