Systems of imprimitivity for wreath products

Let $x,y\in G$ be as follows:

Then $K=\langle x,y\rangle\cong D_{12}$, with $K/[K,K]=\langle\overline{x},\ \overline{y}\rangle\cong C_{2}\times C_{2}$. Let $W$ be the $1$-dimensional $\mathbb{F}_{q}[K]$-module corresponding to the linear character $\theta:K\rightarrow\mathbb{F}_{q}^{\times}$ such that $\theta(x)=1$ and $\theta(y)=-1$. Consider the induced $\mathbb{F}_{q}[G]$-module $V=\operatorname{Ind}_{K}^{G}(W)$. We have $\dim V=[G:K]=4$, and a calculation shows that $V$ is a faithful irreducible $\mathbb{F}_{q}[G]$-module, so claim (i) holds. Since we are inducing a $1$-dimensional module, it is clear that we get a decomposition $V=Z_{1}\oplus Z_{2}\oplus Z_{3}\oplus Z_{4}$ as in (ii), which is nonrefinable since $\dim Z_{i}=1$.

Let $H=\operatorname{SL}_{2}(3)$, so $[G:H]=2$ and $H\trianglelefteq G$. Note that by Maschke’s theorem $\mathbb{F}_{q}[H]$ is semisimple. Thus by examining the ordinary character table of $H$, we can see that $\mathbb{F}_{q}$ is a splitting field for $H$, since it contains a primitive cube root of unity. Then by looking at the character degrees, we conclude that there is no irreducible $\mathbb{F}_{q}[H]$-module of dimension $4$.

In particular, the restriction of $V$ to $H$ is not irreducible. Thus by Clifford theory, the restriction decomposes as

where $W_{1}$, $W_{2}$ are non-isomorphic irreducible $\mathbb{F}_{q}[H]$-modules with $\dim W_{i}=2$. Then $G$ acts on $\{W_{1},W_{2}\}$ and claim (iii) holds.

What remains is to check that $V=W_{1}\oplus W_{2}$ provides a nonrefinable system of imprimitivity for $G$. Equivalently, we need to check that the action of $H$ on $W_{1}$ is primitive, but this is immediate from the fact that $H$ does not have a subgroup of index $2$.∎

Note that the $W_{i}$ are irreducible and pairwise non-isomorphic $\mathbb{F}[M]$-modules, so by Lemma 3.1 any $\mathbb{F}[M]$-submodule of $V$ is a direct sum $W_{i_{1}}\oplus\cdots\oplus W_{i_{\alpha}}$ for some $1\leq i_{1}<\cdots<i_{\alpha}\leq k$. (We will use this fact throughout the proof.)

For the proof of the lemma, we proceed by induction on $k$. In the case $k=1$, if $V=Q_{1}\oplus\cdots\oplus Q_{\ell}$ and $M$ acts on $\{Q_{1},\ldots,Q_{\ell}\}$, then $\ell=1$ since $M=H_{1}$ acts primitively on $V=W_{1}$. Suppose then that $k>1$.

Consider first the case where $M$ is not transitive on $\{Q_{1},\ldots,Q_{\ell}\}$. Let

be the orbits of $M$ on $\{Q_{1},\ldots,Q_{\ell}\}$. Then

where by Lemma 3.1, for all $1\leq i\leq s$ we have

for some subset $\{W_{1}^{(i)},\ldots,W_{\alpha_{i}}^{(i)}\}$ of $\{W_{1},\ldots,W_{k}\}$. Now the action of the direct product $H_{1}^{(i)}\times\cdots\times H_{\alpha_{i}}^{(i)}$ on $W_{1}^{(i)}\oplus\cdots\oplus W_{\alpha_{i}}^{(i)}$ satisfies conditions (i) – (iii) of the lemma, so $d_{i}\leq\alpha_{i}$ for all $i$ by induction. Since

we conclude that $\ell\leq k$.

Thus we can assume that $M$ acts transitively on $\{Q_{1},\ldots,Q_{\ell}\}$. Let $k_{0}>0$ be minimal such that

for some $1\leq j\leq\ell$ and $1\leq j_{1}<\cdots<j_{k_{0}}\leq k$.

For $1\leq i\leq\ell$, set $Q_{i}^{\prime}:=Q_{i}\cap(W_{j_{1}}\oplus\cdots\oplus W_{j_{k_{0}}})$. Then $M$ acts on $\{Q_{1}^{\prime},\ldots,Q_{\ell}^{\prime}\}$, so $Q_{1}^{\prime}\oplus\cdots\oplus Q_{\ell}^{\prime}$ is an $\mathbb{F}[M]$-submodule of $W_{j_{1}}\oplus\cdots\oplus W_{j_{k_{0}}}$. By Lemma 3.1 and the minimality of $k_{0}$, we have in fact

If $k_{0}<k$, then by induction we have $\ell\leq k_{0}$ and so $\ell<k$. Thus we can assume that $k_{0}=k$, so

for all $1\leq i\leq\ell$ and $1\leq j_{1}<\cdots<j_{k-1}\leq k$. In particular, the projection of $Q_{j}$ into any $W_{i}$ is injective, so

for all $i$ and $j$.

Next let $s>0$ be minimal such that

for some $1\leq i\leq k$ and $1\leq i_{1}<\cdots<i_{s}\leq\ell$.

We will show that for all $j\neq i$, the subgroup $H_{j}$ acts on $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}$ nontrivially. Let $h\in H_{j}$. We have $h(Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}})=Q_{i_{1}^{\prime}}\oplus\cdots\oplus Q_{i_{s}^{\prime}}$ for some $1\leq i_{1}^{\prime}<\cdots<i_{s}^{\prime}\leq\ell$. Since $H_{j}$ acts trivially on $W_{i}$, it follows that

Thus $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}=Q_{i_{1}^{\prime}}\oplus\cdots\oplus Q_{i_{s}^{\prime}}$ by the minimality of $s$, so $H_{j}$ acts on $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}$.

To see that the action is nontrivial, let $v\in Q_{i_{1}}$ be nonzero. By (3.1), we have $v=w_{1}+\cdots+w_{k}$ where $w_{r}\in W_{r}$ and $w_{r}\neq 0$ for all $1\leq r\leq k$. Since $W_{j}$ is a nontrivial irreducible $\mathbb{F}[H_{j}]$-module, we have $gw_{j}\neq w_{j}$ for some $g\in H_{j}$. Then $gv\neq v$, so $H_{j}$ acts nontrivially on $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}$. Consequently $W_{j}$ must be contained in $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}$. In particular $W_{j}\cap(Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}})\neq 0$, so by repeating the same arguments we conclude that $W_{i}$ is also contained in $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}$.

Therefore $Q_{i_{1}}\oplus\cdots\oplus Q_{i_{s}}=W_{1}\oplus\cdots\oplus W_{k}$, so $s=\ell$ and $W_{i}\cap(Q_{i_{1}}\oplus\cdots\oplus Q_{i_{\ell-1}})=0$ for all $i$ and $1\leq i_{1}<\cdots<i_{\ell-1}\leq\ell$. Hence the projection of $W_{i}$ into any $Q_{j}$ is injective, so $\dim W_{i}\leq\dim Q_{j}$ for all $i$ and $j$. By (3.2) we conclude that $\dim Q_{j}=\dim W_{i}$ for all $i$ and $j$, from which it follows that $\ell=k$. This completes the proof of the lemma.∎

Let $\{W_{1},\ldots,W_{k}\}$ be the system of imprimitivity defining $G$. Then $V=W_{1}\oplus\cdots\oplus W_{k}$ and $G=(H_{1}\times\cdots\times H_{k})\rtimes K$, where the action of $H_{i}$ is nontrivial irreducible primitive on $W_{i}$, and trivial on $W_{j}$ for $j\neq i$. Furthermore, the action of $K$ on $\{W_{1},\ldots,W_{k}\}$ is faithful and transitive. We denote the base group $H_{1}\times\cdots\times H_{k}$ by $M$.

Suppose that there is another nonrefinable system of imprimitivity, say $V=Z_{1}\oplus\cdots\oplus Z_{\ell}$ such that $\ell>1$ and $G$ acts on $\{Z_{1},\ldots,Z_{\ell}\}$. Since $G$ is irreducible, the action on $\{Z_{1},\ldots,Z_{\ell}\}$ must be transitive. Furthermore, the action of $N_{G}(Z_{i})$ on $Z_{i}$ must be irreducible and primitive [Sup76, Theorem 15.1].

Let $s>0$ be minimal such that $Z_{i}\cap(W_{j_{1}}\oplus\cdots\oplus W_{j_{s}})\neq 0$ for some $1\leq i\leq\ell$ and $1\leq j_{1}<\cdots<j_{s}\leq k$. First consider the case where $s=1$, so $Z_{i}\cap W_{j}\neq 0$ for some $i$ and $j$. Then

is a non-zero $N_{G}(W_{j})$-submodule of $W_{j}$. Since $N_{G}(W_{j})$ acts irreducibly on $W_{j}$, we have $W_{j}=(Z_{1}\cap W_{j})\oplus\cdots\oplus(Z_{\ell}\cap W_{j})$. Furthermore, the action of $N_{G}(W_{j})$ is primitive, so $W_{j}=Z_{i}\cap W_{j}$. Repeating this argument for $Z_{i}$, we see that $Z_{i}=Z_{i}\cap W_{j}$, so $Z_{i}=W_{j}$ and $\{W_{1},\ldots,W_{k}\}=\{Z_{1},\ldots,Z_{\ell}\}$.

Therefore we can suppose that $s>1$ in what follows. Let $\{Z_{i_{1}},\ldots,Z_{i_{r}}\}$ be the orbit of $Z_{i}$ under the base group $M$. For $1\leq t\leq r$, set

Then $Q_{1}\oplus\cdots\oplus Q_{r}$ is an $\mathbb{F}[M]$-submodule of $W_{j_{1}}\oplus\cdots\oplus W_{j_{s}}$, so by Lemma 3.1 and the minimality of $s$ we conclude that

Let $v\in Z_{i}\cap(W_{j_{1}}\oplus\cdots\oplus W_{j_{s}})$ be non-zero, and write $v=w_{1}+\cdots+w_{s}$, where $w_{t}\in W_{j_{t}}$. Note that each $w_{t}$ is non-zero by the minimality of $s$. For $h_{t}\in H_{j_{t}}$ ($1\leq t\leq s-1$), we define

Since $v\in Z_{i}$ and $M$ acts on $W_{j_{1}}\oplus\cdots\oplus W_{j_{s}}\subseteq Z_{i_{1}}\oplus\cdots\oplus Z_{i_{r}}$, each $v_{h_{1},\ldots,h_{s-1}}$ is contained in some $Z_{i_{t}}$.

We claim that $v_{h_{1},\ldots,h_{s-1}}$ and $v_{h_{1}^{\prime},\ldots,h_{s-1}^{\prime}}$ can be contained in the same $Z_{i_{t}}$ only if they are equal. Indeed, if $v_{h_{1},\ldots,h_{s-1}}$ and $v_{h_{1}^{\prime},\ldots,h_{s-1}^{\prime}}$ are both contained in $Z_{i_{t}}$, then

is contained in $Z_{i_{t}}\cap(W_{j_{1}}\oplus\cdots\oplus W_{j_{s-1}})$, and thus must be zero by the minimality of $s$.

It follows then that $r\geq|\Pi_{1}|\cdots|\Pi_{s-1}|$, where $\Pi_{t}$ is the $H_{j_{t}}$-orbit of $w_{t}$. Note that $|\Pi_{t}|\geq 2$ for all $1\leq t\leq s-1$, since each $w_{t}$ is nonzero, and since $H_{j_{t}}$ acts nontrivially. Furthermore, we have $r\leq s$ by Lemma 3.2, so

which forces $s=r=2$ and $|\Pi_{1}|=2$.

Write $\Pi_{1}=\{w_{1},w_{1}^{\prime}\}$. Then $w_{1}+w_{1}^{\prime}$ is fixed by the action of $H_{j_{1}}$ and thus $w_{1}+w_{1}^{\prime}=0$. Hence $hw_{1}=\pm w_{1}$ for all $h\in H_{j_{1}}$. We conclude then from the irreducibility of $H_{j_{1}}$ that $\dim W_{j}=1$ for all $j$, and furthermore $|H|=2$. Note that this also forces $\operatorname{char}\mathbb{F}\neq 2$.

To complete the proof of the theorem, it remains to show that $n$ is even and $K\leq C_{2}\wr S_{n/2}$. To this end, we first adapt an argument from [ST54, p. 276] to show that $\dim Z_{i}=1$. Suppose, for the sake of contradiction, that $\dim Z_{i}>1$. Let $h\in H_{j_{1}}$ be such that $hw_{1}=-w_{1}$. Since $h$ acts trivially on $W_{t}$ for $t\neq j_{1}$, it follows that the fixed point space $V^{h}$ has dimension $n-1$. Thus $Z_{i}$ has nonzero intersection with $V^{h}$, which implies that $hZ_{i}=Z_{i}$ since $G$ acts on $\{Z_{1},\ldots,Z_{\ell}\}$. So then both $v=w_{1}+w_{2}$ and $hv=-w_{1}+w_{2}$ would be contained in $Z_{i}$, which implies that $v+hv=2w_{2}\in Z_{i}$. Thus $w_{2}\in Z_{i}$ since $\operatorname{char}\mathbb{F}\neq 2$, so we have a contradiction due to $Z_{i}\cap W_{j_{2}}=0$.

Therefore we have $\ell=k=n$ and $\dim Z_{j}=1$ for all $1\leq j\leq n$. Note that now $W_{j_{1}}\oplus W_{j_{2}}=Z_{i_{1}}\oplus Z_{i_{2}}$, and $\{Z_{i_{1}},Z_{i_{2}}\}$ is an orbit for the action of $M$ on $\{Z_{1},\ldots,Z_{n}\}$. Since $M$ is a normal subgroup of $G$ and since $G$ acts transitively on $\{Z_{1},\ldots,Z_{n}\}$, every $M$-orbit is of order $2$, and so $n$ is even. By relabeling the summands if necessary, we can assume that the $M$-orbits are

and furthermore that we have

Thus $G$ acts on the set of pairs $\{\{W_{1},W_{2}\},\ldots,\{W_{n-1},W_{n}\}\}$, which shows that $K\leq C_{2}\wr S_{n/2}$.∎

If (ii) holds, then $G$ is not maximal solvable, since $G=H\wr(X\wr Y)=(H\wr X)\wr Y<H_{0}\wr Y$ for some maximal irreducible primitive solvable subgroup $H_{0}$ of $\operatorname{GL}_{d\ell^{\prime}}(\mathbb{F})$. For the other direction, suppose that $G$ is not maximal solvable. By a theorem of Zassenhaus [Zas37, Satz 8], there exists a maximal solvable subgroup $G_{0}\leq\operatorname{GL}_{n}(\mathbb{F})$ that contains $G$.

If $G_{0}$ is primitive, then (ii) holds with $X=K$ and $Y=1$. Suppose then that $G_{0}$ is imprimitive. In this case, by [Sup76, Theorem 15.4] we have $G_{0}=H_{0}\wr K_{0}$ for some $H_{0}\leq\operatorname{GL}_{e}(\mathbb{F})$ maximal irreducible primitive solvable and $K_{0}\leq S_{\ell}$ maximal transitive solvable, where $n=e\ell$ for $\ell>1$.

We assume first that $G$ is not as in the exceptional case of Theorem 1.1. Write $\mathbb{F}^{n}=W_{1}\oplus\cdots\oplus W_{k}=Z_{1}\oplus\cdots\oplus Z_{\ell},$ where $\{W_{1},\ldots,W_{k}\}$ and $\{Z_{1},\ldots,Z_{\ell}\}$ are the systems of imprimitivity defining $G$ and $G_{0}$, respectively. Applying Theorem 1.1 to $G$, it follows that $\{W_{1},\ldots,W_{k}\}$ must be a refinement of $\{Z_{1},\ldots,Z_{\ell}\}$. In other words, we conclude that $\ell$ divides $k$, and for all $1\leq i\leq\ell$ we have

for some subset $B_{i}:=\{W_{1}^{(i)},\ldots,W_{k/\ell}^{(i)}\}$ of $\{W_{1},\ldots,W_{k}\}$.

Therefore the sets $\{B_{1},\ldots,B_{\ell}\}$ form a block system for the action of $K$ on $\{W_{1},\ldots,W_{k}\}$, so $K$ is a subgroup of $X\wr Y$, where $X\leq S_{k/\ell}$ is the action of $N_{K}(B_{1})$ on $B_{1}$, and $Y\leq K_{0}$ is the action of $K$ on $\{Z_{1},\ldots,Z_{\ell}\}$. Furthermore, in this case we have $H\wr X\leq H_{0}$. By the maximality of $K$ we must have $K=X\wr Y$ with $X$ and $Y$ maximal transitive solvable, so (ii) holds.

What remains then is to consider the exceptional case of Theorem 1.1, in which case $n=k$, $d=1$, and $H\leq\operatorname{GL}_{1}(\mathbb{F})$ is cyclic of order $2$. Furthermore, in this case $n$ is even and $K$ is a transitive subgroup of $C_{2}\wr S_{n/2}$. Since $H$ is assumed to be maximal solvable, we have $H=\operatorname{GL}_{1}(\mathbb{F})$, so $\mathbb{F}=\mathbb{F}_{3}$ and $H=\{\pm 1\}$. Now $K$ normalizes the elementary abelian base group $C_{2}^{n/2}$ of $C_{2}\wr S_{n/2}$, so by maximality $K$ must contain $C_{2}^{n/2}$. Thus $K=C_{2}\wr T$ for some maximal transitive solvable subgroup $T$ of $S_{n/2}$. Since $\operatorname{GL}_{2}(\mathbb{F})=\operatorname{GL}_{2}(3)$ is solvable, we conclude that (ii) holds with $X=C_{2}$ and $Y=T$.

∎

If conditions (i) – (iii) hold, it is clear that $H_{1}\wr K_{1}\leq H_{1}\wr(X\wr Y)=(H_{1}\wr X)\wr Y\leq H_{2}\wr K_{2}.$ The other direction of the claim follows from Theorem 1.1, by arguing as in the proof of Corollary 4.1 (paragraphs 3–4).∎