Total Character of a Group G with (G, Z(G)) As a Generalized Camina Pair

Home , G

Canad.Math. Bull. Vol. þ (z), z§Ë@ pp. hz– §z http://dx.doi.org/Ë§. Ëþh/CMB-z§Ëþ-§6 -þ ©Canadian Mathematical Society z§Ë@

Total Character of a Group G with (G, Z(G)) as a Generalized Camina Pair

S. K. Prajapati and R. Sarma

Abstract. We investigate whether the total character of a ûnite group G is a polynomial in a suitable irreducible character of G. When (G, Z(G)) is a generalized Camina pair, we show that the total character is a polynomial in a faithful irreducible character of G if and only if Z(G) is cyclic.

1 Introduction _roughout this article, G denotes a ûnite group. Let Irr(G), nl(G) and lin(G) be the set of all irreducible characters of G, the set of all nonlinear irreducible characters of G and the set of linear characters of G, respectively. Suppose that ρ is the direct sum of all the non-isomorphic irreducible complex representations of G. _e character τ G aòorded by ρ is called the total character of G, that is, τ χ. Since τ is G = ∑χ∈Irr(G) G stable under the action of the Galois group of the splitting ûeld of G, τ g for G ( ) ∈ Z all g ∈ G. _e dimension of the total character of a group seems to have remarkable connection with the geometry of the group. For instance, in the case of the symmetric group G S , τ Ë is the number of involutions of S [], whereas in the case of = n G ( ) n G GL n, q , τ Ë is the number of symmetric matrices in GL n, q [h]. Degree of = ( ) G ( ) ( ) total character is discussed by many authors [ , @, Ëh, Ëþ, Ë@]. A consequence of a well-known theorem due to Burnside and Brauer [þ, _eorem .h] is that the total character of the group G is a constituent of Ë + χ + ⋅ ⋅ ⋅ + χm−Ë if χ is a faithful character which takes exactly m distinct values on G. M. L. Lewis and S. M. Gagola [z] classiûed all the solvable groups for which χz τ for some χ Irr G . = G ∈ ( ) Motivated by this, K. W. Johnson raised the following question (see [Ë ]). Do there exist an irreducible character χ of G and a monic polynomial f (x) ∈ Z[x] such that f χ τ ? ( ) = G _e aim of the article is to solve a weaker version of the same, i.e., to examine the existence of f x x and χ Irr G such that f χ τ . We call such a poly- ( ) ∈ Q[ ] ∈ ( ) ( ) = G nomial f (x) ∈ Q[x], if it exists, a Johnson polynomial of G. _is problem has been studied for dihedral groups D in [Ë ]. In fact, the authors have proved that D has zn zn a Johnson polynomial if and only if Ç ∤ n. To describe the classes of groups to which our results apply, we recall some deû- nitions. A pair (G, N) is said to be a generalized Camina pair (abbreviated GCP) if

Received by the editors April , z§Ë ; revised November þ, z§Ëþ. Published electronically January zÇ, z§Ë@. Author S.K.P. was supported by the Council of Scientiûc and Industrial Research (CSIR), India. AMS subject classiûcation: z§CËþ. Keywords: ûnite groups, group characters, total characters. hz

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N is normal in G and, all nonlinear irreducible characters of G vanish outside N [Ëz]. _ere are a number of equivalent conditions for (G, Z(G)) to be a GCP. An equivalent condition we will refer to is that a pair (G, Z(G)) is a GCP if and only if, for all g ∈ G ∖ Z(G), the conjugacy class of g in G is gG′. In this article, we compute the total character τ of a group G for which G, Z G G ( ( )) is a generalized Camina pair and prove a necessary and suõcient condition for the existence of a Johnson polynomial. Our main results can be stated as follows.

_eorem Ë.Ë Let (G, Z(G)) be a GCP. _en G has a Johnson polynomial if and only if Z(G) is cyclic. In fact, if Z(G) is cyclic, then a Johnson polynomial of G is given by

r m f (x) = dz Q(x~d)l j + d Q(x~d) j, j=Ë j=Ë l∤ j where d = SG~Z(G)SË~z, r = SZ(G)~G′S, m = SZ(G)S, and l = SG′S. In particular, f (x) = dz(x~d)m + d m−Ë(x~d) j when Z(G) is cyclic and Z(G) = G′. ∑ j=Ë In the last section, we apply the above theorem to prove the following.

_eorem Ë.z If G is a non-abelian p-group of order ph or p , then G has a Johnson polynomial if and only if Z(G) is cyclic and G′ ⊆ Z(G).

_e character aòorded by the regular representation shares certain properties with the total character of a group, and so the same question may be asked of it. We say that a group G has a regular-Johnson polynomial f (x) ∈ Q[x] if there is some χ ∈ Irr(G) such that ρ g f χ g , where ρ is the character of the regular representation G ( ) = ( ( )) G of G. In the following theorem a group having a regular-Johnson polynomial is char- acterized.

_eorem Ë.h Let G be a ûnite group. _en G has a regular-Johnson polynomial if and only if G has a faithful irreducible character.

Proof Let G has a regular-Johnson polynomial f ∈ Q[x] with χ ∈ Irr(G) such that ρ g f χ g for all g G, where ρ is the regular character. Now we will show G ( ) = ( ( )) ∈ G that χ is faithful. On the contrary, let g Ë ker χ . _en § ρ g f χ g ≠ ∈ ( ) = G ( ) = ( ( )) = f χ Ë ρ Ë , which is a contradiction. Conversely, let χ Irr G be a faithful ( ( )) = G ( ) ∈ ( ) character of G. Suppose that f (x) = ∏ x−χ(g) . _en f (χ(g)) = ρ (g) for g≠Ë∈G χ(Ë)−χ(g) G all g ∈ G. _e coeõcients of f (x) manifestly lie in the cyclotomic ûeld Q[ξ], where ~ ξ = ezπi n and n = SGS.Next we show that f (x) ∈ Q[x]. Consider the Galois group ∶= Gal( [ξ]~ ). _en × ≅ by r ↦ σ (ξ) ∶= ξr, where × consisting of all G Q Q Zn G r Zn congruence classes mod n of integers coprime to n. _e Galois group G acts on Irr(G) by σ.ϕ(g) = tr(σρ(g)), where ϕ ∈ Irr(G) and ϕ is aòorded by the representation ρ, and σρ is deûned by ûrst realising ρ as matrices over Q[ξ], and then evaluating σρ g σ ρ g entry-wise. _erefore we have σ.ϕ g ϕ gr if σ σ (as ( )( ) = ( ( )) ( ) = ( ) = r described above), where r is coprime to n = SGS. Since g ↦ gr is a permutation of G ûxing Ë, the coeõcients of f (x) are rational.

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2 Further Notation and Preliminaries _roughout this article, C denotes the cyclic group of order n. Suppose G is a û- n nite group. _en Z(G), G′ and Cl(G) denote respectively the center, the commu- tator subgroup and the set of conjugacy classes of G.If a, b ∈ G, then b a = b−Ë ab, [a, b] = a−Ëb−Ë ab.Here cd(G), d(G), and Φ(G) denote the set of irreducible character degrees, the minimal number of generators of G, and the Frattini subgroup of G, respectively. Suppose N is a normal subgroup of G. _en we denote by Irr(GSN) = Irr(G) ∖ Irr(G~N).Here we start by recalling some basic results.

Lemma z.Ë ([þ, _eorem z.hz(a)]) If G has a faithful irreducible character, then Z(G) is cyclic.

Lemma z.z Let G be a non-abelian group. _en χ g § for each g ∑χ∈lin(G) ( ) = ∈ G ∖ G′.

Proposition z.Ë exhibits the relationship between faithful characters and groups having Johnson polynomial.

Proposition z.Ë Let G be a ûnite group. Suppose f (x) ∈ C[x] and χ is a character of G such that f χ τ . _en χ is a faithful character. In particular, an abelian group ( ) = G has a Johnson polynomial if and only if it is cyclic.

Proof Suppose f x x and χ is a character of G such that f χ τ with ( ) ∈ C[ ] ( ) = G ker χ Ë . Since ker ϕ Ë , τ Ë τ g for all g Ë G. Take ( ) ≠ { } ∩ϕ∈Irr(G) ( ) = { } G ( ) ≠ G ( ) ≠ ∈ g Ë ker χ . _en τ Ë f χ Ë f χ g τ g , which is a contradiction. ≠ ∈ ( ) G ( ) = ( ( )) = ( ( )) = G ( ) _is shows that χ must be a faithful character.Hence an abelian group G having a Johnson polynomial implies that G is a cyclic group. For the converse, consider the polynomial f (x) = ∑SGS−Ë x i . i=§

To show that G provides a negative answer to Johnson’s question, we will later in- troduce a speciûc character and then attain a contradiction. For this, we need the following proposition, which is a simple observation.

Proposition z.z Let χ be an irreducible character of G.If g , g G are such that Ë z ∈ χ g χ g but τ g τ g , then there does not exist f x x such that ( Ë) = ( z) G ( Ë) ≠ G ( z) ( ) ∈ C[ ] f χ τ . ( ) = G

3 Groups with (G, Z(G)) a Generalized Camina Pair In this section, we study the total character of a group G for which (G, Z(G)) is a generalized Camina pair (abbreviated as GCP). _e notion of generalized Camina pair was ûrst introduced by Lewis [Ëz]. _e groups with (G, Z(G)) a GCP were studied under the name VZ-groups by Lewis [ËË]. First, we record a couple of lemmata that will be useful.

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Lemma h.Ë ([Ëz, Lemma z.Ë]) Let g ∈ G. _en the following statements are equivalent. (i) _e conjugacy class of g is the coset gG′. (ii) χ(g) = § for all nonlinear χ ∈ Irr(G).

Lemma h.z ([Ëz, Lemma z. ]) Let H be a normal subgroup of a group G such that (G, H) is a GCP. _en G′ is a subgroup of H.

3.1 Remarks on a Group G with (G, Z(G)) a Generalized Camina Pair Let (G, Z(G)) be a GCP. Suppose χ is a nonlinear irreducible character of G. _en χ χ Ë λ ↓Z(G) = ( ) for some λ ∈ Irr(Z(G)). _us SGS = Q Sχ(g)Sz = Q Sχ(g)Sz (since (G, Z(G)) is a GCP) g∈G g∈Z(G) = Q Sχ(Ë)λ(g)Sz g∈Z(G) = χ(Ë)zSZ(G)S.

_erefore the degree of any nonlinear irreducible character of G is SG~Z(G)SË~z. Sup- pose n is the number of nonlinear irreducible characters of G. _en

SGS = Q χ(Ë)z = SG~G′S + n.χ(Ë)z. χ∈Irr(G) _erefore the total number of nonlinear irreducible characters of G is SZ(G)S − SZ(G)~G′S. Let η∶ G → G~G′ be the natural homomorphism and let ϕ∶ Irr(G~G′) → Irr(Z(G)) be deûned by ϕ(λ) ∶= λ ○ η. Suppose X ∶= {λ ∈ Irr(Z(G)) S λ ∉ Image(ϕ)} and Φ̂∶ X → nl(G) deûned by ⎧ ⎪SG~Z(G)SË~z λ(g) if g ∈ Z(G), (h.Ë) Φ̂(λ)(g) ∶= ⎨ ⎪§ otherwise. ⎩⎪ _eorem h.Ë Suppose (G, Z(G)) is a GCP. With the notation in the preceding para- graph, the map Φ̂ is a bijection. In other words,

nl(G) = {Φ̂(λ)S λ ∈ Irr(Z(G)) and G′ ⊈ ker(λ)}.

Proof Clearly Φ̂ is one-to-one.Let χ nl G . _en χ G Z G Ë~z λ, where ∈ ( ) ↓Z(G) = S ~ ( )S λ ∈ Irr(Z(G)). We must show that λ ∈ X. Suppose λ ∉ X. _en G′ ⊆ ker(λ).Hence χ G′ G Z G Ë~z χ Ë . _us χ lin G , which is a contradiction. Hence ↓Z(G)( ) = S ~ ( )S = ( ) ∈ ( ) Φ̂ is a bijection.

In the following proposition we discuss the total character τ of G. G

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Proposition h.Ë Let G, Z G be a GCP. _en the total character τ is given by ( ( )) G ⎪⎧SG~G′S + (SZ(G)S − SZ(G)~G′S)SG~Z(G)SË~z if g = Ë, ⎪ (h.z) τ g G G′ Z G G′ . G Z G Ë~z if g G′ Ë , G ( ) = ⎨S ~ S − S ( )~ S S ~ ( )S ∈ ∖ { } ⎪ ⎩⎪§ otherwise.

Proof Set A ∶= Irr(Z(G)) ∖ Irr(Z(G)~G′). By _eorem h.Ë, nl(G) = {Φ̂(λ)S λ ∈ A} and χ(Ë) = SG~Z(G)SË~z for all χ ∈ nl(G). If g Ë, then τ Ë G G′ Z G Z G G′ G Z G Ë~z.If g G Z G , = G ( ) = S ~ S+(S ( )S−S ( )~ S)S ~ ( )S ∈ ∖ ( ) then by the hypothesis of the proposition and Lemma z.z, we get

τ g χ g χ g §. G ( ) = Q ( ) = Q ( ) = χ∈Irr(G) χ∈lin(G) For g ≠ Ë ∈ Z(G), we have

(h.h) § = Q ψ(g) = Q ϕ(g) + Q λ(g). ψ∈Irr(Z(G)) ϕ∈Irr(Z(G)~G′) λ∈A If g ≠ Ë ∈ G′ ⊆ Z(G), then τ g χ g χ g G ( ) = Q ( ) + Q ( ) χ∈lin(G) χ∈nl(G) = SG~G′S + SG~Z(G)SË~z Q λ(g) λ∈A = SG~G′S − SZ(G)~G′S.SG~Z(G)SË~z (by (h.h)). Finally, if g Z G G′, then by Lemma z.z and (h.h), we get τ g §. _is ∈ ( ) ∖ G ( ) = completes the proof.

With these technical results we give the proof of _eorem Ë.Ë.

Proof of _eorem Ë.Ë Suppose that Z(G) = ⟨g⟩ is a cyclic group of order m. Since (G, Z(G)) is a GCP, by Lemma h.z, G′ ⊆ Z(G).Let G′ = ⟨g k ⟩, SG′S = l, and ′ zπi ∗ SZ(G)~G S = r. Set ζ = e m . _e homomorphism λ ∶ Z(G) → given by g ↦ ζ m ζm C m deûnes a faithful linear character.Hence Z(G) ≅ Irr(Z(G)) = ⟨λ ⟩. _e set of ir- ζm reducible characters of Z(G) whose kernel contains G′ is {λl , λzl , ... , λrl }.Hence ζm ζm ζm nl(G) ∶= {Φ̂(λi )S i = Ë, ... , m and l ∤ i}, where Φ̂ is the map deûned in (h.Ë). ζm Obviously S nl(G)S = SZ(G)S − SZ(G)~G′S.Let

r m f (x) = dz Q(x~d)l j + d Q(x~d) j, j=Ë j=Ë l∤ j

where d = SG~Z(G)SË~z.

Assertion If χ = Φ̂(λ ), then f (χ) = τ . ζm G

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Proof of the Assertion If g = Ë, then

r m f (χ(Ë)) = dz Q(χ(Ë)~d)l j + d Q(χ(Ë)~d) j j=Ë j=Ë l∤ j = dzr + d(m − r) = SG~G′S + SG~Z(G)SË~z(SZ(G)S − SZ(G)~G′S) τ Ë (by (h.z)). = G ( )

Let a ≠ Ë ∈ G′. _en a = g kq where Ë ≤ q ≤ (l − Ë). So

r m f (χ(g kq)) = dz Q(χ(g kq)~d)l j + d Q(χ(g kq)~d) j j=Ë j=Ë l∤ j r m z zπikq j zπikq j = d Q(e r ) + d Q(e m ) j=Ë j=Ë l∤ j = dz.r + d(−SZ(G)~G′S) = SG~G′S − SZ(G)~G′S.SG~Z(G)SË~z τ g kq (by (h.z)). = G ( )

Finally, let gs ∈ Z(G) ∖ G′. _en s is not a integer multiple of k. Now by using the similar arguments as in the above case we get f χ gs § τ gs . _is completes ( ( )) = = G ( ) the assertion.

On the other hand, if Z(G) is non-cyclic, then G has no faithful irreducible character. _erefore, from Proposition z.Ë, G has no Johnson polynomial. _is completes the proof.

Remark h.Ë Since the set of character values of Φ̂(λi ) does not depend on i when ζm (i, m) = Ë, we have f (Φ̂(λi )) = τ . ζm G As a consequence of _eorem Ë.Ë, we get the following:

Corollary h.Ë Every extra-special p-group has a Johnson polynomial.

Proof Suppose G is an extra-special p-group. _en Z(G) = G′ and SZ(G)S = p and by [Ç, _eorem z.ËÇ], (G, Z(G)) is a GCP. _erefore by _eorem Ë.Ë, the polynomial

p−Ë f (x) = pn Q(x~pn) j + pzn(x~pn)p j=Ë is a Johnson polynomial of G and f χ τ for every χ nl G . ( ) = G ∈ ( )

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Table Ë

Group Order Presentation Polynomial f (x) z − G ph ⟨a b S ap = bp = [a b] = ap⟩ f (x) = p ∑p Ë(x~p) j Ë , Ë, , Ë j=Ë + pz(x~p)p G h a b a b az bz a b f x xz x z z ⟨ , S = = Ë, = = [ , ]⟩ z( ) = + G ph a b c ap bp c p a b c f x p p−Ë x p j h odd ⟨ , , S = = = Ë, [ , ] = , h( ) = ∑ = ( ~ ) a c b c j Ë [ , ] = [ , ] = Ë⟩ + pz(x~p)p h z G p ⟨a b S ap = bp = [a b] = ap ⟩ f (x) = pz ∑p (x~d)p j , Ë, , j=Ë z + p ∑p (x~p) j j=Ë, j≠t.p z G p ⟨a b c S ap = bp = c p = [a b] = [a c] = f (x) = pz ∑p (x~d)p j þ , , Ë, , , Ë, þ j=Ë [b, c] = ap , [a, b] = [a, c] = Ë, [b, c] = ap⟩ z + p ∑p (x~p) j j=Ë, j≠t.p G p a b apz bpz a b ap @ ⟨ , S = = Ë, [ , ] = ⟩ Does not exist G p a b c apz bp c p a b ap 6 ⟨ , , S = = = Ë, [ , ] = , Does not exist [a, c] = [b, c] = Ë⟩ G p a b c apz bp c p a b c Ç ⟨ , , S = = = Ë, [ , ] = , Does not exist [a, c] = [b, c] = Ë⟩ G a b c a b cz a b az z ⟨ , , S = = = Ë, [ , ] = , Does not exist az = bz , [a, c] = Ë, [b, c] = Ë⟩ G p a b c d ap bp c p d p a b c Ë§ odd ⟨ , , , S = = = = Ë, [ , ] = , Does not exist [a, c] = [a, d] = [b, c] = [b, d] = [c, d] = Ë⟩

G D a b aÇ bz a b a@ ËË = Ë@ z ⟨ , S = = Ë, [ , ] = ⟩ Does not exist G a b aÇ bz a b az Ëz z ⟨ , S = = Ë, [ , ] = ⟩ Does not exist G a b aÇ b a b a@ a bz Ëh z ⟨ , S = = Ë, [ , ] = , = ⟩ Does not exist G p a b c apz bp c p a b ap Ë odd ⟨ , , S = = = Ë, [ , ] = , Does not exist [a, c] = b, [b, c] = Ë⟩ G p a b c apz bp a b ap Ëþ odd ⟨ , , S = = Ë, [ , ] = , Does not exist ap = c p , [a, c] = b, [b, c] = Ë⟩ G p a b c apz bp c p a b ap Ë@ odd ⟨ , , S = = = Ë, [ , ] = , Does not exist c p = aα p , [a, c] = b, [b, c] = Ë⟩ α denotes a quadratic non-residue mod p G p p a b c d ap bp c p d p Ë6 , > h ⟨ , , , S = = = = Ë, Does not exist [a, b] = c, [b, c] = d, [a, c] = Ë, [a, d] = [b, d] = [c, d] = Ë⟩ G a b c a bh ch a b c ËÇ h ⟨ , , S = = = Ë, [ , ] = , Does not exist [a, c] = Ë, [b, c] = a@⟩

4 An Application 4.1 p-Groups of Order ≤ p We quote some known results that we use in the sequel.

Lemma .Ë ([þ, Lemma z.]) Let H be a subgroup of G. Suppose χ is a character of G. _en χ , χ G H χ, χ with equality if and only if χ g § for all g G H. ⟨ ↓H ↓H⟩ ≤ S ~ S⟨ ⟩ ( ) = ∈ ∖

Lemma .z ([Ë, _eorem z§]) If G is a p-group, then for each χ ∈ Irr(G), χ(Ë)z divides SG ∶ Z(G)S.

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Lemma .h Let G be a non-abelian group of order p . _en cd(G) = {Ë, p}.

Proof Since Z(G) ≠ Ë, SZ(G)S = p or pz. _erefore SG~Z(G)S = ph or pz. So by Lemma .z, the result follows.

_e list of all non-abelian p-groups of order ph and p [Ë§, Table Ë] is displayed in Table Ë along with a Johnson polynomial (if exists). Now we prove _eoremË. z. To prove the theorem, we use the classiûcation of non-abelian p-groups of order ph and p , and follow the notation in Table Ë.

Proof of _eoremË. z Suppose G′ Z G then G G (Ë i Ë§). By Lemmata ⊆ ( ) = i ≤ ≤ .h and .Ë, for these groups G, Z G is a GCP. _erefore, for G (Ë i Ë§) use ( ( )) i ≤ ≤ _eorem Ë.Ë to determine a Johnson polynomial (Z G is cyclic if Ë i þ and ( i ) ≤ ≤ non-cyclic otherwise). Next suppose G′ Z G . _en G G (ËË i ËÇ). We must show that for these ⊈ ( ) = i ≤ ≤ groups there is no Johnson polynomial. For the groups G G (ËË i Ëh), one can = i ≤ ≤ easily check that G has no Johnson polynomial. Next for G G (Ë i ËÇ), the nilpotency class of G is h. _erefore G Z G = i ≤ ≤ ~ ( ) is non-abelian and Z(G) ⊂ G′.Hence SZ(G)S = p. As SG~G′S ≥ pz, we deduce that SG′S = pz. Since there is a normal abelian subgroup N (say) of index p, every nonlinear irreducible characters of G must be induced from N. _erefore, χ(G ∖ N) = § for all χ ∈ nl(G) and cd(G) = {Ë, p}. Since G~Z(G) is an extra-special group of order ph, G~Z(G) has p − Ë nonlinear irreducible characters of degree p which vanish outside Z(G/Z(G)) = G′~Z(G) in G~Z(G). For χ ∈ nl(G~Z(G)) we have ( .Ë) χ pλ ↓Z(G~Z(G)) = for some λ Irr Z G Z G Ë , where Ë is the trivial character ∈ ( ( ~ ( ))) ∖ Z(G~Z(G)) Z(G~Z(G)) of Z(G~Z(G)). In particular, we have all the nonlinear irreducible characters of G having Z(G) in their kernel. Now, let ψ ∈ Irr(GSZ(G)). Since SZ(G)S = p, ψ is faithful and hence ϕ is not G-invariant, where ϕ is an irreducible constituent of ψ↓G . G′ _erefore, by Cliòord’s theorem ψ↓G = p ϕ , where ϕ = ϕ and p is the index of the G′ ∑ i Ë ′ Ë inertia group N of ϕ in G. Now ϕ ↓G = λ, where λ ∈ Irr(Z(G)) ∖ Ë for each i Z(G) Z(G) Ë ≤ i ≤ p. _erefore, by using the fact ψ(Ë) = p, we have

G ψ↓ = βϕ = ρ ′ ϕ , G′ Q Ë G ~Z(G) Ë β∈Irr(G′~Z(G)) where ρ is the regular character of G′ Z G .Hence for each ψ Irr G Z G , G′~Z(G) ~ ( ) ∈ ( S ( )) we have ψ(G′ ∖ Z(G)) = §. Now if g ∈ G′ ∖ Z(G), then ( .z) τ g χ g χ g χ g G ( ) = Q ( ) + Q ( ) + Q ( ) χ∈lin(G) χ∈nl(G~Z(G)) χ∈Irr(GSZ(G)) ′ = SG~G S + Q pλ(g) + Q χ(g) (by ( .Ë))

λ∈Irr(Z(G~Z(G)))∖ËZ(G~Z(G)) χ∈Irr(GSZ(G)) = pz − p + § = pz − p.

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Now suppose G has a Johnson polynomial f x such that f χ τ , where χ ( ) ( ) = G ∈ nl(G). _erefore χ is faithful and χ ∈ Irr(GSZ(G)). By ( .z), we have f § f χ g τ g pz p ( ) = ( ( )) = G ( ) = − for all g G′ Z G . Again for h G N we have, f § f χ h τ h §. ∈ ∖ ( ) ∈ ∖ ( ) = ( ( )) = G ( ) = _erefore, from Proposition z.z, G has no Johnson polynomial for G G (Ë i = i ≤ ≤ ËÇ). _is completes the proof.

4.2 Minimal Non-abelian Groups and p-IFC-groups A non-abelian group G is called a minimal non-abelian group if every proper subgroup of G is abelian. For a prime p and n ≥ z, m ≥ h deûne

− G(n, m) = ⟨a, b S apn = bpm = Ë, [a, b] = apn Ë ⟩. _en G(n, m) is a metacyclic group and its order is pn+m. Again for a prime p and n, m ∈ N deûne G(n, m, Ë) = ⟨a, b S apm = bpn = Ë, [a, b] = c, [a, c] = [b, c] = Ë⟩. _en G(n, m, Ë) is not a metacyclic group and its order is pn+m+Ë. First we recall a result on minimal non-abelian p-groups.

_eorem .Ë ([Ë6, Lemma z.Ë]) Let G be a minimal non-abelian p-group. _en G is isomorphic to Q , G n, m or G n, m, Ë . Ç ( ) ( ) Proposition .Ë Suppose G is a minimal non-abelian p-group. _en G has a Johnson polynomial if and only if G is isomorphic to Q . Ç

Proof Total character of Q a, b a Ë, az bz, b−Ë ab a−Ë is given by Ç = ⟨ S = = = ⟩ τ (Ë) = @, τ (az) = z, and τ (a) = τ (b) = τ (ab) = §.If χ is the faithful QÇ QÇ QÇ QÇ QÇ irreducible character of Q , then one can verify that χz + χ = τ so that xz + x is a Ç QÇ Johnson polynomial of Q . Now observe that Z G n, m ap, bp C C Ç ( ( )) = ⟨ ⟩ ≅ pn−Ë × pm−Ë and Z G n, m, Ë ap, bp , c C C C are non-cyclic. _erefore, they ( ( )) = ⟨ ⟩ ≅ pn−Ë × pm−Ë × p do not have any faithful irreducible character.Hence by Proposition z.Ë, G(n, m) and G(n, m, Ë) have no Johnson polynomials. A p-group G is said to be a p-IFC-group if the Frattini subgroup of every proper subgroup of G is cyclic.

_eorem .z ([Ë6, _eorem h.Ë]) Suppose that G is a p-IFC-group with SG′S ≤ p and p odd. (i) If SG′S = Ë, then G is abelian, and one of the following holds. m (a) G ≅ C n × E , where n, m are non-negative and Φ(G) is a cyclic group of p p order pn−Ë. (b) G ≅ C × C and Φ(G) = Ez . pz pz p (ii) If SG′S = p and d(G) = z, then one of the following holds. − (a) G Mod a, b apn bp Ë, a, b apn Ë , where n z is a positive ≅ pn+Ë = ⟨ S = = [ ] = ⟩ ≥ integer.

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(b) G = ⟨a, b S apz = bpz = c p = Ë, [a, b] = c, [c, a] = [c, b] = Ë⟩, where n ≥ Ë is a positive integer. (c) G = ⟨a, b S apz = bpz = Ë, [a, b] = ap⟩. (iii) If SG′S = p and d(G) ≠ z, then Φ(G) is cyclic.

′ Proposition .z Let p be an odd prime. Suppose G is p-IFC-group with SG S ≤ p. (i) If SG′S = Ë, then G has no Johnson polynomials. (ii) If SG′S = p and d(G) = z, then G is a Johnson polynomial if and only if G ≅ Mod . pn+Ë (iii) If SG′S = p and d(G) ≠ z, then G need not have a Johnson polynomial.

Proof If SG′S = Ë, then by _eorem .z, G is a non-cyclic abelian group and hence by Proposition z.Ë G has no Johnson polynomials. Now suppose G ≅ Mod n+Ë so that p − SG′S = p and d(G) = z.Here Z(G) = ⟨ap⟩, SG~Z(G)S = pz and G′ = ⟨apn Ë ⟩. By Lemma .z, the degree of every nonlinear irreducible character is p and so by Lemma .Ë (G, Z(G)) is GCP. Hence by _eorem Ë.Ë the following polynomial

− − pn z pn Ë f (x) = pz Q (x~p)l j + p Q (x~p) j j=Ë j=Ë p∤ j is a Johnson polynomial. Next suppose G′ p and d G z and G Mod . _en S S = ( ) = ≇ pn+Ë by _eorem .z, either G = ⟨a, b S apz = bpz = c p = Ë, [a, b] = c, [c, a] = [c, b] = Ë⟩ or G = ⟨a, b S apz = bpz = Ë, [a, b] = ap⟩. In the former case, Z(G) = ⟨ap, bp, c⟩, and in the latter, Z(G) = ⟨ap, bp⟩.Hence, in either case the center is non cyclic. _erefore, it does not have faithful irreducible character and hence by Proposition z.Ë, none of these groups has a Johnson polynomial. Finally to justify the third statement of the theorem we will produce examples. Suppose p is an odd prime.Let G a, b, c ap bp c p Ë, a, b c, a, c Ë = ⟨ S = = = [ ] = [ ] = b, c Ë and G a, b, c apz bp c p Ë, a, b ap , a, c b, c Ë . _e [ ] = ⟩ z = ⟨ S = = = [ ] = [ ] = [ ] = ⟩ groups G and G are both p- -groups.Observe that Φ(G ) = G′ = Z(G ) = ⟨c⟩, Ë z IFC Ë Ë Ë Φ(G ) = G′ = ⟨ap⟩, Z(G ) = ⟨ap, c⟩, d(G ) ≠ z, and (G , Z(G )) is a GCP for z z z i i i i Ë, z.Hence by _eorem Ë.Ë, G has a Johnson polynomial but G has no Johnson = Ë z polynomials.

Acknowledgement _e authors thank the anonymous referee for his/her useful comments and suggestions, including the statement of _eoremË. h.

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