Groups That Have the Same Holomorph As a Finite Perfect Group

Groups that have the same holomorph as a ﬁnite perfect group

Andrea Caranti1 & Francesca Dalla Volta2 Lecce, 5 September 2017

1Dipartimento di Matematica Università degli Studi di Trento

2Dipartimento di Matematica e Applicazioni Università degli Studi di Milano–Bicocca

1/22 Four questions Consider the opposite group (G, ◦), where

x ◦ y = yx.

The opposite group of G is isomorphic to G via

inv : G → (G, ◦) x 7→ x−1

It’s just opposite .

Let G be a group.

2/22 The opposite group of G is isomorphic to G via

inv : G → (G, ◦) x 7→ x−1

It’s just opposite .

Let G be a group. Consider the opposite group (G, ◦), where

x ◦ y = yx.

2/22 via

inv : G → (G, ◦) x 7→ x−1

It’s just opposite .

Let G be a group. Consider the opposite group (G, ◦), where

x ◦ y = yx.

The opposite group of G is isomorphic to G

2/22 It’s just opposite .

Let G be a group. Consider the opposite group (G, ◦), where

x ◦ y = yx.

The opposite group of G is isomorphic to G via

inv : G → (G, ◦) x 7→ x−1

2/22 Let (G, ◦) be obtained from G by replacing H with its opposite

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

Is (G, ◦) isomorphic to G? Yes, via

H × K → (H × K, ◦) (x, y) 7→ (x−1, y)

Direct product .

Let G = H × K be a direct product.

3/22 (x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

Is (G, ◦) isomorphic to G? Yes, via

H × K → (H × K, ◦) (x, y) 7→ (x−1, y)

Direct product .

Let G = H × K be a direct product. Let (G, ◦) be obtained from G by replacing H with its opposite

3/22 Is (G, ◦) isomorphic to G? Yes, via

H × K → (H × K, ◦) (x, y) 7→ (x−1, y)

Direct product .

Let G = H × K be a direct product. Let (G, ◦) be obtained from G by replacing H with its opposite

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

3/22 Yes, via

H × K → (H × K, ◦) (x, y) 7→ (x−1, y)

Direct product .

Let G = H × K be a direct product. Let (G, ◦) be obtained from G by replacing H with its opposite

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

Is (G, ◦) isomorphic to G?

3/22 Direct product .

Let G = H × K be a direct product. Let (G, ◦) be obtained from G by replacing H with its opposite

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

Is (G, ◦) isomorphic to G? Yes, via

H × K → (H × K, ◦) (x, y) 7→ (x−1, y)

3/22 , so H, K ⊴ G, and H ∩ K ≤ Z(G). Let (G, ◦) be obtained from G by replacing H with its opposite. Is (G, ◦) isomorphic to G? The answer is possibly not obvious.

Central product .

Let G = HK be a central product

4/22 Let (G, ◦) be obtained from G by replacing H with its opposite. Is (G, ◦) isomorphic to G? The answer is possibly not obvious.

Central product .

Let G = HK be a central product, so H, K ⊴ G, and H ∩ K ≤ Z(G).

4/22 Is (G, ◦) isomorphic to G? The answer is possibly not obvious.

Central product .

Let G = HK be a central product, so H, K ⊴ G, and H ∩ K ≤ Z(G). Let (G, ◦) be obtained from G by replacing H with its opposite.

4/22 The answer is possibly not obvious.

Central product .

Let G = HK be a central product, so H, K ⊴ G, and H ∩ K ≤ Z(G). Let (G, ◦) be obtained from G by replacing H with its opposite. Is (G, ◦) isomorphic to G?

4/22 Central product .

Let G = HK be a central product, so H, K ⊴ G, and H ∩ K ≤ Z(G). Let (G, ◦) be obtained from G by replacing H with its opposite. Is (G, ◦) isomorphic to G? The answer is possibly not obvious.

4/22 , so Hi ⊴ G, and Hi ∩ Hj ≤ Z(G) for i ≠ j.

Assume all Hi are characteristic in G.

Let (G, ◦) be obtained from G by replacing H1 with its opposite.

Are the Hi still characteristic in (G, ◦)?

Characteristic subgroups .

Let G = H1H2 ··· Hn be a central product

5/22 Assume all Hi are characteristic in G.

Let (G, ◦) be obtained from G by replacing H1 with its opposite.

Are the Hi still characteristic in (G, ◦)?

Characteristic subgroups .

Let G = H1H2 ··· Hn be a central product, so Hi ⊴ G, and Hi ∩ Hj ≤ Z(G) for i ≠ j.

5/22 Let (G, ◦) be obtained from G by replacing H1 with its opposite.

Are the Hi still characteristic in (G, ◦)?

Characteristic subgroups .

Let G = H1H2 ··· Hn be a central product, so Hi ⊴ G, and Hi ∩ Hj ≤ Z(G) for i ≠ j.

Assume all Hi are characteristic in G.

5/22 Are the Hi still characteristic in (G, ◦)?

Characteristic subgroups .

Let G = H1H2 ··· Hn be a central product, so Hi ⊴ G, and Hi ∩ Hj ≤ Z(G) for i ≠ j.

Assume all Hi are characteristic in G.

Let (G, ◦) be obtained from G by replacing H1 with its opposite.

5/22 Characteristic subgroups .

Let G = H1H2 ··· Hn be a central product, so Hi ⊴ G, and Hi ∩ Hj ≤ Z(G) for i ≠ j.

Assume all Hi are characteristic in G.

Let (G, ◦) be obtained from G by replacing H1 with its opposite.

Are the Hi still characteristic in (G, ◦)?

5/22 (Multiple) holomorphs Therefore we have connections to • skew right braces, (ab) ◦ c = (a ◦ c)c−1(b ◦ c). which are equivalent to these subgroups. • Hopf Galois extensions, which are linked to these subgroups:

Spoiler/Connections . We will be discussing regular subgroups of the holomorph of a group.

6/22 • skew right braces, (ab) ◦ c = (a ◦ c)c−1(b ◦ c). which are equivalent to these subgroups. • Hopf Galois extensions, which are linked to these subgroups:

Spoiler/Connections . We will be discussing regular subgroups of the holomorph of a group. Therefore we have connections to

6/22 • Hopf Galois extensions, which are linked to these subgroups:

Spoiler/Connections . We will be discussing regular subgroups of the holomorph of a group. Therefore we have connections to • skew right braces, (ab) ◦ c = (a ◦ c)c−1(b ◦ c). which are equivalent to these subgroups.

6/22 C. Greither, B. Pareigis Hopf Galois theory for separable ﬁeld extensions J. Algebra 106 (1987), 239–258 N. P. Byott Uniqueness of Hopf Galois structure of separable ﬁeld extensions Comm. Algebra 24 (1996), 3217–3228

6/22 N. P. Byott Uniqueness of Hopf Galois structure of separable ﬁeld extensions Comm. Algebra 24 (1996), 3217–3228

C. Greither, B. Pareigis Hopf Galois theory for separable ﬁeld extensions J. Algebra 106 (1987), 239–258

6/22 Spoiler/Connections . We will be discussing regular subgroups of the holomorph of a group. Therefore we have connections to • skew right braces, (ab) ◦ c = (a ◦ c)c−1(b ◦ c). which are equivalent to these subgroups. • Hopf Galois extensions, which are linked to these subgroups:

C. Greither, B. Pareigis Hopf Galois theory for separable ﬁeld extensions J. Algebra 106 (1987), 239–258 N. P. Byott Uniqueness of Hopf Galois structure of separable ﬁeld extensions Comm. Algebra 24 (1996), 3217–3228 6/22 If S(G) is the group of permutations on the set G, and ρ : G → S(G) g 7→ (x 7→ xg) is the right regular representation, then

NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G) is (isomorphic to) the holomorph of G. More generally, if N ≤ S(G) is a regular subgroup, then

NS(G)(N) is isomorphic to the holomorph of N.

The Holomorph . The holomorph of a group G is the natural semidirect product Aut(G)G.

7/22 , then

NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G) is (isomorphic to) the holomorph of G. More generally, if N ≤ S(G) is a regular subgroup, then

NS(G)(N) is isomorphic to the holomorph of N.

The Holomorph . The holomorph of a group G is the natural semidirect product Aut(G)G. If S(G) is the group of permutations on the set G, and ρ : G → S(G) g 7→ (x 7→ xg) is the right regular representation

7/22 More generally, if N ≤ S(G) is a regular subgroup, then

NS(G)(N) is isomorphic to the holomorph of N.

NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G) is (isomorphic to) the holomorph of G.

7/22 , then

NS(G)(N) is isomorphic to the holomorph of N.

NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G) is (isomorphic to) the holomorph of G. More generally, if N ≤ S(G) is a regular subgroup

7/22 The Holomorph . The holomorph of a group G is the natural semidirect product Aut(G)G. If S(G) is the group of permutations on the set G, and ρ : G → S(G) g 7→ (x 7→ xg) is the right regular representation, then

NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G) is (isomorphic to) the holomorph of G. More generally, if N ≤ S(G) is a regular subgroup, then

NS(G)(N) is isomorphic to the holomorph of N. 7/22 if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G).

, we may say (warning: abuse of notation ahead) that G and N have the same holomorph

Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular

8/22 if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G).

(warning: abuse of notation ahead) that G and N have the same holomorph

Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say

8/22 if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G).

that G and N have the same holomorph

Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead)

8/22 if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G). Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph

8/22 Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G).

8/22 If G and N have the same holomorph, and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G). Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic)

8/22 , and G and N are isomorphic, then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G). Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph

8/22 , then ρ(G) and N are conjugate under an element of

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(N) = NS(G)(ρ(G)) = Aut(G)ρ(G) = Hol(G). Note that such G and N need not be isomorphic. (Dihedral vs. Quaternionic) If G and N have the same holomorph, and G and N are isomorphic

8/22 , and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G .

Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G

8/22 Same Holomorph . So if N ≤ S(G) is regular, we may say (warning: abuse of notation ahead) that G and N have the same holomorph if

NS(G)(Hol(G)) = NS(G)(NS(G)(ρ(G))), the multiple (double) holomorph of G, and the group

T(G) = NS(G)(Hol(G))/ Hol(G) acts regularly on the set { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G 8/22. { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G ⊆ { } I(G) = N ≤ S(G): N is regular, and NS(G)(N) = Hol(G) ⊆

J (G) = { N ≤ S(G): N is regular, and N ⊴ Hol(G) }.

The latter appears to be easier to compute.

Three sets .

In increasing order

9/22 ⊆ { } I(G) = N ≤ S(G): N is regular, and NS(G)(N) = Hol(G) ⊆

J (G) = { N ≤ S(G): N is regular, and N ⊴ Hol(G) }.

The latter appears to be easier to compute.

Three sets .

In increasing order { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G

9/22 ⊆

J (G) = { N ≤ S(G): N is regular, and N ⊴ Hol(G) }.

The latter appears to be easier to compute.

Three sets .

In increasing order { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G ⊆ { } I(G) = N ≤ S(G): N is regular, and NS(G)(N) = Hol(G)

9/22 The latter appears to be easier to compute.

Three sets .

In increasing order { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G ⊆ { } I(G) = N ≤ S(G): N is regular, and NS(G)(N) = Hol(G) ⊆

J (G) = { N ≤ S(G): N is regular, and N ⊴ Hol(G) }.

9/22 Three sets .

In increasing order { } ∼ H(G) = N ≤ S(G): N is regular, NS(G)(N) = Hol(G) and N = G ⊆ { } I(G) = N ≤ S(G): N is regular, and NS(G)(N) = Hol(G) ⊆

J (G) = { N ≤ S(G): N is regular, and N ⊴ Hol(G) }.

The latter appears to be easier to compute.

9/22 Describing the regular (normal) subgroups of the holomorph • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

, with 1ν(g) = g. Now Aut(G)ρ(G) ∋ ν(g) = γ(g)ρ(g), for some γ : G → Aut(G) which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here

Regular subgroups of the holomorph . An element of the regular subgroup N ≤ Hol(G) = Aut(G)ρ(G) can be written uniquely as ν(g)

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

Now Aut(G)ρ(G) ∋ ν(g) = γ(g)ρ(g), for some γ : G → Aut(G) which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here

Regular subgroups of the holomorph . An element of the regular subgroup N ≤ Hol(G) = Aut(G)ρ(G) can be written uniquely as ν(g), with 1ν(g) = g.

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

for some γ : G → Aut(G) which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here

Regular subgroups of the holomorph . An element of the regular subgroup N ≤ Hol(G) = Aut(G)ρ(G) can be written uniquely as ν(g), with 1ν(g) = g. Now Aut(G)ρ(G) ∋ ν(g) = γ(g)ρ(g),

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

Here

10/22 • to be compared with xρ(y) = xy.

• x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

10/22 • to be compared with xρ(y) = xy.

• ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

Regular subgroups of the holomorph . An element of the regular subgroup N ≤ Hol(G) = Aut(G)ρ(G) can be written uniquely as ν(g), with 1ν(g) = g. Now Aut(G)ρ(G) ∋ ν(g) = γ(g)ρ(g), for some γ : G → Aut(G) which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here • x ◦ y = xγ(y)y is a group operation on G,

10/22 • to be compared with xρ(y) = xy.

• xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y,

10/22 • to be compared with xρ(y) = xy.

10/22 Regular subgroups of the holomorph . An element of the regular subgroup N ≤ Hol(G) = Aut(G)ρ(G) can be written uniquely as ν(g), with 1ν(g) = g. Now Aut(G)ρ(G) ∋ ν(g) = γ(g)ρ(g), for some γ : G → Aut(G) which completely describes N. Such γ are characterized by γ(xγ(y)y) = γ(x)γ(y). Here • x ◦ y = xγ(y)y is a group operation on G, • ν :(G, ◦) → N is an isomorphism, • xν(y) = xγ(y)ρ(y) = xγ(y)y = x ◦ y, • to be compared with xρ(y) = xy. 10/22 and γ(yx) = γ(x)γ(y),

γ is characterized by γ(xβ) = γ(x)β for β ∈ Aut(G), to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

Regular normal subgroups of the holomorph . A regular normal subgroup N ⊴ Hol(G) = Aut(G)ρ(G) is described by the map γ : G → Aut(G) such that N ∋ ν(g) = γ(g)ρ(g), where 1ν(g) = g.

11/22 and γ(yx) = γ(x)γ(y),

γ(xβ) = γ(x)β for β ∈ Aut(G), to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

11/22 and γ(yx) = γ(x)γ(y), to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

11/22 γ(yx) = γ(x)γ(y), to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

Regular normal subgroups of the holomorph . A regular normal subgroup N ⊴ Hol(G) = Aut(G)ρ(G) is described by the map γ : G → Aut(G) such that N ∋ ν(g) = γ(g)ρ(g), where 1ν(g) = g. γ is characterized by γ(xβ) = γ(x)β for β ∈ Aut(G), and

11/22 to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

11/22 For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

11/22 Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please

11/22 .. Skip ring setting, please

11/22 Regular normal subgroups of the holomorph . A regular normal subgroup N ⊴ Hol(G) = Aut(G)ρ(G) is described by the map γ : G → Aut(G) such that N ∋ ν(g) = γ(g)ρ(g), where 1ν(g) = g. γ is characterized by γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y), to be compared with γ(xγ(y)y) = γ(x)γ(y). For the braces: Aut(G) ≤ Aut(G, ◦).

.. Skip ring setting, please 11/22 we have redone the work of

W. H. Mills Multiple holomorphs of ﬁnitely generated abelian groups Trans. Amer. Math. Soc. 71 (1951), 379–392 Here the condition Aut(G) ≤ Aut(G, ◦) translates into the study of the commutative rings (G, +, ·) such that every automorphism of the group (G, +) is also an automorphism of the ring (G, +, ·).

(Aside) Groups and rings . In

A. Caranti and F. Dalla Volta The multiple holomorph of a ﬁnitely generated abelian group J. Algebra 481 (2017), 327–347

12/22 W. H. Mills Multiple holomorphs of ﬁnitely generated abelian groups Trans. Amer. Math. Soc. 71 (1951), 379–392 Here the condition Aut(G) ≤ Aut(G, ◦) translates into the study of the commutative rings (G, +, ·) such that every automorphism of the group (G, +) is also an automorphism of the ring (G, +, ·).

(Aside) Groups and rings . In

A. Caranti and F. Dalla Volta The multiple holomorph of a ﬁnitely generated abelian group J. Algebra 481 (2017), 327–347 we have redone the work of

12/22 Here the condition Aut(G) ≤ Aut(G, ◦) translates into the study of the commutative rings (G, +, ·) such that every automorphism of the group (G, +) is also an automorphism of the ring (G, +, ·).

(Aside) Groups and rings . In

A. Caranti and F. Dalla Volta The multiple holomorph of a ﬁnitely generated abelian group J. Algebra 481 (2017), 327–347 we have redone the work of

W. H. Mills Multiple holomorphs of ﬁnitely generated abelian groups Trans. Amer. Math. Soc. 71 (1951), 379–392

12/22 such that every automorphism of the group (G, +) is also an automorphism of the ring (G, +, ·).

(Aside) Groups and rings . In

A. Caranti and F. Dalla Volta The multiple holomorph of a ﬁnitely generated abelian group J. Algebra 481 (2017), 327–347 we have redone the work of

12/22 (Aside) Groups and rings . In

A. Caranti and F. Dalla Volta The multiple holomorph of a ﬁnitely generated abelian group J. Algebra 481 (2017), 327–347 we have redone the work of

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x)

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

13/22 for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

(Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x)

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular representation.

13/22 for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

Then

• λ(x) = γ(x)ρ(x)

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).)

13/22 for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

• λ(x) = γ(x)ρ(x)

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

13/22 for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x)

13/22 , as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x) for γ(x) = ι(x−1)

13/22 − • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x) for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

13/22 From right to left .

ι : G → Inn(G) ≤ Aut(G) g 7→ (x 7→ g−1xg)

Take N = λ(G) ⊴ Hol(G), where λ is the left regular

representation. (Actually, NS(G)(λ(G)) = Hol(G).) Then

• λ(x) = γ(x)ρ(x) for γ(x) = ι(x−1), as

−1 yι(x )ρ(x) = (xyx−1)x = xy = yλ(x).

− • Here y ◦ x = yγ(x)x = yx 1 x = xy yields the opposite group.

13/22 Commutators and perfect groups , where all values of γ are inner automorphisms.

γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′

.. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and

14/22 , where all values of γ are inner automorphisms.

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′

.. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

14/22 , where all values of γ are inner automorphisms.

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′

.. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

14/22 , where all values of γ are inner automorphisms.

This suggests to look at perfect groups G = G′

.. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

14/22 , where all values of γ are inner automorphisms.

.. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′

14/22 .. Skip Lemma, please

Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′, where all values of γ are inner automorphisms.

14/22 Commutators .

If N ⊴ Hol(G) is regular, then γ : G → Aut(G) satisﬁes

γ(xβ) = γ(x)β for β ∈ Aut(G), and γ(yx) = γ(x)γ(y).

These yield γ([β, g−1]) = [γ(g), β].

When β = ι(h), for some h ∈ G, we obtain

γ([h, g−1]) = ι([γ(g), h]).

This suggests to look at perfect groups G = G′, where all values of γ are inner automorphisms.

.. Skip Lemma, please

14/22 If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

yields [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)). Key fact: central automorphisms of perfect groups are trivial. Lemma If G is a perfect group, then Z(G) ≤ ker(γ).

Proof.

Perfect groups .

γ([h, g−1]) = ι([γ(g), h])

15/22 If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Key fact: central automorphisms of perfect groups are trivial. Lemma If G is a perfect group, then Z(G) ≤ ker(γ).

Proof.

Perfect groups .

γ([h, g−1]) = ι([γ(g), h]) yields [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)).

15/22 If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Lemma If G is a perfect group, then Z(G) ≤ ker(γ).

Proof.

Perfect groups .

γ([h, g−1]) = ι([γ(g), h]) yields [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)). Key fact: central automorphisms of perfect groups are trivial.

15/22 If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Proof.

Perfect groups .

γ([h, g−1]) = ι([γ(g), h]) yields [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)). Key fact: central automorphisms of perfect groups are trivial. Lemma If G is a perfect group, then Z(G) ≤ ker(γ).

15/22 If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Perfect groups .

Proof.

15/22 , the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Perfect groups .

Proof. If g ∈ Z(G)

15/22 yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Perfect groups .

Proof. If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G))

15/22 i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

Perfect groups .

Proof. If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

15/22 , and thus γ(g) = 1.

Perfect groups .

Proof. If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G

15/22 Perfect groups .

Proof. If g ∈ Z(G), the relation [γ(g), h] ≡ [g−1, γ(h−1)] (mod Z(G)) yields [γ(g), h] ∈ Z(G)

i.e. γ(g) is a central automorphism of G, and thus γ(g) = 1.

15/22 The regular normal subgroups of the holomorph of a perfect group , then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless

Theorem Let G be a ﬁnite perfect group.

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

Two decompositions .

16/22 , then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

Two decompositions . Theorem Let G be a ﬁnite perfect group.

16/22 , then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

• If G is centreless

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

Two decompositions . Theorem Let G be a ﬁnite perfect group.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

16/22 , then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

Two decompositions . Theorem Let G be a ﬁnite perfect group.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless

16/22 In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

Two decompositions . Theorem Let G be a ﬁnite perfect group.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless, then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

16/22 follows from γ(x ◦ y) = γ(yx).

Two decompositions . Theorem Let G be a ﬁnite perfect group.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless, then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)),

16/22 Two decompositions . Theorem Let G be a ﬁnite perfect group.

• If N ⊴ Hol(G) is regular, then Inn(G) = γ(G) × ι(ker(γ)).

• If G is centreless, then G = ι−1(γ(G)) × ker(γ)

is a product of two characteristic subgroups.

In the general case, γ(h) = ι(h−1), for h ∈ ι−1(γ(G)), follows from γ(x ◦ y) = γ(yx).

16/22 • T(G) is elementary abelian of order 2n.

• γ is trivial on K = ker(γ) and γ(h) = ι(h−1) on H. ∼ • (G, ◦) = N is obtained from G = H × K by replacing H with its opposite, as in Question 2:

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

G = L1 × · · · × Ln.

Divide the Li in two groups, say, H = L1 × · · · × Lm, K = Lm+1 × · · · × Ln. Now the ordered decomposition G = H × K corresponds to a regular subgroup N ⊴ Hol(G):

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group

17/22 • T(G) is elementary abelian of order 2n.

• γ is trivial on K = ker(γ) and γ(h) = ι(h−1) on H. ∼ • (G, ◦) = N is obtained from G = H × K by replacing H with its opposite, as in Question 2:

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

Divide the Li in two groups, say, H = L1 × · · · × Lm, K = Lm+1 × · · · × Ln. Now the ordered decomposition G = H × K corresponds to a regular subgroup N ⊴ Hol(G):

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

17/22 • T(G) is elementary abelian of order 2n.

• γ is trivial on K = ker(γ) and γ(h) = ι(h−1) on H. ∼ • (G, ◦) = N is obtained from G = H × K by replacing H with its opposite, as in Question 2:

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

Now the ordered decomposition G = H × K corresponds to a regular subgroup N ⊴ Hol(G):

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

Divide the Li in two groups, say, H = L1 × · · · × Lm, K = Lm+1 × · · · × Ln.

17/22 • T(G) is elementary abelian of order 2n.

• γ is trivial on K = ker(γ) and γ(h) = ι(h−1) on H. ∼ • (G, ◦) = N is obtained from G = H × K by replacing H with its opposite, as in Question 2:

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

Divide the Li in two groups, say, H = L1 × · · · × Lm, K = Lm+1 × · · · × Ln. Now the ordered decomposition G = H × K corresponds to a regular subgroup N ⊴ Hol(G):

17/22 • T(G) is elementary abelian of order 2n.

∼ • (G, ◦) = N is obtained from G = H × K by replacing H with its opposite, as in Question 2:

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

Divide the Li in two groups, say, H = L1 × · · · × Lm, K = Lm+1 × · · · × Ln. Now the ordered decomposition G = H × K corresponds to a regular subgroup N ⊴ Hol(G): • γ is trivial on K = ker(γ) and γ(h) = ι(h−1) on H.

17/22 • T(G) is elementary abelian of order 2n.

• All these N are isomorphic to G (see Question 2), so that

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2).

17/22 • T(G) is elementary abelian of order 2n.

Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that

17/22 Centreless groups: how to obtain all regular normal subgroups N of the holomorph . Let the Krull-Remak decomposition of G as an Aut(G)-group be

G = L1 × · · · × Ln.

(x1, x2) ◦ (y1, y2) = (y1x1, x2y2). • All these N are isomorphic to G (see Question 2), so that n • T(G) is elementary abelian of order 2 . 17/22 • Question 4: in this case NS(G)(N) > Hol(G);

∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated.

18/22 • Question 4: in this case NS(G)(N) > Hol(G);

∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

18/22 • Question 4: in this case NS(G)(N) > Hol(G);

∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable.

18/22 • Question 4: in this case NS(G)(N) > Hol(G);

∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites.

18/22 • Question 4: in this case NS(G)(N) > Hol(G);

∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time

18/22 ∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

• Question 4: in this case NS(G)(N) > Hol(G); ∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

18/22 ∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

• Question 4: in this case NS(G)(N) > Hol(G);

18/22 ∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

• Question 4: in this case NS(G)(N) > Hol(G); ∼ • even if they do, (G, ◦) = N need not be isomorphic to G:

18/22 The general case . When G is allowed to have a nontrivial centre, things become more complicated. One still has a central product decomposition

G = L1 ··· Ln,

with the Li ≥ Z(G) characteristic, centrally indecomposable. The regular subgroups N ⊴ Hol(G) are still obtained by replacing

some of the Li with their opposites. But this time ∼ • the groups (G, ◦) = N need not have the same automorphism group of G:

• Question 4: in this case NS(G)(N) > Hol(G); ∼ • even if they do, (G, ◦) = N need not be isomorphic to G: ∼ • Question 3: in this case NS(G)(N) = Hol(G) but N ≠ G.

18/22 Examples even modulo the centre,

• pairwise non-isomorphic, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

19/22 even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic,

19/22 • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre,

19/22 • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect,

19/22 • |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

19/22 • Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

19/22 These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

19/22 by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p)

19/22 via the insertion of a non self-dual module underneath.

Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism

19/22 Killing automorphisms .

For primes p ≡ 1 (mod 3), there is a family of groups Qp which are

• pairwise non-isomorphic, even modulo the centre, • perfect, • centrally indecomposable, and such that

• |Z(Qp)| = 3, and

• Aut(Qp) acts trivially on Z(Qp).

These are obtained from SL(3, p) by killing the transpose inverse automorphism via the insertion of a non self-dual module underneath.

19/22 • thereby inducing the identity on Z(K), and

• which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

(Both have order 3.) (G, ◦), obtained by replacing H with its opposite, is not isomorphic to G. An isomorphism of G to (G, ◦) would induce

• an automorphism of K

• an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

G = HK, a central product of two non-isomorphic Qp, with Z(H) amalgamated with Z(K).

20/22 • thereby inducing the identity on Z(K), and

• which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

(G, ◦), obtained by replacing H with its opposite, is not isomorphic to G. An isomorphism of G to (G, ◦) would induce

• an automorphism of K

• an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

G = HK, a central product of two non-isomorphic Qp, with Z(H) amalgamated with Z(K). (Both have order 3.)

20/22 • thereby inducing the identity on Z(K), and

• which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

An isomorphism of G to (G, ◦) would induce

• an automorphism of K

• an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

G = HK, a central product of two non-isomorphic Qp, with Z(H) amalgamated with Z(K). (Both have order 3.) (G, ◦), obtained by replacing H with its opposite, is not isomorphic to G.

20/22 • thereby inducing the identity on Z(K), and

• which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

• an automorphism of K

• an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

G = HK, a central product of two non-isomorphic Qp, with Z(H) amalgamated with Z(K). (Both have order 3.) (G, ◦), obtained by replacing H with its opposite, is not isomorphic to G. An isomorphism of G to (G, ◦) would induce

20/22 • which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

• thereby inducing the identity on Z(K), and • an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

• an automorphism of K

20/22 • which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

• an anti-automorphism of H

Example 1 (Negative answer to Question 3) .

• an automorphism of K • thereby inducing the identity on Z(K), and

20/22 • which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

Example 1 (Negative answer to Question 3) .

• an automorphism of K • thereby inducing the identity on Z(K), and • an anti-automorphism of H

20/22 • thereby inducing inversion on Z(H).

Example 1 (Negative answer to Question 3) .

• an automorphism of K • thereby inducing the identity on Z(K), and • an anti-automorphism of H • which is just the composition of an automorphism of H with inversion,

20/22 Example 1 (Negative answer to Question 3) .

• an automorphism of K • thereby inducing the identity on Z(K), and • an anti-automorphism of H • which is just the composition of an automorphism of H with inversion, • thereby inducing inversion on Z(H).

20/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

Then every isomorphism L1 → L2 takes z1 to z2.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

• Let z1 generate Z(L1). → ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2.

−1 • Choose the amalgamation so that z2 = z1 . • Then M and the Li are characteristic in G

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

Then every isomorphism L1 → L2 takes z1 to z2.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

→ ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2.

−1 • Choose the amalgamation so that z2 = z1 . • Then M and the Li are characteristic in G

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• Let z1 generate Z(L1).

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

Then every isomorphism L1 → L2 takes z1 to z2. −1 • Choose the amalgamation so that z2 = z1 . • Then M and the Li are characteristic in G

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• Let z1 generate Z(L1). → ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2.

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

−1 • Choose the amalgamation so that z2 = z1 . • Then M and the Li are characteristic in G

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• Let z1 generate Z(L1). → ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2. Then every isomorphism L1 → L2 takes z1 to z2.

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

• Then M and the Li are characteristic in G

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• Let z1 generate Z(L1). → ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2. Then every isomorphism L1 → L2 takes z1 to z2. −1 • Choose the amalgamation so that z2 = z1 .

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

: an automorphism of G • takes M to M

• if it takes L1 to L2, then it takes z1 to z2

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 , and thus ﬁxes the centre elementwise; −1 = z1 , i.e. it inverts the centre.

• takes M to M

• if it takes L1 to L2, then it takes z1 to z2

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 −1 = z1 , i.e. it inverts the centre.

, and thus ﬁxes the centre elementwise;

• if it takes L1 to L2, then it takes z1 to z2

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 −1 = z1 , i.e. it inverts the centre.

• if it takes L1 to L2, then it takes z1 to z2

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• Let z1 generate Z(L1). → ζ ∈ • Fix an isomorphism ζ : L1 L2, and deﬁne z2 = z1 L2. Then every isomorphism L1 → L2 takes z1 to z2. −1 • Choose the amalgamation so that z2 = z1 . • Then M and the Li are characteristic in G: an automorphism of G • takes M to M, and thus ﬁxes the centre elementwise;

21/22 −1 = z1 , i.e. it inverts the centre.

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

• if it takes L1 to L2, then it takes z1 to z2

21/22 , i.e. it inverts the centre.

Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 Example 2 (Negative answer to Question 4) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, with amalgamated centres.

21/22 , acting like ζ inv on L1.

inv −1 = z2 = z2 = z1

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

22/22 , acting like ζ inv on L1.

inv −1 = z2 = z2 = z1

There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite.

22/22 , acting like ζ inv on L1.

inv −1 = z2 = z2 = z1

• is the identity on M,

• takes L1 to L2

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which

22/22 , acting like ζ inv on L1.

inv −1 = z2 = z2 = z1

• takes L1 to L2

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

22/22 inv −1 = z2 = z2 = z1

, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2

22/22 inv −1 = z2 = z2 = z1

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

22/22 inv −1 = z2 = z2 = z1

and ζ inv z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

22/22 inv −1 = z2 = z2 = z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv z1

22/22 −1 = z2 = z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv z1 = z2

22/22 = z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv −1 z1 = z2 = z2

22/22 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv −1 z1 = z2 = z2 = z1

22/22 So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv −1 z1 = z2 = z2 = z1 is compatible with the identity on M.

22/22 , and Aut(G, ◦) is twice as big as Aut(G).

Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv −1 z1 = z2 = z2 = z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦) 22/22 Example 2 (continued) . ∼ ∼ G = L1L2M, a central product of Qp’s, with L1 = L2 ≠ M, −1 amalgamating z2 = z1 .

Get (G, ◦) by replacing L1 with its opposite. There is an automorphism of (G, ◦) which • is the identity on M,

• takes L1 to L2, acting like ζ inv on L1.

In fact for x, y ∈ L1 (x ◦ y)ζ inv = (yx)ζ inv = (yζ xζ )inv = xζ invyζ inv,

and ζ inv inv −1 z1 = z2 = z2 = z1 is compatible with the identity on M.

So the Li are not characteristic in (G, ◦), and Aut(G, ◦) is twice as 22/22 big as Aut(G). Thanks .

That’s All, Thanks!

22/22