# Iterated semi-direct products

Let $G$ be a finite group. Suppose that we can write $G= A \rtimes B$ and also $A = C \rtimes D$. Further suppose that C is normal in $G$ (not just in $A$). Then can we write $G = C \rtimes E$ where $E=G/C$? Of course, if $|C|$ and $[G:C]$ are relatively prime, then this follows from the Schur-Zassenhaus theorem.

We can of course write $G= A \rtimes B = (C \rtimes D) \rtimes B$ and we would like to have some kind of "associativity" of semi-direct product so that $G= C \rtimes (D \rtimes B)$. The question is whether the semi-direct product "$D \rtimes B$" actually makes sense (that $D$ and $B$ act on $C$ is clear).

I am mainly interested in whether this works in the case that we have the following extra hypotheses (i) $G$ is soluble, (ii) $C$ is a $p$-group, (iii) $D$ is cyclic of order prime to $p$, and (iv) $B$ is cyclic (the question is only interesting if $p$ divides $|B|$ due to (iii) and Schur-Zassenhaus). We could also strengthen (i) to (v) $C$ is an elementary abelian $p$-group. I don't know whether any of these extra hypotheses are helpful or not.


$$\begin{array}{c} & & 1 & & 1 & & \\ & & \da{} & & \da{} & & \\ & & C & \ra{=} & C\\ & & \da{} & & \da{} & & \\ 1 & \ra{} & A & \ra{} & G & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{\pi} & & \da{=} \\ 1 & \ra{} & D & \ra{} & E & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{} & & \\ & & 1 & & 1 & & \end{array}$$ in which all rows and columns are exact. We are assuming that all the short exact sequences split (including the last row), apart from the one containing $\pi$; we want to show that this one also splits. It is easy to see what the section $\varepsilon: E \rightarrow G$ should be in terms of the other sections, but unfortunately I can't seem to show that it is actually a homomorphism.

I have the feeling that there is either an easy solution or any easy counterexample satisfying the extra hypotheses above, so I am missing something either way. (I did try some small examples, which seemed to work.)

• No. Pick $G$ dihedral of order 8, $A$ a Klein subgroup of order 4 and exponent 2, $C\subset A$ the center of $G$ (so $G/C$ is also a Klein group). Then $C$ is not part of a semidirect decomposition. – YCor Jan 30 '14 at 20:00
• Thanks. But if we assume hypothesis (iii) that $D=A/C$ is cyclic of order prime to $p=2$ then this doesn't work. I'm looking for a proof or counterexample in the case that all the hypotheses (i)-(v) are satisfied; I just thought that a proof might not required them all. Sorry if I was unclear about this. – Henri Johnston Jan 30 '14 at 20:19
• ok, it was unclear that you required all these extra-hypotheses to be simultaneously fulfilled. – YCor Jan 30 '14 at 22:14

I believe that it must split if $C$ is abelian.

More precisely, I will prove that if $C$ is an abelian $p$-group and $D$ is a $p'$-group, then $C$ has a complement in $G$ and so $G \cong C \rtimes (D \rtimes B)$. We are assuming that $G$ is finite. Note that under these assumptions $C$ is a characteristic subgroup of $A$, and hence normal in $G$.

There is a general result that says that, if $Q$ is a $p'$-group acting on a finite abelian $p$-group $P$, then $P = C_P(Q) \times [P,Q]$. I don't have a reference to hand, but I know it is in Gorenstein's book on Finite Groups, and I can find the precise reference later. Note that this does not hold in general for nonabelian $p$-groups.

So we have $C=C_C(D) \times [C,D]$.

Now $A/[C,D]$ is the direct product $C/[C,D] \times D[C,D]/[C,D]$. Since $A/D[C,D] \cong C/[C,D]$ is abelian, we have $[A,A] \le D[C,D]$ and so $[C,D] = C \cap [A,A]$ is normal in $G$.

Since the direct factors $D[C,D]/[C,D]$ and $C/[C,D]$ of $A/[C,D]$ have coprime orders, they are both characteristic in $A/[C,D]$ and hence normal in $G/[C,D]$. In particular, $D[C,D]/[C,D]$ is normalized by the subgroup $B[C,D]/[C,D]$, which is a complement of $A/[C,D]$ in $G/[C,D]$.

Now the subgroup $BD[C,D]$ of $G$ has contains $B$ and $D$ and intersects $C$ in $[C,D]$. So it satisfies the original hypotheses, but with $C$ replaced by $[C,D]$. So if we now replace $G$ by $BD[C,D]$, we have $C_C(D)=1$.

Now let $N = N_G(D)$. Since, by the Schur-Zassenhaus Theorem, all complements of $C$ in $A$ are conjugate (to $D$), the Frattini Argument shows that $G=AN$, and hence, since $D \le N$, $G=CN$. Also $[C \cap N,D] \le C \cap D = 1$, so $C \cap N \le C_C(D) = 1$, and hence $C \cap N = 1$. So $N$ is a complement of $C$ in $G$. (It is also a complement of the original $C$ that we replaced by $[C,D]$.)

Note that this argument doesn't use the facts that $D$ and $B$ are cyclic, but it does use the assumption that $C$ and $D$ have coprime orders.

I do not know whether such an extension always splits when $C$ is allowed to be a nonabelian $p$-group.

• Many thanks for this. Unfortunately, I'm a bit slow today and my group theory is rather rusty, so there are several steps that I don't understand. Why is it that $C=C_C(D) \times [C,D]$ given that $C$ is abelian? Also, why is $[A,A] \cap C \leq [C,D]$? I do understand the other claims of the first paragraph. I see why $A/[C,D]$ is the direct product of $D[C,D]/[C,D]$ and $C/[C,D]$. However, I am confused by $D[C,D]/[C,D]$ is normalised by a complement $B[C,D]$. Should this be $B[C,D]/[C,D]$? If so, how does one prove this claim? – Henri Johnston Jan 31 '14 at 19:19
• I'm afraid that I don't see why the claims in the sentence starting "Hence..." are true. I do see why the normaliser of $D$ in $G$ does not intersect $C$ is true (given the previous claims), but I don't see why we then get the desired result (what forces $N_G(D)$ to be "big enough"?) Sorry for all these questions on what is probably basic stuff. – Henri Johnston Jan 31 '14 at 19:23
• I have added more detail. Let me know if anything still needs more explanation. – Derek Holt Feb 1 '14 at 9:40
• That's great - many thanks for the further details! – Henri Johnston Feb 1 '14 at 18:59
• Maybe the fact that Gaschütz's Theorem (see my answer below) doesn't hold when $C$ is not abelian will give a counterexample in this case? (But I didn't immediately find the counterexample, even without the cyclic constraints.) – Russ Woodroofe Feb 6 '14 at 2:05

Without answering your question directly, you may be interested in the paper: Multiple semidirect products of associative systems. Richard Steiner. Glasgow Mathematical Journal - GLASGOW MATH J 01/1989; 31(03).

Some of the questions you pose also are discussed in, and are relevant to, work on $cat^n$-groups in homotopy theory. Taking $cat^2$ groups (or equivalently crossed squares) in the sense of Loday, a simple example is a group $G$ together with two specified normal subgroups, $M$ and $N$, (together with their intersection). The commutator between the two normal subgroups yields a map $h: M\times N \to M\cap N$. (Abstracting this a crossed square has crossed modules, $M\to G$, $N\to G$ and then replacing $M\cap N$ by a more general $L$, two more crossed modules $L\to M$, $L\to N$ and an $h$-map, mimicking the earlier examples. From this set up one constructs a `big group' $(L\rtimes M)\rtimes (N\rtimes G)$ unambigously, i.e. you get an isomorphic group if you work out $(L\rtimes N)\rtimes (M\rtimes G)$.

This may seem distant from your specific question but the point is that the axioms of the h-map in a crossed square, encode the compatibility of the actions. This is related to your diagram espeically when $C$ is a central subgroup. Another useful source may be papers on the non-Abelian tensor products of groups, starting probably with:

R. Brown, D. L. Johnson and E. F. Robertson, Some computations of non-abelian tensor products of groups, J. Algebra, 111, (1987), 177 – 202.

Here again questions of the compatibility of the actions is central to the constructions.

Here's a short answer, assuming only that $C$ is an abelian $p$-group and $\vert D \vert$ is relatively prime to $p$.

Remember that $G \cong H\rtimes K$ iff normal subgroup $H$ has a complement in $G$ isomorphic to $K \cong G/H$.

A theorem of Gaschütz says that a normal abelian $p$-group $C$ has a complement in $G$ if and only if $C$ has a complement in a Sylow $p$-subgroup $P$ containing $C$. Since $( \vert D \vert, p)=1$, we have that $C$ is a Sylow $p$-subgroup of $A$. If $P$ is a Sylow $p$-subgroup of $B$, then by counting we get that $CP$ is a Sylow $p$-subgroup of $G$. Then $P$ is a complement to $C$ in $CP$, hence $C$ has a complement in $G$, as desired. $\square$

• By the way, you can find (a slightly more general version of) Gaschütz's Theorem as Theorem 3.3.2 of "The theory of finite groups" by Kurzweil and Stellmacher. The result is very useful in situations like this question, and in my opinion it should be more widely known. – Russ Woodroofe Feb 5 '14 at 0:48
• That's a very neat solution in the case that $C$ is abelian - many thanks for this! – Henri Johnston Feb 5 '14 at 10:20

Having looked at Derek's answer, I believe that there is another solution that does not assume that $C$ is abelian but does assume some further hypotheses (which may not be necessary, but are true for the application I have in mind).

To be clear, here is what I am assuming. Suppose that $G$ is finite, $C$ and $D$ are of relatively prime orders and that $B$ is cyclic. Also suppose that all the short exact sequences in the commutative diagram of the question are split, apart from the one containing $\pi$, which is the sequence that we want to show is split. Let $b$ be any generator of $B$. For clarity, let $D'$ denote a subgroup of $A$ that is the imaging of a splitting $D \rightarrow A$, and similarly $E'$ and $B'$ (i.e. $D', B'$ and $E'$ are all subgroups of $G$). EDIT: Suppose that any lift $b'$ of $b$ under the quotient map $G \rightarrow B$ has the same order as $b$. Then we can take $B'=\langle b' \rangle$ END EDIT (maybe this hypothesis always holds for some reason; I know that it holds in the application of interest to me).

Let $N=N_G(D')$. By the same reasoning as in Derek's answer, we have $G=AN$ (note that because we are using the part of the Schur-Zassenhaus Theorem regarding conjugate complements, we should assume that either $A$ or $D$ are soluble, but I am fine with this). So there exists $b' \in N$ that maps to $b$ under the quotient map $G \rightarrow B$. Now take $B'=\langle b' \rangle$, which we can do by the above hypothesis. Thus $B' \leq N=N_G(D')$ and so $E':=D'B'$ is a subgroup of $G$. But $A \cap B'=1$ and $D' \leq A$, so we have $D' \cap B'=1$ and so $|E'| = |B||D|$. Now $G=AB'=(CD')B'=C(D'B')=CE'$. Furthermore, $|G|=|C||E'|$, which with $G=CE'$ forces $C \cap E'=1$ . Thus $E'$ is the desired complement of $C$ in $G$.

Does this seem right? Can we get rid of the hypothesis about the choice of splitting $B \rightarrow G$?

• I don't see why $A\cap B' = 1$. We know that $b'$ and $b$ have the same image in $G/A$, but how do you know that the order of $b'$ is not larger than the order of $b$? Since $G/C$ is a split extension, we can assume that $A \cap B' \le C \cap N = C_N(D')$. – Derek Holt Feb 3 '14 at 22:21
• I agree that if $b'$ and $b$ don't have the same order, then this doesn't work. However, if we make the (strong) assumption that any lift $b'$ of $b$ to $G$ has the same order, then everything should work. This assumption does hold when one uses (the proof of) Lemma 2.6 in this preprint arxiv.org/abs/1204.2133. We take $G=Gal(L'/K)$, $A=G_0=Gal(L'/K')$, $B=Gal(K'/K)$, and $b=\sigma$ and $b'=\tau$ (any lift of $\sigma$), then $b$ and $b'$ both have the same order $d$. – Henri Johnston Feb 4 '14 at 10:40
• Sorry, I should have mentioned this motivation earlier, but I was hoping that there may be some purely group-theoretic proof; it seems that extra arithmetic input of the lemma is necessary. The whole reason for this is so that I can prove a stronger version of the lemma, which I need to fill a gap that I discovered later in the preprint. – Henri Johnston Feb 4 '14 at 10:43
• I realised that my assumptions weren't clearly stated - now edited in the above answer. – Henri Johnston Feb 4 '14 at 11:02