On the capabillity of p -groups of class two and prime exponent - - PowerPoint PPT Presentation

On the capabillity of p-groups of class two and prime exponent

Arturo Magidin

University of Louisiana at Lafayette

Groups St Andrews 2013

Arturo Magidin

Capability

Definition A group G is capable if and only if there exists K such that G ∼ = K/Z(K).

Arturo Magidin

The gold standard theorem

Theorem (Baer) Let G be a finitely generated abelian group, and write G = Ca1 ⊕ · · · ⊕ Car , with a1|a2| · · · |ar. Then G is capable if and only if r ≥ 2 and ar−1 = ar.

Arturo Magidin

p-groups of class 2

Specific p-group of class 2 can be checked for capability easily enough.

Arturo Magidin

p-groups of class 2

Specific p-group of class 2 can be checked for capability easily enough. But no full characterization of capability for p-groups of class 2 (similar to Baer’s for abelian) exists, and current techniques do not seem sufficient.

Arturo Magidin

p-groups of class 2

Specific p-group of class 2 can be checked for capability easily enough. But no full characterization of capability for p-groups of class 2 (similar to Baer’s for abelian) exists, and current techniques do not seem sufficient. For the subclass of p-groups of class 2 and exponent p, there is some hope: there are known sufficient and known necessary conditions for capability. Sufficient conditions are notoriously hard to come by.

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Some known necessary conditions for groups of class 2 and exponent p

If G is a nontrivial central product, G = AB with A ⊆ C(B), B ⊆ C(A), [A, A] ∩ [B, B] = {1}, then G is not capable.

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Some known necessary conditions for groups of class 2 and exponent p

If G is a nontrivial central product, G = AB with A ⊆ C(B), B ⊆ C(A), [A, A] ∩ [B, B] = {1}, then G is not capable. In particular, if G is extraspecial, then G is capable if and

• nly if |G| = p3 and G is of exponent p. (Beyl, Felgner,

Schmidt; 1979).

Arturo Magidin

Some known necessary conditions for groups of class 2 and exponent p

If G is a nontrivial central product, G = AB with A ⊆ C(B), B ⊆ C(A), [A, A] ∩ [B, B] = {1}, then G is not capable. In particular, if G is extraspecial, then G is capable if and

• nly if |G| = p3 and G is of exponent p. (Beyl, Felgner,

Schmidt; 1979). If G is capable, and rank([G, G]) = k, then rank(G/Z(G)) ≤ 2k + k

2

• (Heineken and Nikolova; 1996).

Arturo Magidin

Some known necessary conditions for groups of class 2 and exponent p

If G is a nontrivial central product, G = AB with A ⊆ C(B), B ⊆ C(A), [A, A] ∩ [B, B] = {1}, then G is not capable. In particular, if G is extraspecial, then G is capable if and

• nly if |G| = p3 and G is of exponent p. (Beyl, Felgner,

Schmidt; 1979). If G is capable, and rank([G, G]) = k, then rank(G/Z(G)) ≤ 2k + k

2

• (Heineken and Nikolova; 1996).

(Intuitively: if G is capable, then its commutator subgroup cannot be too small)

Arturo Magidin

Some known necessary conditions for groups of class 2 and exponent p

If g1, . . . , gn generated G, Z(G) = G′, and

n

• i=1

[CG(gi), CG(gi)] = 1, then G is not capable (follows easily from work of Ellis of 1996).

Arturo Magidin

Some known necessary conditions for groups of class 2 and exponent p

If g1, . . . , gn generated G, Z(G) = G′, and

n

• i=1

[CG(gi), CG(gi)] = 1, then G is not capable (follows easily from work of Ellis of 1996). Conjecture Let G be a p-group of exponent p with Z(G) = G′. Then G is capable if and only if for all minimal generating sets g1, . . . , gn

N

• i=1
• CG(gi), CG(gi)
• = 1.

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Some known sufficient conditions for groups of class 2 and exponent p

If g1, . . . , gn project onto a basis for Gab, and the nontrivial commutators among [gj, gi] are distinct and form a basis for [G, G], then G is capable. (Ellis, 1996)

Arturo Magidin

Some known sufficient conditions for groups of class 2 and exponent p

If g1, . . . , gn project onto a basis for Gab, and the nontrivial commutators among [gj, gi] are distinct and form a basis for [G, G], then G is capable. (Ellis, 1996) (Intuitively: if the relations among basic commutators are very simple, then G is capable)

Arturo Magidin

Some known sufficient conditions for groups of class 2 and exponent p

If g1, . . . , gn project onto a basis for Gab, and the nontrivial commutators among [gj, gi] are distinct and form a basis for [G, G], then G is capable. (Ellis, 1996) (Intuitively: if the relations among basic commutators are very simple, then G is capable) If G is a coproduct, G = A ∐N2 B with A and B nontrivial, then G is capable (2008).

Arturo Magidin

Some known sufficient conditions for groups of class 2 and exponent p

If g1, . . . , gn project onto a basis for Gab, and the nontrivial commutators among [gj, gi] are distinct and form a basis for [G, G], then G is capable. (Ellis, 1996) (Intuitively: if the relations among basic commutators are very simple, then G is capable) If G is a coproduct, G = A ∐N2 B with A and B nontrivial, then G is capable (2008). (A ∐N2 B is the most general group of class 2 and exponent p that contains a copy of A and a copy of B)

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Other results

If G is a p-group of class two and exponent p, then there exist groups G1 and G2 such that: G ≤ G1, G2;

Arturo Magidin

Other results

If G is a p-group of class two and exponent p, then there exist groups G1 and G2 such that: G ≤ G1, G2; Gp

1 = Gp 2 = {1};

Arturo Magidin

Other results

If G is a p-group of class two and exponent p, then there exist groups G1 and G2 such that: G ≤ G1, G2; Gp

1 = Gp 2 = {1};

Z(Gi) = [Gi, Gi], i = 1, 2;

Arturo Magidin

Other results

If G is a p-group of class two and exponent p, then there exist groups G1 and G2 such that: G ≤ G1, G2; Gp

1 = Gp 2 = {1};

Z(Gi) = [Gi, Gi], i = 1, 2; Neither is a nontrivial central product;

Arturo Magidin

Other results

If G is a p-group of class two and exponent p, then there exist groups G1 and G2 such that: G ≤ G1, G2; Gp

1 = Gp 2 = {1};

Z(Gi) = [Gi, Gi], i = 1, 2; Neither is a nontrivial central product; G1 is capable and G2 is not capable.

Arturo Magidin

Counterpart to Heineken-Nikolova

I want to talk about a counterpart to: Theorem (Heineken, Nikolova) If G is capable, and rank([G, G]) = k, then rank(G/Z(G)) ≤ 2k + k

2

• .

Arturo Magidin

Counterpart to Heineken-Nikolova

I want to talk about a counterpart to: Theorem (Heineken, Nikolova) If G is capable, and rank([G, G]) = k, then rank(G/Z(G)) ≤ 2k + k

2

• .

Namely, a result that says that if [G, G] is “sufficiently large”, then G is capable.

Arturo Magidin

Set-up

Let G be a group of class 2 and exponent p, Z(G) = [G, G], and g1, . . . , gn projecting onto a basis for Gab.

Arturo Magidin

Set-up

Let G be a group of class 2 and exponent p, Z(G) = [G, G], and g1, . . . , gn projecting onto a basis for Gab. Let F = x1, . . . , xn be the relatively free group of rank n, class 3, and exponent p.

Arturo Magidin

Set-up

Let G be a group of class 2 and exponent p, Z(G) = [G, G], and g1, . . . , gn projecting onto a basis for Gab. Let F = x1, . . . , xn be the relatively free group of rank n, class 3, and exponent p. If ψ: F → G is induced by xi → gi, then ψ factors through F/F3, and we have: F

/Z(F)

− − − − − − − − − − − → F/F3

/[X,F]

 

• 

 /XF3 F/[X, F] − − − − − − − − − − − →

/(XF3/[X,F])

G

Arturo Magidin

Example

Say G = g1, g2, g3 | [g1, g2] = [g1, g3], G3 = 1, Gp = 1.

Arturo Magidin

Example

Say G = g1, g2, g3 | [g1, g2] = [g1, g3], G3 = 1, Gp = 1. We can take X = [x1, x2][x1, x3]−1 ≤ F. Then F/XF3 ∼ = G.

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Example

Say G = g1, g2, g3 | [g1, g2] = [g1, g3], G3 = 1, Gp = 1. We can take X = [x1, x2][x1, x3]−1 ≤ F. Then F/XF3 ∼ = G. In any witness to the capability of G, we would need to have

• [x1, x2][x1, x3]−1, xj
• = 1,

j = 1, 2, 3; i.e., [X, F] would have to be trivial.

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Connection to epicenter

Z ∗(G) is the epicenter of G. It is the smallest normal subgroup

• f G for which G/N is capable.

Arturo Magidin

Connection to epicenter

Z ∗(G) is the epicenter of G. It is the smallest normal subgroup

• f G for which G/N is capable.

F

/Z(F)

− − − − − − − − − − − → F/F3

/[X,F]

 

• 

 /XF3 F/[X, F] − − − − − − − − − − − →

/(XF3/[X,F])

G

Arturo Magidin

Connection to epicenter

Z ∗(G) is the epicenter of G. It is the smallest normal subgroup

• f G for which G/N is capable.

F

/Z(F)

− − − − − − − − − − − → F/F3

/[X,F]

 

• 

 /XF3 F/[X, F] − − − − − − − − − − − →

/(XF3/[X,F])

G Proposition The image of Z(F/[X, F]) in G is Z ∗(G).

Arturo Magidin

Connection to epicenter

Z ∗(G) is the epicenter of G. It is the smallest normal subgroup

• f G for which G/N is capable.

F

/Z(F)

− − − − − − − − − − − → F/F3

/[X,F]

 

• 

 /XF3 F/[X, F] − − − − − − − − − − − →

/(XF3/[X,F])

G Proposition The image of Z(F/[X, F]) in G is Z ∗(G). Corollary G is capable if and only Z(F/[X, F]) = XF3/[X, F].

Arturo Magidin

Idea

How many relations in F do we need to specify that the elements of X are central?

Arturo Magidin

Idea

How many relations in F do we need to specify that the elements of X are central? If S generates X, then in principle we need n|S| relations: [s, xi], i = 1, . . . , n, s ∈ S.

Arturo Magidin

Idea

How many relations in F do we need to specify that the elements of X are central? If S generates X, then in principle we need n|S| relations: [s, xi], i = 1, . . . , n, s ∈ S. But there may be commutator relations that reduce this number.

Arturo Magidin

Idea

How many relations in F do we need to specify that the elements of X are central? If S generates X, then in principle we need n|S| relations: [s, xi], i = 1, . . . , n, s ∈ S. But there may be commutator relations that reduce this

• number. E.g., if [x2, x1], [x3, x1], and [x3, x2] are in S,

Arturo Magidin

Idea

How many relations in F do we need to specify that the elements of X are central? If S generates X, then in principle we need n|S| relations: [s, xi], i = 1, . . . , n, s ∈ S. But there may be commutator relations that reduce this

• number. E.g., if [x2, x1], [x3, x1], and [x3, x2] are in S, then

[x2, x1, x3] and [x3, x1, x2] already give [x3, x2, x1], since [x3, x2, x1][x2, x1, x3][x1, x3, x2] = 1.

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Idea II

Suppose that rank(X) = k, and for every X ′ with rank(X ′) > k, the number of relations needed to specify the elements of X ′ are central is strictly larger than those needed for X.

Arturo Magidin

Idea II

Suppose that rank(X) = k, and for every X ′ with rank(X ′) > k, the number of relations needed to specify the elements of X ′ are central is strictly larger than those needed for X. Since the relations that specify X also specify the pullback of Z ∗(G), this would imply that Z ∗(G) is trivial; i.e., that G is capable.

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Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m.

Arturo Magidin

Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m. Let X be the relations that specify X is central.

Arturo Magidin

Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m. Let X be the relations that specify X is central. Clearly rank(X) ≤ nm.

Arturo Magidin

Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m. Let X be the relations that specify X is central. Clearly rank(X) ≤ nm. How much smaller than nm can it be?

Arturo Magidin

Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m. Let X be the relations that specify X is central. Clearly rank(X) ≤ nm. How much smaller than nm can it be? Turns out that any decrease in size must be a consequence of the Hall-Witt identity [x, y, z][y, z, x][z, x, y] = 1.

Arturo Magidin

Bounding the size

Say X is minimal with F/XF3 ∼ = G, and let rank(X) = m. Let X be the relations that specify X is central. Clearly rank(X) ≤ nm. How much smaller than nm can it be? Turns out that any decrease in size must be a consequence of the Hall-Witt identity [x, y, z][y, z, x][z, x, y] = 1. (In particular, if rank(X) ≤ 2, then rank(X) = nrank(X), and all such groups are capable.)

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The theorem

Definition Fix n, and let m be such that 0 ≤ m ≤ n

2

• . Define f(m) as

f(m) = max{nm − rank(X) | rank(X) = m}.

Arturo Magidin

The theorem

Definition Fix n, and let m be such that 0 ≤ m ≤ n

2

• . Define f(m) as

f(m) = max{nm − rank(X) | rank(X) = m}. That is, f(m) is the most we can “save” in specifying that X is central.

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The theorem

Definition Fix n, and let m be such that 0 ≤ m ≤ n

2

• . Define f(m) as

f(m) = max{nm − rank(X) | rank(X) = m}. That is, f(m) is the most we can “save” in specifying that X is central. Theorem Let G be a group such that the corresponding X has rank m, and X has rank k. If f(m + 1) < n(m + 1) − k, then G is capable.

Arturo Magidin

But what is f?

For each triple of indices, we may hope to save one relation using the Hall-Witt identity.

Arturo Magidin

But what is f?

For each triple of indices, we may hope to save one relation using the Hall-Witt identity. (e.g., if we have [x2, x1], [x3, x1], [x3, x2] in X, we need only specify that [x2, x1, x3] and [x3, x1, x2] are trivial to get that [x3, x2, x1] is trivial).

Arturo Magidin

But what is f?

For each triple of indices, we may hope to save one relation using the Hall-Witt identity. (e.g., if we have [x2, x1], [x3, x1], [x3, x2] in X, we need only specify that [x2, x1, x3] and [x3, x1, x2] are trivial to get that [x3, x2, x1] is trivial). In addition, if we have all [xj, xi], 1 ≤ i < j ≤ N, but we use larger indices, we may save one relation for any two other indices.

Arturo Magidin

But what is f?

For each triple of indices, we may hope to save one relation using the Hall-Witt identity. (e.g., if we have [x2, x1], [x3, x1], [x3, x2] in X, we need only specify that [x2, x1, x3] and [x3, x1, x2] are trivial to get that [x3, x2, x1] is trivial). In addition, if we have all [xj, xi], 1 ≤ i < j ≤ N, but we use larger indices, we may save one relation for any two other indices.E.g., if [xN+1, x1] and [xN+1, x2] are in X, then specifying [xN+1, x1, x2] will give [xN+1, x2, x1], because we already have [x2, x1, xN+1] as a consequence of [x2, x1] ∈ X.

Arturo Magidin

But what is f?

For each triple of indices, we may hope to save one relation using the Hall-Witt identity. (e.g., if we have [x2, x1], [x3, x1], [x3, x2] in X, we need only specify that [x2, x1, x3] and [x3, x1, x2] are trivial to get that [x3, x2, x1] is trivial). In addition, if we have all [xj, xi], 1 ≤ i < j ≤ N, but we use larger indices, we may save one relation for any two other indices.E.g., if [xN+1, x1] and [xN+1, x2] are in X, then specifying [xN+1, x1, x2] will give [xN+1, x2, x1], because we already have [x2, x1, xN+1] as a consequence of [x2, x1] ∈ X. Theorem Fix n > 1. If 0 ≤ m ≤ n

2

• , and we write m =

r

2

• + s, 0 ≤ s < r,

then f(m) = r 3

• +

s 2

• .

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable.

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2)

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable.

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable. (However, there are G with |[G, G]| = p2, p3 that are capable, and some that are not)

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable. (However, there are G with |[G, G]| = p2, p3 that are capable, and some that are not) (iii) If |G/Z| = p6 and |[G, G]| ≥ p8, then G is capable.

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable. (However, there are G with |[G, G]| = p2, p3 that are capable, and some that are not) (iii) If |G/Z| = p6 and |[G, G]| ≥ p8, then G is capable. (iv) If |G/Z| = p7 and |[G, G]| ≥ p14, then G is capable.

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable. (However, there are G with |[G, G]| = p2, p3 that are capable, and some that are not) (iii) If |G/Z| = p6 and |[G, G]| ≥ p8, then G is capable. (iv) If |G/Z| = p7 and |[G, G]| ≥ p14, then G is capable. (v) . . .

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Consequences

Corollary Let G be of class 2 and odd prime exponent p. (i) If |G/Z| = p4 and |[G, G]| ≥ p2, then G is capable. (In fact, G is capable if and only if [G, G] ≥ p2) (ii) If |G/Z| = p5 and |[G, G]| ≥ p4, then G is capable. (However, there are G with |[G, G]| = p2, p3 that are capable, and some that are not) (iii) If |G/Z| = p6 and |[G, G]| ≥ p8, then G is capable. (iv) If |G/Z| = p7 and |[G, G]| ≥ p14, then G is capable. (v) . . . (Intuitively: if [G, G] is large enough, then G is capable.)

Arturo Magidin

Thank you for your attention.

Arturo Magidin