Fusion systems and amalgams

Abstract

We study reduced fusion systems from the point of view of their essential subgroups, using the classification by Goldschmidt and Fan of amalgams of prime index to analyze certain pairs of such subgroups. Our results are applied here to study reduced fusion systems over $2$-groups of order at most 64, and also reduced fusion systems over $2$-groups having abelian subgroups of index two. More applications are given in later papers.

Appendices

Appendix A: Background on groups

We list here some elementary results about finite groups which are needed throughout the paper, beginning with a well known property of $p$-groups.

Lemma 6.1

If $Q<P$ are finite $p$-groups for some prime $p$, then $Q<N_P(Q)$.

Proof

See, e.g., [29, Theorem 2.1.6].$\square$

We next look at automorphisms of finite $p$-groups.

Lemma 6.2

Fix a prime $p$, a finite $p$-group $P$, a subgroup $P_0\le \mathrm{Fr }(P)$, and a sequence of subgroups

$$\begin{aligned} P_0\trianglelefteq {}P_1\trianglelefteq \cdots \trianglelefteq P_k=P. \end{aligned}$$

Set

$$\begin{aligned} \mathcal{A }= \bigl \{\alpha \in \mathrm{Aut }(P)\,\big |\, x^{-1}\alpha (x)\in {}P_{i-1}, \text{ all} x\in {}P_i, \text{ all} i=1,\dots ,k \bigr \} \le \mathrm{Aut }(P)\!: \end{aligned}$$

the group of automorphisms which leave each $P_i$ invariant, and which induce the identity on each quotient group $P_i/P_{i-1}$. Then $\mathcal{A }$ is a $p$-group. If the $P_i$ are all characteristic in $P$, then $\mathcal{A }\trianglelefteq \mathrm{Aut }(P)$, and hence $\mathcal{A }\le {}O_p(\mathrm{Aut }(P))$.

Proof

See, for example, [19, Theorems 5.1.4 & 5.3.2].$\square$

As an easy exercise, Lemma 6.2 implies the following corollary, which contains a list of some 2-groups whose automorphism groups are 2-groups. Note, in the last case, that $P\cong C_2\times D_8$ contains a unique (hence characteristic) subgroup $Q\cong C_2\times C_4$, so that $\mathrm{Fr }(P)<Z(P)<Q<P$ is a chain of characteristic subgroups.

Corollary 6.3

For a finite $2$-group $P$, $\mathrm{Aut }(P)$ is a $2$-group if at least one of the following hold:

• (a) $P/[P,P]\cong {}C_{2^{k_1}}\times {}C_{2^{k_2}}\times \cdots \times C_{2^{k_r}}$, where $k_1,\ldots ,k_r$ are pairwise distinct.

• (b) $P$ is dihedral of order at least $8$, or semidihedral or quaternion of order at least $16$.

• (c) $P\cong C_2\times D_8$.

The next two results involve the intersection of a Sylow subgroup with the commutator subgroup.

Lemma 6.4

For any finite group $G$ with $S\in \mathrm{Syl }_{p}(G)$, $S\cap {}O^p(G)\le S\cap [G,G]$, with equality if $G/O^p(G)$ is abelian.

Proof

Set $G^{\prime }=[G,G]$. Since $G/G^{\prime }$ is abelian, its largest $p$-group quotient $G/O^p(G)G^{\prime }$ is isomorphic to its Sylow $p$-subgroup $SG^{\prime }/G^{\prime }\cong {}S/(S\cap {}G^{\prime })$ (and the isomorphism is induced by the inclusion $S\le {}G$). Hence $S\cap {}G^{\prime }=S\cap {}O^p(G)G^{\prime }$, and so $S\cap {}O^p(G)\le S\cap {}G^{\prime }$ with equality if $G^{\prime }\le O^p(G)$.$\square$

The following proposition goes essentially back to Schur [28, IX–X].

Proposition 6.5

Fix a finite group $G$ with $S\in \mathrm{Syl }_{p}(G)$. Let $Z\le {}Z(G)$ be a $p$-subgroup. Then $Z\cap [G,G]=Z\cap [S,S]$.

Proof

This follows as an application of the transfer in (co)homology. See, e.g., [22, Satz IV.2.2].$\square$

The next lemma describes nonabelian 2-groups with abelian subgroup of index two.

Lemma 6.6

Let $S$ be a finite nonabelian $2$-group containing an abelian subgroup $A\trianglelefteq {}S$ of index two. Then the following hold.

• (a) $[S,S]\cong {}A/Z(S)$, and all elements of $(S/Z(S)){\backslash }(A/Z(S))$ have order two.

• (b) If $|[S,S]|=2$, then $S/Z(S)\cong {}C_2^2$, and $S$ contains exactly three abelian subgroups of index two.

• (c) If $|[S,S]|\ge 4$, then $|S/Z(S)|\ge 8$, and $A$ is the unique abelian subgroup of index two in $S$.

• (d) If $S$ contains three abelian subgroups of index two which are permuted transitively by some automorphism of $S$, then either $Z(S)$ is not a direct factor of $A$, or $[S,S]\le \mathrm{Fr }(Z(S))$.

Proof

(a) For each $x\in {}S{\backslash }A$, $x^2\in {}C_A(x)=Z(S)$, and thus $xZ(S)$ has order two in $S/Z(S)$. Also, $[S,S]=[x,A]$ is the image of $\mathrm{Id }{-}c_x$ as a homomorphism from $A$ to itself, and $Z(S)$ is its kernel. Hence $[S,S]\cong {}A/Z(S)$.

(b) Assume $|[S,S]|=2$. Then by (a), $|S/Z(S)|=4$ and $S/Z(S)\cong C_2^2$. Each abelian subgroup of index two in $S$ contains $Z(S)$, and since $|S:Z(S)|=4$, each subgroup of index two in $S$ which contains $Z(S)$ is abelian. So there are exactly three abelian subgroups of index two in $S$.

(c) Now assume $|[S,S]|\ge 4$, so $|S/Z(S)|\ge 8$ by (a). If $B\ne {}A$ is another abelian subgroup of index two in $S$, then $AB=S$, so $Z(S)\ge A\cap {}B$, and $|S/Z(S)|\le 4$, a contradiction. Thus $A$ is the only abelian subgroup of index two.

(d) Assume $S$ contains three abelian subgroups $A=A_1,A_2,A_3$ of index two which are permuted transitively by some automorphism of $S$. Thus $|[S,S]|=2$ and $S/Z(S)\cong {}C_2^2$ by (b,c). Fix a generator $z\in [S,S]$. If $Z(S)$ is a direct factor of $A=A_1$, then it is a direct factor of each $A_i$ since $\mathrm{Aut }(S)$ acts transitively on the $A_i$, and there are elements $a_i\in {}A_i{\backslash }Z(S)$ of order two. Then $a_1a_2a_3\in {}Z(S)$ since the $a_i$ represent the three nonidentity elements in $S/Z(S)\cong {}C_2^2$, $[a_i,a_j]=z$ for distinct $i,j\in \{1,2,3\}$ since $S=\langle {a_i,a_j,Z(S)}\rangle$ is nonabelian, and so $(a_1a_2a_3)^2=z^3=z$. Thus $z\in \mathrm{Fr }(Z(S))$. $\square$

The following result about actions on abelian 2-groups is very useful in certain situations.

Lemma 6.7

Fix a finite abelian $2$-group $A$ and a subgroup $G\le \mathrm{Aut }(A)$ with Sylow subgroup $S\in \mathrm{Syl }_{2}(G)$ of order two. Assume $S\nleq {}Z(G)$, and $[S,A]\cong {}C_{2^n}$ for some $n\ge 1$. Then there are unique factorizations $A=A_0\times {}A_1$ and $G=G_0\times {}G_1$ such that $|G_0|$ is odd, $G_1\cong \Sigma _3$, $A_1\cong {}C_{2^n}\times {}C_{2^n}$, and for $i=1,2$, $G_i$ sends $A_i$ to itself and centralizes (i.e., acts trivially on) $A_{3-i}$.

Proof

See, e.g., [24, Proposition 2.3].$\square$

Lemma 6.8

Fix a prime $p$, and a finite group $G$ with subgroups $A\trianglelefteq {}B\trianglelefteq {}G$, both normal in $G$, such that $A$ is an abelian $p$-group, $B/A$ has order prime to $p$, and $C_A(B)=1$. Then $G$ splits as a semidirect product $G=A\overset{{}}{\rtimes }H$, where $H\cong G/A$.

Proof

By the spectral sequence for the extension $1\rightarrow B/A\rightarrow G/A \rightarrow G/B\rightarrow 1$, $H^i(G/A;A)=0$ for each $i\ge 0$ since $H^0(B/A;A)=C_A(B)=0$ (and since $(|B/A|,|A|)=1$). In particular, $G\cong A\overset{{}}{\rtimes }(G/A)$.

Alternatively, since $|B/A|$ is prime to $|A|$, by the Schur-Zassenhaus theorem [19, Theorem 6.2.1], there is $K\le B$ such that $KA=B$ and $K\cap {}A=1$, and each such subgroup is $B$-conjugate, and hence $A$-conjugate, to $K$. So $G=AN_G(K)$ by a Frattini argument, and since $N_A(K)=C_A(K)=1$, $N_G(K)$ is a complement to $A$ in $G$.$\square$

Appendix B: Finite $2$-groups with normal dihedral or quaternion subgroups

We prove here some elementary results about certain finite 2-groups which have normal dihedral or quaternion subgroups, and their automorphisms. We begin by stating the following general proposition about automorphisms of products of $p$-groups.

Proposition 7.1

[24, Proposition 3.2(a)] Fix a pair of finite $p$-groups $S_1$ and $S_2$, set $S=S_1\times {}S_2$, and let ${\mathrm{pr}}_i\in \mathrm{Hom }(S,S_i)$ be the projection. Let $\alpha \in \mathrm{Aut }(S)$ be such that $\alpha (\Omega _1(Z(S_1)))=\Omega _1(Z(S_1))$. Then for $i=1,2$, ${\mathrm{pr}}_i(\alpha (S_i))=S_i$ and $\alpha (S_iZ(S_{3-i}))=S_iZ(S_{3-i})$.

We next look at semidirect products with normal dihedral subgroup.

Lemma 7.2

Fix a finite $2$-group $S$, and subgroups $\Delta \trianglelefteq {}S$ and $T\le {}S$. Assume that $S=T\Delta$, $T\cap {}\Delta =1$, $\Delta$ is dihedral of order at least $8$, and $[T,\Delta _0]=1$ for some dihedral subgroup $\Delta _0\le {}\Delta$ of order $8$. Let $A,\Delta _1,\Delta _2\le {}\Delta$ be the three subgroups of index two where $A$ is cyclic, and let $A_0=[\Delta ,\Delta ]$ be the subgroup of index two in $A$. Set $Z=Z(\Delta )$ for short. Fix $b\in {}\Delta _0{\backslash }A$.

• (a) Assume $\varphi \in \mathrm{Hom }(\Delta ,S)$ is such that $\varphi (Z)=Z$ and $\varphi (A)\trianglelefteq {}S$. Then $\varphi (A_0)=A_0$, $\varphi (A)\le {}TA$, and $\varphi (b)\in {}TAb$.

• (b) Assume $\alpha \in \mathrm{Aut }(S)$ is such that $\alpha (Z)=Z$. Then either $\alpha$ sends each of the subgroups $T\Delta _1$ and $T\Delta _2$ to itself or it exchanges them.

Proof

Fix a generator $a\in {}A$; thus $A_0=\langle {a^2}\rangle$. Since $\Delta \trianglelefteq {}S$ and $A$ is characteristic in $\Delta$, $A\trianglelefteq {}S$. For all $t\in {}T$, $[t,\Delta _0]=1$ by assumption, so $[t,b]=1$, and $[t,a^k]=1$ if $|a^k|\le 4$. Hence $tat^{-1}=a^{4j+1}$ for some $j$, so $[t,a]\in \langle {a^4}\rangle$. Thus $[T,\Delta ]\le \langle {a^4}\rangle$, and so $\langle {T,a^4}\rangle \trianglelefteq {}S$.

(a) Assume $\varphi \in \mathrm{Hom }(\Delta ,S)$ is such that $\varphi (Z)=Z$ and $\varphi (A)\trianglelefteq {}S$. Since $A_0=[\Delta ,\Delta ]$, $\varphi (A_0)\le [S,S]\le {}TA_0$. If $\varphi (A_0)\le \langle {T,a^4}\rangle$, then there are distinct elements $a^i\ne {}a^j$ in $A_0$ such that $\varphi (a^i),\varphi (a^j)\in {}Ta^k$ for some $k$. Then $1\ne {}a^{i-j}$ and $\varphi (a^{i-j})\in {}T$, which is impossible since $Z\le \langle {a^{i-j}}\rangle$ and $\varphi (Z)=Z\nleq T$.

Thus $\varphi (a^2)\notin \langle {T,a^4}\rangle$, and the image of $\varphi (a)$ in $S/\langle {T,a^4}\rangle \cong {}D_8$ has order four. Hence $\varphi (a)=ta^i$ for some $t\in {}T$ and some odd $i$. Also, the image of $\varphi (b)$ in $S/\langle {T,a^4}\rangle$ must invert that of $\varphi (a)$, and so $\varphi (b)\in {}TAb$. Since $\varphi (A)\trianglelefteq {}S$, $b(ta^i)b^{-1}=ta^{-i}$, and $(ta^{-i})^{-1}ta^{i}=a^{2i}$, we have $a^{2i}\in \varphi (A)$. Thus $A_0=\langle {a^{2i}}\rangle \le \varphi (A)$, so $\varphi (A_0)=A_0$. This finishes the proof of (a).

(b) Now assume $\alpha \in \mathrm{Aut }(S)$ is such that $\alpha (Z)=Z$. By (a), applied with $\varphi =\alpha |_{\Delta }$, $\alpha (A)\le TA$ and $\alpha (A_0)=A_0$. Hence $\alpha (\langle {a^4}\rangle )=\langle {a^4}\rangle$, and $\alpha$ induces an automorphism $\bar{\alpha }$ of $S/\langle {a^4}\rangle \cong {}T\times {}D_8$ which sends $Z(\Delta /\langle {a^4}\rangle )=A_0/\langle {a^4}\rangle$ to itself. By Proposition 7.1, $\bar{\alpha }(TA_0/\langle {a^4}\rangle )=TA_0/\langle {a^4}\rangle$, and thus $\alpha (TA_0)=TA_0$. Since $\alpha (TA)=TA$, it now follows that $\bar{\alpha }$ either sends the two subgroups $T\Delta _i/TA_0$ of $S/TA_0\cong {}C_2^2$ to themselves ($i=1,2$) or switches them.$\square$

The next lemma involves a similar situation.

Lemma 7.3

Fix a finite $2$-group $S$ with a normal dihedral or quaternion subgroup $\Delta \trianglelefteq {}S$ of order at least $8$. Assume two of the three subgroups of index two in $\Delta$ are $S$-conjugate. Let $\Delta _0\le \Delta$ be a dihedral or quaternion subgroup of order $8$, and set $T=C_S(\Delta _0)$. Let $A\trianglelefteq \Delta$ be the cyclic subgroup of index two: the one which is normal in $S$ if $\Delta \cong {}Q_8$. Fix a generator $a\in {}A$, and choose $b\in \Delta _0{\backslash }A$. Let $Z\le {}A_0<A$ be the subgroups of order two and index two, respectively. Then the following hold.

• (a) $A\le [S,S]$.

• (b) $[S:T\Delta ]=2$. For each $g\in {}S$, $gbg^{-1}=a^ib$ where $i$ is even if $g\in {}T\Delta$ and $i$ is odd if $g\in {}S{\backslash }T\Delta$.

• (c) $TA_0\trianglelefteq {}S$, $S/TA_0\cong {}D_8$, and $Z(S/TA_0)=TA/TA_0$.

• (d) If $T$ is abelian, then $[S,S]$ is abelian.

• (e) Assume $b_1,b_2\in {}S$ are in distinct cosets of $TA$, $[b_1,A]=A_0=[b_2,A]$, $b_i^2\in {}A$, and $\langle {A,b_i}\rangle \trianglelefteq {}S$. Then $\langle {A,b_1,b_2}\rangle$ is dihedral, semidihedral, or quaternion.

• (f) Assume that $\Delta$ is dihedral and $Z$ is a direct factor of $T$. Let $\alpha \in \mathrm{Aut }(S)$ be such that $\alpha ^2\in \mathrm{Inn }(S)$, $\alpha (Z)=Z$, and $\alpha (T\Delta )\ne {}T\Delta$. Set $\widehat{\Delta }=\alpha (\Delta ){\cdot }\Delta$. Then $\widehat{\Delta }$ is dihedral, $\widehat{\Delta }\trianglelefteq S$, $[\widehat{\Delta }:\Delta ]=2$, $T\cap \widehat{\Delta }=Z$, and $T\widehat{\Delta }=S$.

Proof

By assumption, $\Delta \trianglelefteq S$. Hence $A\trianglelefteq S$ since $A$ is characteristic in $\Delta$, except when $\Delta \cong {}Q_8$ in which case $A$ was chosen to be the unique subgroup of index two in $\Delta$ normal in $S$.

(a) By assumption, the subgroups $\langle {A_0,b}\rangle$ and $\langle {A_0,ba}\rangle$ are $S$-conjugate. Hence there is $x\in {}S$ such that $c_x(bA_0)=baA_0$, so $xbx^{-1}=ba^i$ for some odd $i$, and $A=\langle {a^i}\rangle \le [S,S]$.

(b) By definition (and since $A=\langle {a}\rangle \trianglelefteq S$),

$$\begin{aligned} T=C_S(\Delta _0) = \bigl \{ g\in {}S \,\big |\, gbg^{-1}=b,~ gag^{-1}=a^{4j+1} \text{ some} j \bigr \}. \end{aligned}$$

(15)

Hence

$$\begin{aligned} T\Delta = \bigl \{ g\in {}S\,|\,c_g\in \mathrm{Aut }_T(\Delta )\mathrm{Inn }(\Delta )\} = \{g\in {}S \,\big |\, gbg^{-1}=a^ib \text{ some} i\equiv 0 \text{(mod} \text{2)} \bigr \}. \end{aligned}$$

Since $\Delta \trianglelefteq {}S$, this proves that $gbg^{-1}=a^ib$ with $i$ odd whenever $g\in {}S{\backslash }T\Delta$. Also, $T\Delta <S$ since $\langle {A_0,b}\rangle$ and $\langle {A_0,ab}\rangle$ are $S$-conjugate, and hence $[S:T\Delta ]=2$.

(c) If $\Delta =\Delta _0$ has order $8$, then $TA_0=T=C_S(\Delta )$ is normal in $S$ since $\Delta \trianglelefteq {}S$. So assume $|\Delta |>8$. Since $A\trianglelefteq {}S$, $1\ne \langle {a^4}\rangle \trianglelefteq {}S$. Also, $[T,A]\le \langle {a^4}\rangle$ by (1), and hence $T\Delta /\langle {a^4}\rangle \cong {}(T/Z)\times {}D_8$. For $x\in {}S$, $c_x(T\Delta )=T\Delta$ by (b), so $c_x(TA_0/\langle {a^4}\rangle )=TA_0/\langle {a^4}\rangle$ by Proposition 7.1 applied with $\alpha =c_x\in \mathrm{Aut }(T\Delta /\langle {a^4}\rangle )$. Thus $TA_0\trianglelefteq {}S$.

Throughout the rest of the proof of the lemma, we set $\bar{S}=S/TA_0$, and let $\bar{P}\le \bar{S}$ or $\bar{g}\in {}\bar{S}$ be the image of $P\le {}S$ or $g\in {}S$. Thus $\bar{T\Delta }=\langle {\bar{a},\bar{b}}\rangle \cong {}C_2^2$.

For $x\in {}S{\backslash }T\Delta$, $c_x(\bar{a})=\bar{a}$ and $c_x(\bar{b})=\bar{ab}$ by (b). Thus $\bar{S}\cong {}D_8$, with center $\langle {\bar{a}}\rangle =\bar{TA}$.

(d) Assume $T$ is abelian. Since $A$ is cyclic, $\mathrm{Aut }(A)$ is abelian, and hence $[S,S]$ is in the kernel of the map $S\overset{}{\longrightarrow }\mathrm{Aut }(A)$ induced by conjugation. Thus $[S,S]\le C_S(A)$. Also, $[S,S]\le {}TA$ since $TA$ is normal of index four in $S$. Since $T$ is abelian, $C_{TA}(A)$ is also abelian, and so is $[S,S]$.

(e) Fix $b_1,b_2$ as above, and set $\Delta _i=\langle {A,b_i}\rangle \trianglelefteq {}S$ and $\widehat{\Delta }=\Delta _1\Delta _2$. Then $\widehat{\Delta }/A\cong {}C_2^2$ since the $\Delta _i$ are normal and distinct and contain $A$ with index two. Also, $(\bar{b_1}\bar{b_2})^2=\bar{a}$ since $\bar{S}\cong D_8$ and $\bar{b_1}$ and $\bar{b_2}$ are in distinct nonidentity cosets of $\bar{A}=Z(\bar{S})$ and have order two in $\bar{S}$. Since $(b_1b_2)^2\in {}A$ (recall $\widehat{\Delta }/A\cong C_2^2$), we have $(b_1b_2)^2=a^j$ for some odd $j$. Thus $\widehat{A}\mathop {=}\limits ^{\mathrm{def}}\langle {b_1b_2}\rangle$ is cyclic of index two in $\widehat{\Delta }$, conjugation by $b_1$ inverts the subgroup $A$ of index two in $\widehat{A}$, and hence $\widehat{\Delta }$ is dihedral, semidihedral, or quaternion (cf. [19, Theorem 5.4.4]).

(f) Now assume that $\Delta$ is dihedral, and that $T=T_0Z$ where $T_0\cap {}Z=1$. Let $\alpha \in \mathrm{Aut }(S)$ be such that $\alpha ^2\in \mathrm{Inn }(S)$, $\alpha (Z)=Z$, and $\alpha (T\Delta )\ne {}T\Delta$. Set $\Delta ^*=\alpha (\Delta )$, and $\widehat{\Delta }=\Delta \Delta ^*$. Since $\Delta$ and $\Delta ^*$ are both normal in $S$ and are exchanged by $\alpha$ (since $\alpha ^2\in \mathrm{Inn }(S)$), $\widehat{\Delta }$ is normal in $S$.

Set $a_*=\alpha (a)$ and $b_*=\alpha (b)$. By (c), $TA_0\trianglelefteq {}S$, and $\bar{S}=S/TA_0$ is dihedral of order 8 with center $TA/TA_0$.

Since $A\le [S,S]$ by (a), $\alpha (A)\le [S,S]\le {}TA$. If $b_*\in {}T\Delta$, then $\alpha (\Delta )\le {}T\Delta$, and by Lemma 7.2(a) (applied with $\varphi =\alpha |_{\Delta }$ and $T$ replaced by $T_0$), $b_*\in {}T_0Ab$. So $\bar{b_*}\in \{\bar{b},\bar{ab}\}$. Since $[T,b]=1$ (recall $b\in \Delta _0$), $[\bar{\alpha (T)},\bar{b_*}]=1$, so $\bar{\alpha (T)}\le \langle {\bar{a},\bar{b}}\rangle =\bar{T\Delta }$. Thus $\alpha (T)\le {}T\Delta$, so $\alpha (T\Delta )=T\Delta$, and this contradicts our original assumption about $\alpha$.

Thus $b_*\notin {}T\Delta$. Hence by (b), $b_*bb_*^{-1}=a^ib$ for some odd $i$. Set $\widehat{a}=b_*b$; then $\widehat{a}^2=a^i$, and thus $\langle {b_*,b}\rangle =\langle {\widehat{a},b}\rangle$ is a dihedral group which contains $\Delta$ with index two. Since $\alpha ^2\in \mathrm{Inn }(S)$, $\alpha (b_*)=\alpha ^2(b)=a^jb$ for some $j$, and so $\alpha (\widehat{a})=a^jbb_*=a^j(b_*b)^{-1}=a^j\widehat{a}^{-1}$ is in $\langle {\widehat{a}}\rangle$. So $a_*\in \langle {a}\rangle$, $\langle {b_*,b}\rangle =\Delta \Delta ^*$ is dihedral, and it contains $\Delta$ with index two. Also, $T\widehat{\Delta }=S$ since $[S:T\Delta ]=2$ and $\widehat{\Delta }\nleq {}T\Delta$, and $T\cap \widehat{\Delta }=T\cap \Delta =Z$ since $T\Delta \cap \widehat{\Delta }=\Delta$.$\square$

This will now be applied to prove the following lemma.

Lemma 7.4

Fix a finite $2$-group $S$, and subgroups $T,\Delta _0\le {}S$ such that $\Delta _0$ is dihedral of order $8$, $T\cap {}\Delta _0=1$, and $[T,\Delta _0]=1$. Let $U,V\le {}\Delta _0$ be the two noncyclic subgroups of order four, and set $Z=U\cap {}V=Z(\Delta _0)$. Assume $N_S(TU)=T\Delta _0$, $N_S(\Delta _0)>T\Delta _0$, and either $Z\le {}Z(S)$ or $T$ contains no subgroup isomorphic to $D_8$. Let $\Delta \trianglelefteq {}S$ be the normal closure of $\Delta _0$ in $S$. Then $\Delta$ is dihedral, $T\cap {}\Delta =1$, $[S:T\Delta ]=2$, and all noncentral involutions in $\Delta$ are $S$-conjugate.

Proof

Case 1: Assume first that $Z\le {}Z(S)$. Set $\Delta _{-1}=U$. We will construct subgroups $\Delta _0<\Delta _1<\cdots <\Delta _m<S$, all normalized by $T$, such that $[S:T\Delta _m]=2$, and such that for each $0\le {}i\le {}m$,

• (i) $\Delta _i$ is dihedral of order $2^{3+i}$;

• (ii) $T\cap {}\Delta _i=1$ and $T\Delta _i=N_S(T\Delta _{i-1})=N_S(\Delta _{i-1})$; and

• (iii) $N_S(\Delta _i)>T\Delta _i$.

To simplify notation, we set $S_i=T\Delta _i$ whenever $\Delta _i$ has been defined.

When $i=0$, the only condition which is not immediate from the hypotheses is that $T\Delta _0=N_S(U)$. One inclusion is clear: $N_S(U)\ge {}T\Delta _0$ since $[U,T]=1$ and $U\trianglelefteq \Delta _0$. If $g\in {}N_{N_S(U)}(T\Delta _0)$, then since $g$ normalizes $U$ and $T\Delta _0$ and $C_{T\Delta _0}(U)=TU$, $g\in {}N_S(TU)=T\Delta _0$. Thus $N_{N_S(U)}(T\Delta _0)=T\Delta _0$, and so $N_S(U)=T\Delta _0$ by Lemma 6.1.

Assume, for some $i\ge 0$, that we have constructed $\Delta _i$ which satisfies (i)–(iii). If $[S:S_i]=2$ (recall $S_i=T\Delta _i$), then $\Delta _i\trianglelefteq {}S$ since $N_S(\Delta _i)>S_i$. For $g\in {}S{\backslash }S_i$, $g\notin {}N_S(\Delta _{i-1})=S_i$ implies that $g$ exchanges the two conjugacy classes of noncentral involutions in $\Delta _i$, and hence the noncentral involutions in $\Delta _i$ are all $S$-conjugate. If $i>0$, then the normal closure of $\Delta _0$ in $\Delta _i$ is $\Delta _{i-1}$ which is not normal in $S$, and thus $\Delta _i$ is the normal closure of $\Delta _0$ in $S$ (this is trivial if $i=0$). So the subgroup $\Delta =\Delta _i$ satisfies the conditions in the statement of the lemma.

Now assume $[S:S_i]\ge 4$. Set $S_{i+1}=N_S(\Delta _i)>S_i$. Let $\Delta _{i-1}$ and $\Delta _{i-1}^*$ be the two dihedral subgroups of index two in $\Delta _i$. Since $S_i=N_S(\Delta _{i-1})$ and $\Delta _i\trianglelefteq {}S_{i+1}$, conjugation by any element of $S_{i+1}{\backslash }S_i$ exchanges $\Delta _{i-1}$ with $\Delta _{i-1}^*$. The product of any two elements of $S_{i+1}{\backslash }S_i$ thus lies in $S_i$, so $[S_{i+1}:S_i]=2$, and $S_i\trianglelefteq {}S_{i+1}<S$.

For each $x\in {}N_S(S_i)$, since $c_x(Z)=Z$ by assumption ($Z\le {}Z(S)$), Lemma 7.2(b) implies that $c_x$ either leaves $T\Delta _{i-1}$ and $T\Delta _{i-1}^*$ invariant or exchanges them; and leaves them invariant only if $x\in {}N_S(T\Delta _{i-1})=S_i$. Thus $[N_S(S_i):S_i]=2$. Since $S_i\trianglelefteq {}S_{i+1}$ with index two, this implies $N_S(S_i)=S_{i+1}$.

Choose any $g\in {}N_S(S_{i+1}){\backslash }S_{i+1}$ such that $g^2\in {}S_{i+1}$, and set $\alpha =c_g\in \mathrm{Aut }(S_{i+1})$. Then $\alpha (Z)=Z$ since $Z\le {}Z(S)$, and $\alpha (S_i)\ne {}S_i$ since $g\notin {}S_{i+1}=N_S(S_i)$. Also, $C_S(\Delta _0)=TZ$ since $N_S(U)=T\Delta _0$ by (ii) when $i=0$ (and since $C_S(\Delta _0)\le C_S(U)\le N_S(U)$). The hypotheses of Lemma 7.3(f) thus hold (but where $T$ in Lemma 7.3 corresponds to $TZ$ here). So if we define $\Delta _{i+1}=\Delta _i{\cdot }\alpha (\Delta _i)$, then $\Delta _{i+1}\trianglelefteq S_{i+1}$ is dihedral, $[\Delta _{i+1}:\Delta _i]=2$, $T\cap \Delta _{i+1}=1$, and $T\Delta _{i+1}=S_{i+1}$. Thus (i) and (ii) hold, (iii) holds since $g\in {}N_S(\Delta _{i+1})$, and this finishes the induction step in the proof.

Case 2: Now assume $Z\nleq {}Z(S)$, and set $S_*=C_S(Z)<S$. By Case 1, $S_*$ has the form described in the lemma: it contains a normal dihedral subgroup $\Delta \trianglelefteq {}S_*$ (the normal closure of $\Delta _0$ in $S_*$), $T\cap \Delta =1$, and $[S_*:T\Delta ]=2$. We prove that $T$ contains a subgroup isomorphic to $D_8$, contradicting the assumptions.

Choose any $g\in {}N_S(S_*){\backslash }S_*$ such that $g^2\in {}S_*$, and set $\beta =c_g\in \mathrm{Aut }(S_*)$. Then $\beta (Z)\ne {}Z$. Assume first that $\beta (\Delta )\nleq {}T\Delta$, and choose $x\in \beta (\Delta ){\backslash }T\Delta$. Let $A\trianglelefteq \Delta$ be the cyclic subgroup of index two ($A\trianglelefteq {}S_*$), and choose $b\in U{\backslash }Z$. (Recall $C_2^2\cong {}U\le \Delta _0$.) If $c_x(b)\in \Delta {\backslash }A$ is $\Delta$-conjugate to $b$, then $c_{ax}(U)=U$ for some $a\in {}A$, which is impossible since we showed in the proof of Case 1 that $N_S(U)=T\Delta _0$. Thus $b$ is not $\Delta$-conjugate to $xbx^{-1}$, so $[x,b]$ generates $A$, and hence $A\le \beta (\Delta )$ since $x\in \beta (\Delta )\trianglelefteq {}S_*$. Since $A$ is cyclic of order at least four, this is possible only if $\beta (A)=A$, which is impossible since $\beta (Z)\ne {}Z$.

Thus $\beta (\Delta )\le {}T\Delta$. Let $\psi \in \mathrm{Hom }(\Delta ,T)$ be the composite $\Delta \overset{\beta }{\longrightarrow }T\Delta \twoheadrightarrow T\Delta /\Delta \cong {}T$. Since $\beta (Z)\ne {}Z$ and $\beta (Z)\le Z(T\Delta )$, $\beta (Z)\nleq \Delta$, and hence $Z\nleq \mathrm{Ker }(\psi )$. Since any nontrivial normal subgroup of $\Delta$ contains $Z$, this implies that $\psi$ is injective, and thus that $T$ contains a subgroup isomorphic to $D_8$.$\square$

We also need the following corollary to Lemma 7.4.

Lemma 7.5

Fix a finite $2$-group $S$, and an abelian subgroup $P\le {}S$ such that $|N_S(P)/P|=2$ and $P\not \trianglelefteq {}S$. Assume either

• (a) $\mathrm{Aut }_S(P)\nleq O_2(\mathrm{Aut }(P))$, or

• (b) there is a direct factor $U\le P$ such that $U\cong C_2^2$ and $1\ne [N_S(P),U]\le U$.

Then there are subgroups $\Delta _0\le \Delta \trianglelefteq S$ such that $\Delta$ is dihedral, $\Delta _0\cong D_8$, $C_S(\Delta _0)\le {}P$, and the noncentral involutions in $\Delta$ are all $S$-conjugate. In case (b), $\Delta$ can be taken to be the normal closure of $U$ in $S$.

Proof

Set $\widehat{P}=N_S(P)<S$, fix $x\in {}N_S(\widehat{P}){\backslash }\widehat{P}$ such that $x^2\in {}\widehat{P}$, and set $Q=xPx^{-1}$. Set $Z=[\widehat{P},\widehat{P}]$. Since $\widehat{P}$ is nonabelian, and $P,Q$ are distinct abelian subgroups of index two, Lemma 6.6(c,a) implies that $|Z|=2$, and $Z(\widehat{P})=P\cap {}Q$ has index two in $P$ and in $Q$. Also, $Z\le Z(\widehat{P})$.

We first show that (a) implies (b). Consider the subgroups

$$\begin{aligned} \Theta _1=\bigl \{\alpha \in \mathrm{Aut }(P) \,\big |\, [\alpha ,P]\le \mathrm{Fr }(P)\bigr \} \quad \text{ and}\quad \Theta _2=\bigl \{\alpha \in \mathrm{Aut }(P) \,\big |\, \alpha |_{\Omega _1(P)}=\mathrm{Id }\bigr \} \,. \end{aligned}$$

Both are normal in $\mathrm{Aut }(P)$, and they are $2$-subgroups by Lemma 6.2 and [19, Theorem 5.2.4], respectively. Thus $\Theta _1\Theta _2\le O_2(\mathrm{Aut }(P))$. Since $\mathrm{Aut }_{\widehat{P}}(P)=\mathrm{Aut }_S(P)\nleq O_2(\mathrm{Aut }(P))$ by assumption, $\mathrm{Aut }_{\widehat{P}}(P)\nleq \Theta _1\Theta _2$. Hence for $y\in {}\widehat{P}{\backslash }P$, $Z=[y,P]\nleq \mathrm{Fr }(P)$ and $Z(\widehat{P})=C_{P}(y)\ngeq \Omega _1(P)$.

Thus there is $g\in {}P{\backslash }Z(\widehat{P})$ such that $|g|=2$ and $P=Z(\widehat{P})\times \langle {g}\rangle$, and also $T<Z(\widehat{P})$ such that $Z(\widehat{P})=Z\times {}T$ ($T=\mathrm{Ker }(f)$ for any $f:Z(\widehat{P})\rightarrow Z(\widehat{P})/\mathrm{Fr }(Z(\widehat{P}))\rightarrow C_2$ with $Z\nleq \mathrm{Ker }(f)$). Set $U=\langle {Z,g}\rangle$. Then $U\cong {}C_2^2$, $P=TU$, and $T\cap {}U=1$. So (b) holds.

Now assume (b). Thus $P=TU$ where $U\cong C_2^2$, $T\cap {}U=1$, $U\nleq Z(\widehat{P})$, and $Z=[\widehat{P},\widehat{P}]\le U$. In particular, $Z=U\cap {}Z(\widehat{P})$, and $Z\nleq \mathrm{Fr }(Z(\widehat{P}))$ since $U\cap \mathrm{Fr }(P)=1$. Hence $Z(\widehat{P})=T^*\times Z$ for some $T^*$, so $P=Z(\widehat{P})U=T^*\times {}U$, and so we can assume $T=T^*\le Z(\widehat{P})$.

Set $V=xUx^{-1}$ and $\Delta _0=UV$. Then $V\nleq P$ since $x$ normalizes $Z(\widehat{P})$ but not $P=Z(\widehat{P})U$. Thus $[U,V]=Z$, so $\Delta _0\cong {}D_8$ since $U\cong V\cong C_2^2$. Also, $T\cap \Delta _0=T\cap {}Z=1$, $[T,\Delta _0]\le [T,\widehat{P}]=1$ since $T\le Z(\widehat{P})$, $N_S(TU)=N_S(P)=\widehat{P}=T\Delta _0$, and $N_S(\Delta _0)>T\Delta _0$ since $x\in {}N_S(\Delta _0){\backslash }T\Delta _0$. Since $T$ is abelian, it contains no subgroup isomorphic to $D_8$. So by Lemma 7.4, the normal closure $\Delta$ of $\Delta _0$ in $S$ is dihedral, and all noncentral involutions in $\Delta$ are $S$-conjugate.

Set $R=N_{C_S(\Delta _0)}(Z(\widehat{P}))$. Then $R$ normalizes $Z(\widehat{P})U=P$, so $R\le N_S(P)=\widehat{P}$. Hence $R\le C_{\widehat{P}}(\Delta _0) = Z(\widehat{P})$, so by Lemma 6.1, $C_S(\Delta _0)=Z(\widehat{P})\le P$.$\square$