An involutory automorphism of a double cover of A8 with central fixed involutions

Problem

Suppose $G$ is finite, $F^*(G)$ is quasisimple, and $\alpha\in\operatorname{Aut}(G)$ has order two. Unless $F^*(G)$ has generalized quaternion Sylow $2$ -subgroups, must $C_G(\alpha)$ contain an involution outside $Z(F^*(G))$ ?

\exists t\in C_G(\alpha)\setminus Z(F^*(G)):\quad t^2=1,\ t\ne1\;?

The fixed-involution condition

An involution is an element of order two. A group is quasisimple if it is perfect and its quotient by its centre is nonabelian simple. For a finite group $G$ , let $F(G)$ be its Fitting subgroup and let $L(G)$ be the subgroup generated by its components, namely its subnormal quasisimple subgroups. The generalized Fitting subgroup is

F^*(G)=F(G)L(G).

For $\alpha\in\operatorname{Aut}(G)$ , write $C_G(\alpha)=\{g\in G:\alpha(g)=g\}$ .

Problem 7.15 of the Kourovka Notebook, due to R. Griess, asks whether, if $F^*(G)$ is quasisimple and $\alpha$ has order two, $C_G(\alpha)$ contains an involution outside $Z(F^*(G))$ , except when $F^*(G)$ has quaternion Sylow $2$ -subgroups (3). Here generalized quaternion groups are understood to have order at least eight.

Theorem 1. There exist a finite group $E$ , an element $z\in E$ , and $\alpha\in\operatorname{Aut}(E)$ such that

\begin{align*} E'&=E,& Z(E)&=\langle z\rangle\cong C_2, & E/Z(E)&\cong A_8,\tag{2a}\\ |\alpha|&=2,& \{g\in C_E(\alpha):g^2=1\}&=\{1,z\}.\tag{2b} \end{align*}

Moreover, $E$ contains a subgroup isomorphic to $C_2\times C_2$ . Consequently $F^*(E)=E$ , and no Sylow $2$ -subgroup of $F^*(E)$ is generalized quaternion.

Clifford realizations of the double covers of symmetric groups are classical; the cover in which transpositions lift to involutions is described by Brinkman (1, Section 3). The order calculation for products of disjoint transposition lifts appears in (1, Proposition 3.2). For the alternating covers, the corresponding lifting criterion is recalled in (2, Lemma 2.2). We give the construction and the required calculations over $\mathbb F_7$ .

The Clifford group and its permutation action

Put $k=\mathbb F_7$ , let $V=k^8$ have basis $e_1,\ldots,e_8$ , and set

\tag{3} q(v)=\sum_{i=1}^8 v_i^2,\qquad B(v,w)=q(v+w)-q(v)-q(w)=2\sum_{i=1}^8 v_iw_i.

Let $\mathcal C=\operatorname{Cl}(V,q)$ . We identify $V$ with its canonical image in $\mathcal C$ , so

\tag{4} v^2=q(v)1,\qquad vw+wv=B(v,w)1.

The injectivity of the canonical map also follows from the basis construction below. Scalars are identified with their images in $\mathcal C$ .

For $i\ne j$ , define

\tag{5} u_{ij}=5(e_i-e_j),\qquad z=-1.

All constants in these formulas belong to $k$ . Since $2\cdot5^2=1$ , the elements $u_{ij}$ satisfy

\begin{align*} u_{ij}^2&=1,& u_{ji}&=-u_{ij},\tag{6a}\\ u_{ij}u_{rs}&=-u_{rs}u_{ij} &&\text{if }\{i,j\}\cap\{r,s\}=\varnothing.\tag{6b} \end{align*}

In particular, they are units. Define

\tag{7} E=\big\langle u_{ij}u_{rs}:i\ne j,\ r\ne s\big\rangle \le\mathcal C^\times.

Every element of $E$ is a product of an even number of the $u_{ij}$ : the inverse of $u_{ij}u_{rs}$ is $u_{rs}u_{ij}$ .

For $\sigma\in S_8$ , let $P_\sigma e_i=e_{\sigma(i)}$ . Using (4), we obtain

\tag{8} u_{ij}vu_{ij}^{-1} =B(u_{ij},v)u_{ij}-v=-P_{(ij)}v.

Indeed, $B(u_{ij},v)=10(v_i-v_j)$ , and $10\cdot5=1$ in $k$ . It follows that conjugation by a generating pair in (7) acts on $V$ as

\tag{9} (u_{ij}u_{rs})v(u_{ij}u_{rs})^{-1} =P_{(ij)(rs)}v.

Lemma 2. There is a surjective homomorphism $\pi:E\longrightarrow A_8$ determined by

\tag{10} gvg^{-1}=P_{\pi(g)}v\quad(v\in V),\qquad \pi(u_{ij}u_{rs})=(ij)(rs).

Proof. Equation (9) gives the asserted action for the generators. Products and inverses preserve the condition that conjugation act as an even permutation. A permutation is determined by its action on the basis of $V$ , so this defines $\pi$ uniquely and proves that it is a homomorphism. Every even permutation is a product of an even number of transpositions. Pairing these transpositions and using (9) proves surjectivity. ◻

Let $\gamma$ be the algebra automorphism of $\mathcal C$ given by $\gamma(v)=-v$ for $v\in V$ , and let $\rho$ be the anti-automorphism fixing $V$ pointwise. Thus $\rho$ reverses products. From (6a) and (7),

\tag{11} \gamma(g)=g,\qquad \rho(g)=g^{-1},\qquad \rho(g)g=1 \quad(g\in E).

The scalar kernel

For $d\in V^*$ , denote exterior contraction by $i_d$ . On a decomposable element it is given by

\tag{12} i_d(v_1\wedge\cdots\wedge v_m) =\sum_{r=1}^m(-1)^{r-1}d(v_r) v_1\wedge\cdots\wedge\widehat{v_r}\wedge\cdots\wedge v_m.

In particular, $i_d$ vanishes on scalars.

Lemma 3. Let $W$ be a finite-dimensional vector space with basis $f_1,\ldots,f_n$ and dual basis $d_1,\ldots,d_n$ . If $\xi\in\bigwedge W$ satisfies $i_{d_j}\xi=0$ for all $j$ , then $\xi$ is a scalar.

Proof. For $e\in W$ and $d\in W^*$ , put $P(v)=v-d(v)e$ . Expanding a decomposable exterior product gives

\tag{13} (\bigwedge P)(\xi)=\xi-e\wedge i_d\xi.

Every term involving two occurrences of $e$ vanishes. The terms involving one occurrence of $e$ , moved to the first position, are precisely the terms subtracted on the right of (13). Decomposable elements span $\bigwedge W$ , so the identity holds for every $\xi$ .

Apply this identity to $P_j(v)=v-d_j(v)f_j$ . The hypothesis gives $(\bigwedge P_j)(\xi)=\xi$ for each $j$ . Successively deleting every coordinate yields $P_n\cdots P_1=0$ , whence

\xi=(\bigwedge P_n)\cdots(\bigwedge P_1)(\xi) =(\bigwedge 0)(\xi).

The map $\bigwedge 0$ is the identity on scalars and zero on every positive exterior degree. Therefore $\xi$ is scalar. ◻

There is a linear isomorphism

\tag{15} \Phi:\mathcal C\longrightarrow\bigwedge V,\qquad \Phi(e_{i_1}\cdots e_{i_m}) =e_{i_1}\wedge\cdots\wedge e_{i_m} \quad(i_1<\cdots<i_m),

including $\Phi(1)=1$ . For completeness, (4) rewrites every product as a linear combination of the ordered products in (15). Their independence follows by letting $e_i$ act on $\bigwedge V$ as

T_i(\xi)=e_i\wedge\xi+i_{e_i^*}\xi.

These operators satisfy $T_i^2=1$ and $T_iT_j=-T_jT_i$ for $i\ne j$ , so they define a representation of $\mathcal C$ . Applying $T_{i_1}\cdots T_{i_m}$ to $1$ gives $e_{i_1}\wedge\cdots\wedge e_{i_m}$ . The latter elements form the exterior basis, proving the asserted independence and isomorphism.

The same alternating deletion formula as (12), with products in place of exterior products, defines a linear map $D_d:\mathcal C\to\mathcal C$ . Equivalently,

\tag{17} D_d(1)=0,\qquad D_d(vx)=d(v)x-vD_d(x).

It descends from the tensor algebra because its graded derivation rule sends every defining relation $v^2-q(v)1$ to $d(v)v-vd(v)=0$ . On the ordered basis,

\tag{18} \Phi D_d=i_d\Phi.

Lemma 4. For $v\in V$ and $x\in\mathcal C$ ,

\tag{19} D_{B(v,-)}(x)=vx-\gamma(x)v.

Consequently, if $\gamma(x)=x$ and $x$ commutes with every vector in $V$ , then $x$ is scalar.

Proof. The identity vanishes on scalars and is linear in $x$ . If it holds for $x$ , then

\begin{align*} D_{B(v,-)}(wx) &=B(v,w)x-w\bigl(vx-\gamma(x)v\bigr)\\ &=vwx+w\gamma(x)v\\ &=v(wx)-\gamma(wx)v. \end{align*}

Here (4) was used in the second line. Induction on products of vectors proves (19).

Under the stated hypotheses, $D_{B(v,-)}(x)=0$ for all $v$ . Equation (3) and $2\cdot4=1$ in $k$ give

B(4e_i,-)=e_i^*\qquad(1\le i\le8).

Thus (18) gives $i_{e_i^*}\Phi(x)=0$ for every $i$ . Lemma 3 makes $\Phi(x)$ scalar. Since $\Phi$ fixes scalars, so is $x$ . ◻

Proposition 5. The sequence

\tag{22} 1\longrightarrow\langle z\rangle \longrightarrow E\xrightarrow{\ \pi\ }A_8 \longrightarrow1

is exact, and $Z(E)=\langle z\rangle=\{1,z\}$ . In particular, $E$ is finite.

Proof. If $g\in\ker\pi$ , then (10) says that $g$ commutes with every vector. Also $\gamma(g)=g$ by (11). Lemma 4 therefore gives $g=r1$ for some $r\in k$ . Reversion fixes scalars, so (11) implies $r^2=1$ . Hence $g=1$ or $g=-1$ .

The disjoint roots $u_{12}$ and $u_{34}$ anticommute, and therefore

\tag{23} (u_{12}u_{34})^2=-1=z.

Thus $z\in E$ . The two scalars $1$ and $z$ are distinct and central, and both belong to $\ker\pi$ . This proves exactness. The image of $Z(E)$ under the surjection $\pi$ is central in $A_8$ , whose centre is trivial. Consequently $Z(E)=\ker\pi$ . Finally, every fibre of $\pi$ has two elements and $A_8$ is finite. ◻

Perfectness of the cover

Write $[x,y]=xyx^{-1}y^{-1}$ . We prove perfectness directly from the root relations.

Lemma 6. The element $z$ belongs to $E'$ .

Proof. Put $a=u_{12}$ , $b=u_{34}$ and $c=u_{56}$ . These are pairwise anticommuting involutions. Thus $x=ab$ and $y=ac$ belong to $E$ and satisfy

xy=-bc=cb,\qquad yx=-cb,\qquad [x,y]=-1=z.\quad\Box

◻

Lemma 7. For $a\ne b$ and $a\ne c$ , the element $u_{ab}u_{ac}$ belongs to $E'$ .

Proof. The assertion is immediate if $b=c$ . Otherwise set $x=u_{ab}u_{ac}$ . The polar form gives

u_{ab}u_{ac}+u_{ac}u_{ab}=1,

because $2\cdot5^2=1$ . Since both roots square to one,

\tag{26} x^2=x-1,\qquad x^3=-1=z.

Choose distinct $d,e$ outside $\{a,b,c\}$ and put $h=u_{bc}u_{de}$ . By (10), conjugation by $h$ fixes coordinate $a$ and interchanges coordinates $b,c$ . Consequently

hxh^{-1}=u_{ac}u_{ab}=x^{-1},\qquad [h,x]=x^{-2}.

Thus $x^2\in E'$ . Equations (26) and Lemma 6 give $x^3\in E'$ , so $x=x^3(x^2)^{-1}\in E'$ . ◻

Proposition 8. The group $E$ is perfect and quasisimple.

Proof. Consider a generating pair $u_{ab}u_{cd}$ of $E$ . Choose $t\notin\{a,b,c,d\}$ . Inserting two pairs of equal roots gives

\tag{28} u_{ab}u_{cd} =(u_{ab}u_{at})(u_{at}u_{ct})(u_{ct}u_{cd}).

The first and third factors belong to $E'$ by Lemma 7. For the middle factor, $u_{at}u_{ct}=u_{ta}u_{tc}$ , so the same lemma applies. Thus every generator in (7) belongs to $E'$ , proving $E'=E$ . By Proposition 5, $E/Z(E)\cong A_8$ . The alternating group $A_8$ is nonabelian simple, so $E$ is quasisimple. ◻

The automorphism and its fixed involutions

Set $a=u_{12}$ and define

\tag{29} \alpha(g)=aga^{-1}=aga\qquad(g\in E).

Conjugation by $a$ preserves $E$ : for any two roots $u,v$ ,

a(uv)a=(au)(va)\in E.

It follows that $\alpha$ is an automorphism and $\alpha^2=1$ . Moreover,

\tag{31} \alpha(u_{12}u_{34})=u_{34}u_{12}=-u_{12}u_{34} \ne u_{12}u_{34}.

The inequality follows by multiplying by the inverse of $u_{12}u_{34}$ and using $-1\ne1$ . Hence $|\alpha|=2$ .

Lemma 9. If $\alpha(g)=g$ , then $\pi(g)$ fixes $1$ and $2$ individually.

Proof. Put $\sigma=\pi(g)$ . The equality $\alpha(g)=g$ says that $g$ commutes with $u_{12}$ . By (10),

u_{\sigma(1),\sigma(2)}=gu_{12}g^{-1}=u_{12}.

The coefficients $5,-5,0$ are pairwise distinct in $k$ . Comparison of the two nonzero coordinates gives $\sigma(1)=1$ and $\sigma(2)=2$ . ◻

Lemma 10. If $\pi(g)$ is a double transposition, then $g^2=z$ .

Proof. All double transpositions are conjugate in $A_8$ . Indeed, they are conjugate in $S_8$ ; an odd conjugating permutation can be multiplied on the right by a transposition of two points fixed by $(12)(34)$ , which centralizes $(12)(34)$ and changes its parity. Choose $\sigma\in A_8$ with $\pi(g)=\sigma(12)(34)\sigma^{-1}$ , and choose $h\in E$ with $\pi(h)=\sigma$ . Put $y=h(u_{12}u_{34})h^{-1}$ . Then $\pi(y)=\pi(g)$ , so Proposition 5 gives $g=zy$ or $g=y$ . Since $z$ is central and $z^2=1$ ,

g^2=y^2=h(u_{12}u_{34})^2h^{-1}=hzh^{-1}=z.\quad\Box

◻

Proposition 11. The only elements of $C_E(\alpha)$ whose square is one are $1$ and $z$ .

Proof. Suppose $\alpha(g)=g$ and $g^2=1$ . Then $\sigma=\pi(g)$ is an even permutation with $\sigma^2=1$ . By Lemma 9, its support lies in $\{3,\ldots,8\}$ . Every permutation whose square is one is a product of disjoint transpositions. Their number is even because $\sigma\in A_8$ , and is at most three because its support has size at most six. Thus $\sigma=1$ or $\sigma$ is a double transposition. The second case is impossible by Lemma 10, since $g^2=1\ne z$ . Hence $\sigma=1$ , and Proposition 5 gives $g\in\{1,z\}$ . Conversely, both central elements $1,z$ are fixed by $\alpha$ and have square one. ◻

The generalized Fitting subgroup and the Sylow exception

Proof of Theorem 1. Propositions 5 and 8 prove (2a). The automorphism defined in (29) has order two by (31). Proposition 11 therefore proves (2b).

Consider

\tag{34} t=u_{12}u_{34}u_{56}u_{78}\in E.

Moving each of the four pairwise anticommuting roots past the others gives

t^2=(-1)^{4\cdot3/2}=1,\qquad \pi(t)=(12)(34)(56)(78)\ne1.

Therefore $t\ne1,z$ . Since $z$ is a central involution, $\langle z,t\rangle=\{1,z,t,zt\}\cong C_2\times C_2$ .

For $m\ge3$ , the generalized quaternion group of order $2^m$ has the presentation

Q_{2^m}=\langle r,s\mid r^{2^{m-1}}=1,\; s^2=r^{2^{m-2}},\ srs^{-1}=r^{-1}\rangle.

Its elements outside $\langle r\rangle$ have square $r^{2^{m-2}}$ , and its cyclic subgroup $\langle r\rangle$ has the unique involution $r^{2^{m-2}}$ . Thus $Q_{2^m}$ has exactly one involution. A Sylow $2$ -subgroup of $E$ containing $\langle z,t\rangle$ cannot be generalized quaternion. All Sylow $2$ -subgroups of the finite group $E$ are conjugate, so none is generalized quaternion.

Finally, $E$ is quasisimple and is subnormal in itself. It is therefore a component of itself, giving $L(E)=E$ and $F^*(E)=E$ . Its fixed subgroup has the sole involution $z\in Z(F^*(E))$ , while the quaternion exception does not apply. ◻

Theorem 1 and Corollary 2 of Guralnick and Robinson (2) give an elementary abelian subgroup meeting every involution class of a finite quasisimple group, and imply that two involutions of that group have commuting conjugates. Their hypotheses concern involutions in the group; they impose no fixed-point condition for each prescribed automorphism of order two.

References

Preprint · Lean (GitHub)

J. Brinkman, Generators of order two for $S_n$ and its two double covers, Beitr. Algebra Geom. 41 (2000), no. 1, 223–231. Full text.

R. M. Guralnick and G. R. Robinson, Commuting involutions and elementary abelian subgroups of simple groups, J. Algebra 607 (2022), Part A, 300–314. doi:10.1016/j.jalgebra.2021.04.041.

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 7.15. arXiv:1401.0300.