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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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.


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

share|cite|improve this answer

Your Answer


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.