𝑝-groups with exactly four codegrees

Theorem 1.1.

Let G be a finite p-group such that

If any of the following holds, then G has nilpotence class at most 4:

1. |cd⁡(G)|=2,

2. cd⁡(G)={1,p,p2},

3. |G:G′|=p2.

Theorem 1.2.

Let G be a p-group such that (1.1) holds. If G has coclass at most 3, then G has nilpotence class at most 4, and |G|≤p7.

Hypothesis ${\boldsymbol{(*)}}$.

If G is a p-group with nilpotence class n such that |G|≥p2⁢n, then |Z2⁢(G)|≠p2.

Theorem 1.3.

Let G be a finite p-group such that (1.1) holds. If G has coclass at most 6, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G|≤p10.

Theorem 1.4.

If G is a finite p-group such that (1.1) holds, then c⁢(G)≤a+1.

Theorem 1.5.

If G is a finite p-group such that cod⁡(G)={1,p,pa-1,pa} for a≥4, then c⁢(G)≤a.

Lemma 2.1 ([5, Lemma 3.1]).

Let G be a group, and suppose that χ∈Irr⁡(G) is faithful. If χ≠1G, then |G|=χ(1)cod(χ)<cod(χ)2.

Lemma 2.2.

Let G be a p-group with p2∉cod⁡(G). If χ is an irreducible character of G such that cod⁡(χ)>p, then χ is nonlinear.

Proof.

Let χ be an irreducible character of G such that cod⁡(χ)=pa for some a>2, and suppose that χ is linear. Then ker⁡(χ)≥G′, so G/ker⁡(χ) is abelian. Since χ is a faithful irreducible character of G/ker⁡(χ), this quotient must be cyclic, and |G/ker⁡(χ)|=χ⁢(1)⁢cod⁡(χ)=pa>p2. By [5, Corollary 2.5], G/G′ is elementary abelian, which is impossible since ker⁡(χ)≥G′ and G/ker⁡(χ) is cyclic with order greater than p2. ∎

Lemma 2.3.

If a finite p-group G has a faithful irreducible character of degree p, then G has a normal abelian subgroup of index p and cd⁡(G)={1,p}.

Lemma 2.4 ([1, Theorem 22.5]).

If G is a nonabelian p-group with

then one and only one of the following holds:

1. G has an abelian subgroup of index p,

2. G/Z⁢(G) is of order p3 and exponent p.

Lemma 2.5 ([1, Lemma 1.1]).

Let A be an abelian subgroup of index p of a nonabelian p-group G. Then |G|=p⁢|G′|⁢|Z⁢(G)|.

Lemma 2.6 ([6, Satz III.2.13]).

Let G be a p-group with nilpotence class n. Then

where i=1,…,n-1.

Lemma 2.7.

Let G be a finite p-group such that (1.1) holds. If |G| is minimal such that G has nilpotence class n≥3, then G has a faithful irreducible character.

Proof.

Suppose G does not have a faithful irreducible character. Let

and let |G| be minimal such that G has class n. Let K be the kernel of an irreducible character of G, and note that K>1. Either cod⁡(G/K)⊊cod⁡(G), in which case c⁢(G/K)≤2, or cod⁡(G/K)=cod⁡(G), and since |G| is minimal, we have c⁢(G/K)≤n-1. Hence there is a central series of G with the n-1 term contained in K for all kernels of irreducible characters of G, and since the intersection of these kernels is trivial, G has class at most n-1, a contradiction. Thus G must have a faithful irreducible character.∎

Proof of Theorem 1.4.

Let G be a minimal counterexample, and notice that

for otherwise c⁢(G)≥a+3 implies c⁢(G/Z)≥a+2≥5, and by [5, Theorem 1.2], cod⁡(G/Z)={1,p,pb,pa}, contradicting that |G| is minimal. By Lemma 2.7, G has a faithful character of codegree at most pa and by Lemma 2.1, |G|≤p2⁢a-1. Using [5, Theorem 1.2] and c⁢(G/Za-1)=3, we have cod⁡(G/Za-1)=cod⁡(G). Let χ be an irreducible character of G/Za-1 such that cod⁡(χ)=pa. Then we have pa≤χ(1)pa=|G:ker(χ)|≤|G:Za-1|≤pa, so χ is a faithful linear character of G/Za-1, which is impossible since G/Za-1 has class 3. Hence c⁢(G)≤a+1. ∎

Proof of Theorem 1.5.

Let G be a minimal counterexample. Then c⁢(G)=a+1, and by Lemma 2.7, G has a faithful character with codegree at most pa, giving |G|≤p2⁢a-1. The quotient G/Za-2 has class 3 and therefore has the same set of codegrees as G. Let χ be an irreducible character of G/Za-2 such that cod⁡(χ)=pa. By Lemma 2.2, χ is nonlinear, so we have

which shows that χ is faithful, χ⁢(1)=p and, by Lemma 2.3, cd⁡(G/Za-2)={1,p}. Let μ be an irreducible character of G/Za-2 with codegree pa-1. Then

which shows that |ker(μ):Za-2|=p and hence ker⁡(μ)≤Za-1.

Suppose |Za-1:Za-2|=p. Then μ is a faithful character of G/Za-1, and Za/Za-1 is cyclic. Notice that

so |Za-2|=pa-2. Hence |Za-1|=pa-1, and Zi+1/Zi is the unique minimal normal subgroup of G/Zi for i≤a-1. This implies Za-1≤G′≤Za. Since p2∉cod⁡(G), we have that G/G′ is elementary abelian and hence |Za:G′|≤p. Notice that Z⁢(μ)=Za and, by [7, Lemma 2.31], |G:Z(μ)|=p2. We also have |G′:Za-1|=p as G′≥Za-1≥G3 and G′/G3 is elementary abelian while G′/Za-1 is cyclic. If G′=Za, then |G/Za-2|=p4. Since |G:Za-2|=pa+1, this implies that a=3, and we obtain a contradiction to the fact that χ is a faithful nonlinear character with codegree pa≥p4. Thus |Za:G′|=p. Since Za/Za-1 is cyclic (of order p2), there is an element g so that Za=〈a,Za-1〉. Observe that [gp,h]⁢Za-2=[g,h]p⁢Za-2 for all h∈G. This implies that gp∈Za-1, which contradicts g⁢Za-1 having order p2.

We may now assume that |Za-1:Za-2|>p. Let G/Za-2=H, so H has class 3, |Z⁢(H)|>p and μ is an irreducible character of H with |ker⁡(μ)|=p. Since χ is a faithful irreducible character of H, Z⁢(H) is cyclic, and hence ker⁡(μ) is the unique subgroup of Z⁢(H) of order p. As G/G′ is elementary abelian, the subgroup H3=[H′,H] in the lower central series of H is also elementary abelian by Lemma 2.6. Since H3 is also contained in the cyclic subgroup Z⁢(H), we see that |H3|=p and hence H3=ker⁡(μ). As H/H3 has class 2, we have H′/H3≤Z⁢(H/H3)=Z⁢(μ)/H3, which shows that H/Z⁢(μ) is abelian and hence |H:Z(μ)|=p2. Then Z⁢(μ)=Z2⁢(H) as Z⁢(μ)≤Z2⁢(H). By [7, Lemma 2.27], we have that Z⁢(μ)/ker⁡(μ)=Z2⁢(H)/H3 is cyclic. We also have H3≤H′≤Z2⁢(H), so H′/H3 is cyclic as well as elementary abelian, and hence |H′:H3|=p. This is impossible since |Z(H):H3|=p and |H′:Z(H)|≥p. Thus c⁢(G)≤a. ∎

Lemma 2.8.

Let G be a finite p-group such that c⁢(G)≥4 and (1.1) holds. If χ is a faithful irreducible character of G, then cod⁡(χ)=pa.

Proof.

Suppose first that cod⁡(χ)=p or p2. Then |G|≤p3 by Lemma 2.1, which is impossible since c⁢(G)≥4. Suppose cod⁡(χ)=pb≥p3. Now p2∉cod⁡(G), and hence G/G′ is elementary abelian.

Let c⁢(G)=n. Note that G/Zn-2 has nilpotence class 2. Let φ∈Irr⁡(G/Zn-2) be nonlinear, and let cod⁡(φ)=pr. Note that, since φ is nonlinear, r is equal to either b or a. Put ker⁡(φ)=K and Z⁢(φ)=Y. By definition, φ(1)pr=|G:K|, and since G/Zn-2 has nilpotence class 2, |G:Y|=φ(1)2 by [7, Theorem 2.31]. Combining these equations yields prφ(1)=φ(1)2|Y:K|, and hence we have pr=φ(1)|Y:K|. Now |G|=χ(1)pb≤χ(1)pr=χ(1)φ(1)|Y:K|, so

This shows that pr>χ⁢(1)≥φ⁢(1)⁢|K|, so φ(1)|K|<pr=φ(1)|Y:K|. Finally, this implies |K|<|Y:K|.

Let G¯=G/K. Clearly, c⁢(G¯)=2 and Y¯=G′¯. Since G¯/G′¯ is elementary abelian, G′¯ is elementary abelian by Lemma 2.6. Since Y¯ is cyclic, we have Y¯/G′¯≤ℤp and G′¯≅ℤp. Hence |Y:K|≤p2. This implies that |K|≤p. Since n is at least 4, we see that Zn-2≤K has order at least p2, a contradiction. Hence cod⁡(χ)=pa, as desired. ∎

Proof of Theorem 1.1 i.

Let G be a minimal counterexample. By Lemmas 2.7 and 2.8, G has at least one faithful irreducible character χ, and such a character must have codegree pa. Let |G|=pn, and notice that χ⁢(1)=pn-a. Since c⁢(G/Z)=4, [5, Theorem 1.2] implies cod⁡(G/Z)=cod⁡(G). Let φ∈Irr⁡(G/Z) have codegree pa. If φ is nonlinear, then φ⁢(1)=pn-a, and hence

This shows that φ is a faithful irreducible character of G, contradicting that the kernel of φ contains Z. Hence φ must be linear. By Lemma 2.2, p2∈cod⁡(G), and by a similar argument, pa=p3. By Lemma 2.1, we have |G|<cod(χ)2=p6, implying that |G|≤p5, and hence G has nilpotence class at most 4. ∎

Proof of Theorem 1.1 ii.

Let G be a minimal counterexample. Notice that, for a nontrivial kernel K of an irreducible character of G, if cod⁡(G/K)=cod⁡(G), then cd⁡(G/K)=cd⁡(G), for otherwise we have a contradiction with Theorem 1.1 i. As |G| is minimal, we must have c⁢(G/K)<c⁢(G), and we can apply Lemmas 2.7 and 2.8. Now c⁢(G)=5, and G has a faithful irreducible character χ with codegree pa. By Lemma 2.3, χ must have degree p2, for otherwise |cd⁡(G)|={1,p}, a contradiction. This implies that pa+2=χ⁢(1)⁢cod⁡(χ)=|G|, and since G has class 5, we see that a is at least 4.

The quotient G/Z2 has nilpotence class 3, hence cod⁡(G/Z2)=cod⁡(G), and we can find γ∈Irr⁡(G/Z2) with codegree pa. Since a≥4 and |cod⁡(G)|=4, either p2 or p3 is not in cod⁡(G). Hence γ is nonlinear, and we have

which is impossible. Thus G can have nilpotence class at most 4. ∎

Lemma 2.9 ([3, Theorem 2.4]).

Let G be a maximal class p-group such that |G|≥p4. Then p3∈cod⁡(G).

Lemma 2.10.

Let G be a nonabelian p-group such that |G:G′|=p2. Then |G:G3|=p3, and |G:G4|≤p5.

Proof.

Since |G:G′|=p2, G is two-generated, and we can write G=〈x,y〉. Then G′/G3=〈[x,y]〉⁢G3/G3, and since G/G′ is elementary abelian, we have |G′:G3|=p. For G3/G4, we have G3/G4=〈[x,y,x],[x,y,y]〉⁢G4/G4, and hence |G3:G4|=p or p2. ∎

Proof of Theorem 1.1 iii.

Let G be a minimal counterexample. Then G has nilpotence class 5, and by Lemmas 2.7 and 2.8, G has a faithful character χ with codegree pa. Let φ∈Irr⁡(G/G3) be nonlinear. As G/G3 has class 2, G/Z⁢(φ) is abelian and |G:Z(φ)|=φ(1)2. Since Z⁢(φ) contains G′ which has index p2 in G, we see that φ⁢(1)=p. Put cod⁡(φ)=pr. Then

By Lemma 2.10, |G:G3|=p3, which implies r=2 and cod⁡(G/G3)={1,p,p2}.

The nilpotence class of G/G4 is 3, which implies cod⁡(G/G4)={1,p,p2,pa}. Let θ∈Irr⁡(G/G4) be nonlinear with codegree pa. Then

By Lemma 2.10, |G:G4|≤p5, which implies a≤4. Recalling that G has a faithful character with codegree pa, we have that |G|≤p7. If |G|=p6, then by Lemma 2.9, a=3, implying that |G| is actually at most p5, which contradicts that G has nilpotence class 5. Hence |G|=p7 and a=4.

We now have

If |G5|=p2, then G/G5 has maximal class, and by Lemma 2.9, p3∈cod⁡(G/G5), a contradiction. Hence |G5|=p. Suppose |G:G4|=p4, and let N be a normal subgroup of G such that G5<N<G4. Then G/N has order p5 and class 4, and we can again obtain a contradiction from Lemma 2.9. Now let |G:G4|=p5, and let N be a normal subgroup of G such that G4<N<G3. Then G/N has order p4 and class 3, and Lemma 2.9 yields the desired contradiction. Thus, no minimal counterexample exists. ∎

Lemma 2.11.

Let G be a finite p-group such that (1.1) holds. If G has coclass 2, then the nilpotence class of G is at most 4, and |G|≤p6.

Proof.

Let G be a minimal counterexample, and suppose c⁢(G)=c>5. The nilpotence class of G/Z is c-1>4, so G/Z must be maximal class since |G| was minimal with coclass 2, but this contradicts Theorem 1.1 iii. Hence we may assume G has nilpotence class 5, which implies that |G|=p7.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible character of G, then either c⁢(G/K)≤4, or c⁢(G/K)=5 and G/K has maximal class. Since the latter contradicts Theorem 1.1 iii, c⁢(G/K)≤4 for all irreducible characters of G, which contradicts that G has class 5. Hence G has a faithful irreducible character χ, and by Lemma 2.8, χ has codegree pa. By Lemma 2.1, we have pa≥p4.

Let H=G/Z2. As H has class 3 and order p4 or p5, by [5, Theorem 1.2], cod⁡(H)=cod⁡(G). If |Z2|≥p3, then |H|≤p4≤pa, which implies that H is cyclic, a contradiction. Hence |Z2|=p2 and |H|=p5. Let χ∈Irr⁡(H) have codegree pa. If χ is linear, then |H/ker⁡(χ)|=pa, which implies H is abelian, a contradiction. Hence χ is nonlinear, and we have

This shows that χ is a faithful character of H with degree p, and pa=p4.

Notice that Z2≤G′, and |G:G′|≥p3 by Theorem 1.1 iii. This shows that |H:H′|≥p3 and thus |H:H′|=p3. By Lemma 2.3 and Lemma 2.5, we have |H|=p⁢|H′|⁢|Z⁢(H)|, which implies that Z⁢(H)=Z3/Z2 is a cyclic group of order p2. However, we have exp(Z3/Z2)∣exp(Z)=p, a contradiction. ∎

Lemma 2.12.

Let G be a finite p-group such that (1.1) holds. If G has coclass 3, then the nilpotence class of G is at most 4, and |G|≤p7.

Proof.

Let G be a minimal counterexample, and suppose c⁢(G)=c>5. The nilpotence class of G/Z is c-1>4, so G/Z must be maximal class or have coclass 2 since |G| was minimal with coclass 3, but this contradicts Theorem 1.1 iii and Lemma 2.11. Hence we may assume G has nilpotence class 5.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible character of G, then either c⁢(G/K)≤4, or c⁢(G/K)=5 and G/K has maximal class or coclass 2, contradicting Theorem 1.1 iii and Lemma 2.11, respectively. Hence we must have c⁢(G/K)≤4 for all irreducible characters of G, which contradicts that G has class 5. Thus G has a faithful irreducible character χ, and by Lemma 2.8, χ has codegree pa.

As G has coclass 3 and nilpotence class 5, |G|=p8, and by Lemma 2.1, 5≤a≤7. If a=7, then χ⁢(1)=p. By Lemma 2.3 and Theorem 1.1 i, this is impossible, so a is at most 6. Theorem 1.1 of [4] implies cod⁡(G)={1,p,p2,pa}.

Suppose p2≤|Z|≤p3. Let θ∈Irr⁡(G/Z) have codegree pa. Since a is at least 5, θ is nonlinear, and pa+1≤θ(1)cod(θ)≤|G:Z|≤p6. This shows that a=5 and |Z|=p2. Since G/Z2 has class 3, we have p5∈cod⁡(G/Z2). This implies |G:Z2|≥p6, which is impossible since |G:Z|=p6. Hence Z=G5 has order p.

The quotient G/Z2 has class 3, and hence pa∈cod⁡(G/Z2). As |G:Z2|≤p6 and pa=p5 or p6 is the codegree of a nonlinear character of G/Z2, we must have |G:Z2|=p6 and a=5. Let γ∈Irr⁡(G/Z2) have codegree p5. Then γ is nonlinear, so

which shows that γ is a faithful character of G/Z2 with degree p. By Lemma 2.3, this implies that cd⁡(G/Z2)={1,p}.

Suppose |Z3|≥p4. As G/Z2 has a faithful character, Z3/Z2 is cyclic, and we can write Z3=〈x,Z2〉, where xp∉Z2. Let g∈G such that [xp,g]∉Z. Putting G¯ for G/Z, we have 1¯≠[xp,g]¯=[x,g]p¯, and since [x,g]∈Z2, we have [x,g]p¯=1¯, a contradiction. Hence |Z3|=p3.

Since cd⁡(G/Z2)={1,p} and |G:Z3|=p5, Lemma 2.4 implies that G/Z2 has an abelian subgroup of index p. By Lemma 2.5, we have

which shows that |G′|=p6, and hence |Z4|=p6. Applying Lemma 2.5 to G/Z3 gives p5=p⋅p3⋅p3, which is impossible. Therefore, G must have nilpotence class at most 4. Since G has coclass 3, n-3=c≤4, and hence |G|≤p7. ∎

Proof of Theorem 1.3.

We assume that the result is not true, and we take G to be a counterexample having coclass n so that n is minimal with 4≤n≤6, and then choose G so that G and all of its quotients satisfy Hypothesis (∗) and let |G| be minimal such that G has coclass n and |cod⁡(G)|=4. Suppose c⁢(G)=c>5. The nilpotence class of G/Z is c-1>4, so G/Z has coclass at most n-1 since |G| was minimal with coclass n, but either this contradicts Theorem 1.2 when n=4 or it contradicts the choice of G when n is 5 or 6. Hence we may assume G has nilpotence class 5, and hence |G|=p5+n.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible character of G, then either c⁢(G/K)≤4, or c⁢(G/K)=5 and G/K has coclass at most n-1, contradicting either Theorem 1.2 when n=4 or the choice of G when n is 5 or 6. Hence we must have c⁢(G/K)≤4 for all irreducible characters of G, which contradicts the assumption that G has class 5. Thus, G has a faithful irreducible character χ, and by Lemma 2.8, χ has codegree pa. By Lemma 2.1, 2⁢a>n+5. When n=4, this implies a≥5, and when n=5 or 6, this implies a≥6.

As c⁢(G/Z2)=3, we have pa∈cod⁡(G/Z2), and hence |G/Z2|≥pa+1. As G satisfies Hypothesis (∗), the order of Z2 is at least p3, which implies the order of G/Z2 is at most pn+2. This shows that a≤n+1. When n=4 or 5, we see that we must have a=n+1. When n=6, we have either a=n or a=n+1. Suppose a=n+1, and observe that this forces Z2 to have order p3. Also, notice that G/Z2 now has a faithful character with degree p and codegree pa.

Since Z2 has order p3 and exp(Z2/Z)∣exp(Z) by Lemma 2.6, Z2/Z is necessarily elementary abelian. Similarly, Z3/Z2 is elementary abelian. Since G/Z2 has a faithful irreducible character, we have |Z3/Z2|=p. Hence |Z3|=p4 and |G:Z3|=pn.

Applying Lemma 2.5 to G/Z2, we have |G/Z2|=p⁢|Z3/Z2|⁢|(G′⁢Z2)/Z2|. This yields |G:G′Z2|=p2. By Lemma 2.10, this implies that |G:G3Z2|=p3. Since G3 and Z2 are both contained in Z3, this implies that |G:Z3|≤p3, which is a contradiction. This proves the result when a=n+1.

We now have a=n. This implies that n=6. Let χ∈Irr⁡(G) be faithful with codegree p6. Then χ⁢(1)=p5, so |G:Z|≥χ(1)2=p10. This implies |Z|=p. By Lemma 2.6, Z2/Z and Z3/Z2 have exponent p.

Since p6∈cod⁡(G/Z2), we have |G:Z2|≥p7, so |Z2|=p3 or p4. Suppose |Z2|=p4. Then G/Z2 has a faithful character of degree p and Z3/Z2 is cyclic, showing that |Z3/Z2|=p. Let N⊲G such that |Z2/N|=p, and put X/N=Z⁢(G/N). If X=Z2, then Z2⁢(G/N)=Z3/N, contradicting Hypothesis (*), so we must have X=Z3. This shows that [Z3,G]≤N for all such N, and hence [Z3,G]≤Z, which is impossible. Hence |Z2|=p3.

Suppose |Z3|=p4. Let N⊲G such that Z<N<Z2. Put Z⁢(G/N)=X/N, and notice that Z2≤X≤Z3. If X=Z2, then

Since G/N has class 3 or 4, this contradicts Hypothesis (*). Hence X=Z3. Now Z2⁢(G/N)=Z4/N, so G/N has class 3, which implies N≥G4. Since G4>Z, we have p2≤|G4|≤|N|=p2, which implies N=G4 for all normal subgroups of G lying between Z and Z2. This shows that Z2/Z is cyclic, contradicting that it is elementary abelian of order p2. Hence |Z3|≥p5.