# Some sufficient conditions for the Taketa inequality.

1. Introduction. A long standing open problem in the character
theory of finite solvable groups is whether the derived length dl(G) of
a solvable group G is bounded above by the cardinality of cd(G), the set
of irreducible character degrees of that group, i.e. whether the
so-called Taketa inequality dl(G) [less than or equal to] [absolute
value of cd(G)] is true for every finite solvable group G. This
inequality appeared first in the proof of the fact that all M-groups are
solvable. This proof was given by Taketa by establishing that an M-group
has to satisfy the Taketa inequality. The famous Isaacs-Seitz conjecture
claims that the Taketa inequality is true not only for M-groups but for
any finite solvable group. In the literature we know only some classes
of solvable groups besides M-groups for which the conjecture is true.
For example, T. R. Berger has shown that all finite groups of odd order
satisfy the Taketa inequality [1]. In their paper, "Irreducible
character degrees and normal subgroups" I. M. Isaacs and G. Knutson
[5] have proved that if N is a normal nilpotent subgroup of G then dl(N)
[less than or equal to] [absolute value of cd(G|N)] where cd(G|N) is the
set of degrees of irreducible characters of G whose kernels do not
contain N. They also remark that the inequality dl(N) [less than or
equal to] [absolute value of cd(G|N)] includes the Taketa inequality as
a special case when N is replaced by G'. As a corollary, it turns
out that they prove that dl(G) [less than or equal to] [absolute value
of cd(G)] when G' is nilpotent. Some of the other sufficient
conditions refer to the cardinality of cd(G). I. M. Isaacs has shown
that the condition [absolute value of cd(G)] [less than or equal to] 3
is sufficient for the Taketa inequality [4] (or Corollary 12.6 and
Theorem 12.15 of [6]). In his Ph.D. thesis, S. Garrison has obtained
that [absolute value of cd (G)] = 4 is another sufficient condition for
the conjecture which is later generalized by I. M. Isaacs and Greg
Knutson (see Theorem C of [5]). The last known sufficient condition for
the Taketa inequality regarding the cardinality of the set of the
irreducible character degrees is [7] due to Mark Lewis dealing with the
case [absolute value of cd (G)] = 5. The problem is still open for
solvable groups with six irreducible character degrees.

Motivated by these results we obtain in this paper some further sufficient conditions for the conjecture.

2. Main theorems. We start with the following proposition.

Proposition 2.1. Let G be a finite group and let N be a normal Hall subgroup of G. Suppose that both G/N' and N satisfy the Taketa inequality. Then G satisfies the Taketa inequality.

The proof of this Proposition 2.1 is essentially the same as the proof of Lemma 12.16 of [6]. But for the sake of completeness and as a short reminder we repeat a condensed form of the proof here.

Proof. Let [pi] be the set of primes dividing [absolute value of N]. Since N is a normal Hall subgroup of G, cd(N) is exactly the set of [pi] -parts of the elements of cd(G) and every degree in cd(G/N') divides the index [absolute value of G : N] by Theorem 6.15 of [6]. This yields that [absolute value of cd (N)] + [absolute value of cd (G/N')]-1 [less than or equal to] [absolute value of cd (G)].Now we have dl(G) [less than or equal to] + dl(G/N') + dl(N') [less than or equal to] dl(G/N') + dl(N)- 1 [less than or equal to] [absolute value of cd (G/N')] + [absolute value of cd (N)]-1 [less than or equal to] [absolute value of cd (G)] as desired.

As a corollary of this proposition, we give a generalization of the fact that super-solvable groups satisfy the Taketa inequality (see Theorem 6.22 of [6]).

Theorem 2.2. Let G be a finite group and p be the smallest prime divisor of the order of G. If G has a normal p-complement then G satisfies the Taketa inequality.

Proof. Since all finite groups of odd order satisfy the Taketa inequality by [1], we may assume that the order of G is even so that p = 2. Let N be the normal 2-complement of G. Since the order of N is odd, N satisfies the Taketa inequality. Also, N/N' is an abelian normal subgroup of G/N' and the factor group is a 2-group. So G/N' is an M-group by Theorem 6.22, Theorem 6.23 of [6] and satisfies the Taketa inequality. Thus G itself satisfies the Taketa inequality by Proposition 2.1. []

Corollary 2.3. Let M be a normal subgroup of a group G, where M is super-solvable and G/M is a p-group where p is the smallest prime number dividing [absolute value of G].Then G satisfies the Taketa inequality.

Proof. We know that M has a normal p-complement and since G/M is a p-group G has also a normal p-complement. So we are done by Theorem 2.2.

Corollary 2.4. Let G be a rational group with super-solvable derived subgroup. Then G satisfies the Taketa inequality.

Proof. Since G is rational and factor groups of a rational group are still rational, G/G' is a rational group which is also abelian. It is well known that only abelian rational groups are elementary abelian 2-groups. So we are done by Corollary 2.3.

Theorem 2.5. Let N be a normal subgroup of a group G, where N has an abelian normal p-complement for some prime number p. Then dl(N) [less than or equal to] [absolute value of cd(G/N)]. In particular, if G' has an abelian normal p-complement, then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. We will induct on [absolute value of N].If [absolute value of N] = 1, then dl(N) = 0 and the result holds. Assume N > 1. We have dl(N) = 1 + dl(N') [less than or equal to] 1 + [absolute value of cd (G/N')] [less than or equal to] [absolute value of cd (G/N)], where the first inequality holds by the inductive hypothesis since N' < N, and the second inequality holds by Theorem 3.1 of [5]. To establish the second claim of the theorem, replace N with G'.

Let us consider the following condition for a solvable group G:

[pi](H') < [pi](H)

for every nontrivial Hall subgroup H of G. Under this condition, all Sylow subgroups of G are abelian and so G is an M-group by Theorem 6.23 of [6]. Thus the condition above is sufficient for the Taketa inequality. In the next theorem, we will provide a slightly weaker sufficient condition:

Theorem 2.6. Let G be a solvable group. Assume that [pi](H') < [pi](H) for every Hall subgroups H of G satisfying 2 [less than or equal to] [absolute value of [pi](H)]. Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. We will induct on the order of G. Since Taketa inequality holds for p-groups, we may assume that 2 [less than or equal to] [absolute value of [pi] (G)]. This starts the induction and also allows us to conclude [pi] (G') < [pi] (G) by the fact that every group is a Hall subgroup of itself.

Thus there exists a prime number q dividing the order of G but fails to divide the order of G'. Thus G' is a q'-group and so a Hall q'-subgroup H of G contains G'. Since H is a Hall subgroup of G, the hypothesis is satisfied for H and so dl(H) [less than or equal to] [absolute value of cd (H)] by induction argument (Here H is a proper subgroup of G, since q does not divide the order of H). Now we have a normal Hall subgroup H for which Taketa inequality holds and the factor group G/H is a q-group. So by Corollary 12.16 of [6] we have dl(G) [less than or equal to] [absolute value of cd (G)].

As a preparation for the proof of the following theorems we prove the following proposition:

Proposition 2.7. Let P be a class of finite solvable groups which is closed with respect to taking quotients. Suppose there exists a group in P for which the Taketa inequality is not true and let G be such a group of smallest possible order. Then the following hold:

(i) [G.sup.(n-1)] is the unique minimal normal subgroup of G where n = dl(G),

(ii) cd(G/[G.sup.(n-1)]) = cd(G),

(iii) dl(G)= [absolute value of cd (G)] + 1,

(iv) F(G), the Fitting subgroup of G, is a p-group for some prime p.

Furthermore if G" is nilpotent, then

(v) p divides the index [absolute value of G : G'].

Proof. First assume that G has two distinct minimal normal subgroups M and N. Thus G is isomorphic to a subgroup of G/M x G/N since M [intersection] N = 1. As the Taketa inequality is true for both G/M and G/N we get dl(G) [less than or equal to] max{dl(G/M), dl(G/N)} [less than or equal to] max{[absolute value of cd (G/M)], [absolute value of cd (G/N)]} [less than or equal to] [absolute value of cd (G)]. But this is a contradiction. So G has a unique minimal normal subgroup and consequently F(G) is a p-group for some prime p. This completes the proof of (iv).

Let M be the unique minimal normal subgroup of G. In this case, M is abelian by the solvability of G and so dl(G) [less than or equal to] dl(M) + dl(G/M) = 1 + dl(G/M) [less than or equal to] 1 + [absolute value of cd (G/M)] [less than or equal to] 1 + [absolute value of cd (G)] [less than or equal to] dl(G).

So we have dl(G) = [absolute value of cd (G)] + 1, [absolute value of cd (G/M)] = [absolute value of cd (G)], dl(G/M) = d/(G)- 1 = n-1.

Since [G.sup.(n-1)] is non-trivial normal subgroup of G, M is contained in G(n-1). The equation dl(G/M)_= n-1 yields that 1 = [[bar.G].sup.(n-1)] = [[bar.G].sup.[bar.(n-1)]] where [bar.G] = G/M. So we have M = [G.sup.(n-1)]. This gives the proof of (i), (ii), (iii).

Now suppose that G" is nilpotent. In this case, G" [subset or equal to] F(G) and so G" is a p-group. To prove (v),we will assume that p does not divide the index [absolute value of G : G'] and show that dl(G) [less than or equal to] [absolute value of cd (G)] which is a contradiction. This will complete the proof. Since G" is a p-group there exists a Sylow p-subgroup P of G' containing G". It follows that P is normal in G. Since we assume that p does not divide the index [absolute value of G : G'], P is a normal Hall subgroup of G for which Taketa inequality holds. Clearly we may assume that 1 [not equal to] P' since G" [subset or equal to] P and Taketa inequality holds for groups dl(G) [less than or equal to] 3. Since 1 [not equal to] P', we have dl(G/P') [less than or equal to] [absolute value of cd (G/P')]. Finally we have dl(G) [less than or equal to] [absolute value of cd (G)] by Proposition 2.1.

Theorem 2.8. Let G be a solvable group. Assume that for all [chi], [psi] [member of] Irr(G), [ker.sub.[chi]] = ker [psi] if 1 < [chi] (1) = [psi] (1). Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Since Irr (G/N) [subset or equal to] Irr (G), the hypothesis is inherited by factor groups. Suppose the theorem is false and let G be a minimal counter example to this theorem. Then by Proposition 2.7, G has a unique minimal normal subgroup M and cd G/M)=cd G).

Clearly we may assume 1 [not equal to] G' so that M [subset or equal to] G' and cd(G|M) [subset or equal to] cd(G|G') = cd(G)-{1}.Let k [member of] cd(G|M).In this case, 1 = k and there exists an irreducible character [chi] of G such that [chi] (1) = k and M [??] [ker.sub.[chi]] On the other hand, k [member of] cd(G) = cd(G/M) and so there exists an irreducible character [psi] of G such that M [subset or equal to] ker [psi] and [psi](1) = k. But by hypothesis, ker [chi] = ker [psi] which is a contradiction. So we are done.

Y. Berkovich, D. Chillag and M. Herzog have classified the finite groups in which the degrees of nonlinear irreducible characters are distinct and shown that such groups have at most three distinct irreducible character degrees [2]. So these groups satisfy the Taketa inequality. In the following corollary we have the same conclusion without exploring the structure of these groups.

Corollary 2.9. Let G be a solvable group in which distinct nonlinear irreducible characters have distinct degrees. Then, dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. This is an immediate corollary of Theorem 2.8. []

Let G be a finite group and [absolute value of G]= [p.sup.[alpha]1.sub.1 ... [p.sup.[alpha]r.sub.r] where [p.sub.1], ...,[p.sub.r] are distinct primes and [[alpha].sub.1], ...,[[alpha].sub.r] are non negative integers. We will denote the maximum of the [[alpha].sub.i]'s by [delta] (G). Suppose that [delta] (G) [less than or equal to] 2. Then all Sylow subgroups of G are abelian and so dl(G) [less than or equal to] [absolute value of cd (G)] as mentioned above. The next theorem gives a slightly better bound by putting an additional hypothesis:

Theorem 2.10. Let G be a group and k 2 {1, 2, 3, 4, 5}.If [G.sup.(k)] is nilpotent and [delta] (G) [less than or equal to] 13-2k then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Fix a k [member of] {1, 2, 3, 4, 5} and suppose [G.sup.(k)] is nilpotent, [delta](G) [less than or equal to] 13-2k. We will assume that the assertion is false and look for a contradiction. Let G be a minimal counter example to the assertion. In this case 6 [less than or equal to] [absolute value of cd (G)] by [4], [3] and [7]. Clearly the condition is inherited by factor groups and so we can apply Proposition 2.7. Then n = dl(G) = [absolute value of cd (G)] + 1 [greater than or equal to] 7 and F(G) is a p-group for some prime number p. By hypothesis [G.sup.(k)] is nilpotent and so G(k) [subset or equal to] F(G). Thus [G.sup.(k)] is a p-group and so contained in a Sylow p-subgroup P of [G.sup.(k-1)] and P has to be normal in G. If P' = 1 ,then [G.sup.(k+1)] [subset or equal to] P' = 1 and so 7 [less than or equal to] dl(G) [less than or equal to] k + 1 [less than or equal to] 6 which is a contradiction. So P' is nontrivial so that [absolute value of G/P'] < [absolute value of G] and hence dl(G/P') [less than or equal to] [absolute value of cd (G/P')] by the minimality of G. Thus we see by Proposition 2.1 that P is not a Sylow p-subgroup of G.

When we consider the hypothesis [delta](G) [less than or equal to] 13-2k together with the last paragraph, we have that the order of P which is the p-part of the order of [G.sup.(k-1)] divides [p.sup.12-2k] so that cd(P) [subset or equal to] {1 = [p.sup.0], p, ...,[p.sup.5-k]}. Thus n-k = dl([G.sup.(k)]) [less than or equal to] dl(P) [less than or equal to] [absolute value of cd (P)] [less than or equal to] 6-k and so n [less than or equal to] 6. But this is a contradiction since 7 [less than or equal to] n by the first paragraph. []

Corollary 2.11. Let G be a group. If G' is super-solvable and [delta] (G) [less than or equal to] 9 then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. This is an immediate consequences of Theorem 2.10 since the derived subgroup of a super-solvable group is nilpotent.

Theorem 2.12. Let G be a group with super-solvable derived subgroup. Suppose that G/G' is a p-group for some prime p and [2.sup.k] [not equivalent to] 1(p) for k = 1, ...,n where [[absolute value of G].sup.2] = [2.sup.n]. Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Let G be a minimal counter example to the Theorem. Since the conjecture is true for groups of odd order by [1], the order of G is even and p = 2 by Corollary 2.3. Let H be the unique 2'-Hall subgroup of G' so that H [??] G and let S [member of] [Syl.sub.p](G). Then, SH is a proper subgroup of G. If SH [??] G then G/SH [congruent to] G'/(G' [intersection] SH),but as G" is a p-group by Proposition 2.7 we have G" [less than or equal to] (G' [intersection] SH).Therefore G/SH is abelian which implies that G' [less than or equal to] SH and hence G = SH which is not the case. So SH/H is a Sylow p-subgroup of G/H which is not normal. Thus we conclude that 1 < [G/H : [N.sub.G/H](SH/H)] [equivalent to] 1 (p).But [G/H : [N.sub.G/H] (SH/H)] divides [G' : H] which is a power of 2. This contradiction completes the proof.

doi: 10.3792/pjaa.89.103

Acknowledgements. We would like to thank to the referee for his/her suggestions which improved our original paper. The statements of Proposition 2.1 and Theorem 2.2 belong essentially to him/her.

The work of the first author was supported by Scientific Research Projects Coordination Unit of Istanbul University. The project number is 4383.

References

[1] T. R. Berger, Characters and derived length in groups of odd order, J. Algebra 39 (1976), no. 1, 199-207.

[2] Y. Berkovich, D. Chillag and M. Herzog, Finite groups in which the degrees of the nonlinear irreducible characters are distinct, Proc. Amer. Math. Soc. 115 (1992), no. 4, 955-959.

[3] S. C. Garrison, III, On groups with a small number of character degrees, Pro-Quest LLC, Ann Arbor, MI, 1973.

[4] I. M. Isaacs, Groups having at most three irreducible character degrees, Proc. Amer. Math. Soc. 21 (1969), 185 188.

[5] I. M. Isaacs and G. Knutson, Irreducible character degrees and normal subgroups, J. Algebra 199 (1998), no. 1, 302 326.

[6] I. M. Isaacs, Character theory of finite groups, corrected reprint of the 1976 original [Academic Press, New York], AMS Chelsea Publishing, Providence, RI, 2006.

[7] M. L. Lewis, Derived lengths of solvable groups having five irreducible character degrees. I, Algebr. Represent. Theory 4 (2001), no. 5, 469 489.

Utku YILMAZTURK, *) Temha ERKOC *) and Ismail S. GULOGLU **)

(Communicated by Masaki Kashiwara, m.j.a., Oct. 15, 2013)

2000 Mathematics Subject Classification. Primary 20C15.

*) Department of Mathematics, Faculty of Science, Istanbul University, 34134, Vezneciler, Istanbul, Turkey.

**) Department of Mathematics, Faculty of Arts and Science, Dogus University, Acibadem, Kadikoy, 34722, Istanbul, Turkey.

Motivated by these results we obtain in this paper some further sufficient conditions for the conjecture.

2. Main theorems. We start with the following proposition.

Proposition 2.1. Let G be a finite group and let N be a normal Hall subgroup of G. Suppose that both G/N' and N satisfy the Taketa inequality. Then G satisfies the Taketa inequality.

The proof of this Proposition 2.1 is essentially the same as the proof of Lemma 12.16 of [6]. But for the sake of completeness and as a short reminder we repeat a condensed form of the proof here.

Proof. Let [pi] be the set of primes dividing [absolute value of N]. Since N is a normal Hall subgroup of G, cd(N) is exactly the set of [pi] -parts of the elements of cd(G) and every degree in cd(G/N') divides the index [absolute value of G : N] by Theorem 6.15 of [6]. This yields that [absolute value of cd (N)] + [absolute value of cd (G/N')]-1 [less than or equal to] [absolute value of cd (G)].Now we have dl(G) [less than or equal to] + dl(G/N') + dl(N') [less than or equal to] dl(G/N') + dl(N)- 1 [less than or equal to] [absolute value of cd (G/N')] + [absolute value of cd (N)]-1 [less than or equal to] [absolute value of cd (G)] as desired.

As a corollary of this proposition, we give a generalization of the fact that super-solvable groups satisfy the Taketa inequality (see Theorem 6.22 of [6]).

Theorem 2.2. Let G be a finite group and p be the smallest prime divisor of the order of G. If G has a normal p-complement then G satisfies the Taketa inequality.

Proof. Since all finite groups of odd order satisfy the Taketa inequality by [1], we may assume that the order of G is even so that p = 2. Let N be the normal 2-complement of G. Since the order of N is odd, N satisfies the Taketa inequality. Also, N/N' is an abelian normal subgroup of G/N' and the factor group is a 2-group. So G/N' is an M-group by Theorem 6.22, Theorem 6.23 of [6] and satisfies the Taketa inequality. Thus G itself satisfies the Taketa inequality by Proposition 2.1. []

Corollary 2.3. Let M be a normal subgroup of a group G, where M is super-solvable and G/M is a p-group where p is the smallest prime number dividing [absolute value of G].Then G satisfies the Taketa inequality.

Proof. We know that M has a normal p-complement and since G/M is a p-group G has also a normal p-complement. So we are done by Theorem 2.2.

Corollary 2.4. Let G be a rational group with super-solvable derived subgroup. Then G satisfies the Taketa inequality.

Proof. Since G is rational and factor groups of a rational group are still rational, G/G' is a rational group which is also abelian. It is well known that only abelian rational groups are elementary abelian 2-groups. So we are done by Corollary 2.3.

Theorem 2.5. Let N be a normal subgroup of a group G, where N has an abelian normal p-complement for some prime number p. Then dl(N) [less than or equal to] [absolute value of cd(G/N)]. In particular, if G' has an abelian normal p-complement, then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. We will induct on [absolute value of N].If [absolute value of N] = 1, then dl(N) = 0 and the result holds. Assume N > 1. We have dl(N) = 1 + dl(N') [less than or equal to] 1 + [absolute value of cd (G/N')] [less than or equal to] [absolute value of cd (G/N)], where the first inequality holds by the inductive hypothesis since N' < N, and the second inequality holds by Theorem 3.1 of [5]. To establish the second claim of the theorem, replace N with G'.

Let us consider the following condition for a solvable group G:

[pi](H') < [pi](H)

for every nontrivial Hall subgroup H of G. Under this condition, all Sylow subgroups of G are abelian and so G is an M-group by Theorem 6.23 of [6]. Thus the condition above is sufficient for the Taketa inequality. In the next theorem, we will provide a slightly weaker sufficient condition:

Theorem 2.6. Let G be a solvable group. Assume that [pi](H') < [pi](H) for every Hall subgroups H of G satisfying 2 [less than or equal to] [absolute value of [pi](H)]. Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. We will induct on the order of G. Since Taketa inequality holds for p-groups, we may assume that 2 [less than or equal to] [absolute value of [pi] (G)]. This starts the induction and also allows us to conclude [pi] (G') < [pi] (G) by the fact that every group is a Hall subgroup of itself.

Thus there exists a prime number q dividing the order of G but fails to divide the order of G'. Thus G' is a q'-group and so a Hall q'-subgroup H of G contains G'. Since H is a Hall subgroup of G, the hypothesis is satisfied for H and so dl(H) [less than or equal to] [absolute value of cd (H)] by induction argument (Here H is a proper subgroup of G, since q does not divide the order of H). Now we have a normal Hall subgroup H for which Taketa inequality holds and the factor group G/H is a q-group. So by Corollary 12.16 of [6] we have dl(G) [less than or equal to] [absolute value of cd (G)].

As a preparation for the proof of the following theorems we prove the following proposition:

Proposition 2.7. Let P be a class of finite solvable groups which is closed with respect to taking quotients. Suppose there exists a group in P for which the Taketa inequality is not true and let G be such a group of smallest possible order. Then the following hold:

(i) [G.sup.(n-1)] is the unique minimal normal subgroup of G where n = dl(G),

(ii) cd(G/[G.sup.(n-1)]) = cd(G),

(iii) dl(G)= [absolute value of cd (G)] + 1,

(iv) F(G), the Fitting subgroup of G, is a p-group for some prime p.

Furthermore if G" is nilpotent, then

(v) p divides the index [absolute value of G : G'].

Proof. First assume that G has two distinct minimal normal subgroups M and N. Thus G is isomorphic to a subgroup of G/M x G/N since M [intersection] N = 1. As the Taketa inequality is true for both G/M and G/N we get dl(G) [less than or equal to] max{dl(G/M), dl(G/N)} [less than or equal to] max{[absolute value of cd (G/M)], [absolute value of cd (G/N)]} [less than or equal to] [absolute value of cd (G)]. But this is a contradiction. So G has a unique minimal normal subgroup and consequently F(G) is a p-group for some prime p. This completes the proof of (iv).

Let M be the unique minimal normal subgroup of G. In this case, M is abelian by the solvability of G and so dl(G) [less than or equal to] dl(M) + dl(G/M) = 1 + dl(G/M) [less than or equal to] 1 + [absolute value of cd (G/M)] [less than or equal to] 1 + [absolute value of cd (G)] [less than or equal to] dl(G).

So we have dl(G) = [absolute value of cd (G)] + 1, [absolute value of cd (G/M)] = [absolute value of cd (G)], dl(G/M) = d/(G)- 1 = n-1.

Since [G.sup.(n-1)] is non-trivial normal subgroup of G, M is contained in G(n-1). The equation dl(G/M)_= n-1 yields that 1 = [[bar.G].sup.(n-1)] = [[bar.G].sup.[bar.(n-1)]] where [bar.G] = G/M. So we have M = [G.sup.(n-1)]. This gives the proof of (i), (ii), (iii).

Now suppose that G" is nilpotent. In this case, G" [subset or equal to] F(G) and so G" is a p-group. To prove (v),we will assume that p does not divide the index [absolute value of G : G'] and show that dl(G) [less than or equal to] [absolute value of cd (G)] which is a contradiction. This will complete the proof. Since G" is a p-group there exists a Sylow p-subgroup P of G' containing G". It follows that P is normal in G. Since we assume that p does not divide the index [absolute value of G : G'], P is a normal Hall subgroup of G for which Taketa inequality holds. Clearly we may assume that 1 [not equal to] P' since G" [subset or equal to] P and Taketa inequality holds for groups dl(G) [less than or equal to] 3. Since 1 [not equal to] P', we have dl(G/P') [less than or equal to] [absolute value of cd (G/P')]. Finally we have dl(G) [less than or equal to] [absolute value of cd (G)] by Proposition 2.1.

Theorem 2.8. Let G be a solvable group. Assume that for all [chi], [psi] [member of] Irr(G), [ker.sub.[chi]] = ker [psi] if 1 < [chi] (1) = [psi] (1). Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Since Irr (G/N) [subset or equal to] Irr (G), the hypothesis is inherited by factor groups. Suppose the theorem is false and let G be a minimal counter example to this theorem. Then by Proposition 2.7, G has a unique minimal normal subgroup M and cd G/M)=cd G).

Clearly we may assume 1 [not equal to] G' so that M [subset or equal to] G' and cd(G|M) [subset or equal to] cd(G|G') = cd(G)-{1}.Let k [member of] cd(G|M).In this case, 1 = k and there exists an irreducible character [chi] of G such that [chi] (1) = k and M [??] [ker.sub.[chi]] On the other hand, k [member of] cd(G) = cd(G/M) and so there exists an irreducible character [psi] of G such that M [subset or equal to] ker [psi] and [psi](1) = k. But by hypothesis, ker [chi] = ker [psi] which is a contradiction. So we are done.

Y. Berkovich, D. Chillag and M. Herzog have classified the finite groups in which the degrees of nonlinear irreducible characters are distinct and shown that such groups have at most three distinct irreducible character degrees [2]. So these groups satisfy the Taketa inequality. In the following corollary we have the same conclusion without exploring the structure of these groups.

Corollary 2.9. Let G be a solvable group in which distinct nonlinear irreducible characters have distinct degrees. Then, dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. This is an immediate corollary of Theorem 2.8. []

Let G be a finite group and [absolute value of G]= [p.sup.[alpha]1.sub.1 ... [p.sup.[alpha]r.sub.r] where [p.sub.1], ...,[p.sub.r] are distinct primes and [[alpha].sub.1], ...,[[alpha].sub.r] are non negative integers. We will denote the maximum of the [[alpha].sub.i]'s by [delta] (G). Suppose that [delta] (G) [less than or equal to] 2. Then all Sylow subgroups of G are abelian and so dl(G) [less than or equal to] [absolute value of cd (G)] as mentioned above. The next theorem gives a slightly better bound by putting an additional hypothesis:

Theorem 2.10. Let G be a group and k 2 {1, 2, 3, 4, 5}.If [G.sup.(k)] is nilpotent and [delta] (G) [less than or equal to] 13-2k then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Fix a k [member of] {1, 2, 3, 4, 5} and suppose [G.sup.(k)] is nilpotent, [delta](G) [less than or equal to] 13-2k. We will assume that the assertion is false and look for a contradiction. Let G be a minimal counter example to the assertion. In this case 6 [less than or equal to] [absolute value of cd (G)] by [4], [3] and [7]. Clearly the condition is inherited by factor groups and so we can apply Proposition 2.7. Then n = dl(G) = [absolute value of cd (G)] + 1 [greater than or equal to] 7 and F(G) is a p-group for some prime number p. By hypothesis [G.sup.(k)] is nilpotent and so G(k) [subset or equal to] F(G). Thus [G.sup.(k)] is a p-group and so contained in a Sylow p-subgroup P of [G.sup.(k-1)] and P has to be normal in G. If P' = 1 ,then [G.sup.(k+1)] [subset or equal to] P' = 1 and so 7 [less than or equal to] dl(G) [less than or equal to] k + 1 [less than or equal to] 6 which is a contradiction. So P' is nontrivial so that [absolute value of G/P'] < [absolute value of G] and hence dl(G/P') [less than or equal to] [absolute value of cd (G/P')] by the minimality of G. Thus we see by Proposition 2.1 that P is not a Sylow p-subgroup of G.

When we consider the hypothesis [delta](G) [less than or equal to] 13-2k together with the last paragraph, we have that the order of P which is the p-part of the order of [G.sup.(k-1)] divides [p.sup.12-2k] so that cd(P) [subset or equal to] {1 = [p.sup.0], p, ...,[p.sup.5-k]}. Thus n-k = dl([G.sup.(k)]) [less than or equal to] dl(P) [less than or equal to] [absolute value of cd (P)] [less than or equal to] 6-k and so n [less than or equal to] 6. But this is a contradiction since 7 [less than or equal to] n by the first paragraph. []

Corollary 2.11. Let G be a group. If G' is super-solvable and [delta] (G) [less than or equal to] 9 then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. This is an immediate consequences of Theorem 2.10 since the derived subgroup of a super-solvable group is nilpotent.

Theorem 2.12. Let G be a group with super-solvable derived subgroup. Suppose that G/G' is a p-group for some prime p and [2.sup.k] [not equivalent to] 1(p) for k = 1, ...,n where [[absolute value of G].sup.2] = [2.sup.n]. Then dl(G) [less than or equal to] [absolute value of cd (G)].

Proof. Let G be a minimal counter example to the Theorem. Since the conjecture is true for groups of odd order by [1], the order of G is even and p = 2 by Corollary 2.3. Let H be the unique 2'-Hall subgroup of G' so that H [??] G and let S [member of] [Syl.sub.p](G). Then, SH is a proper subgroup of G. If SH [??] G then G/SH [congruent to] G'/(G' [intersection] SH),but as G" is a p-group by Proposition 2.7 we have G" [less than or equal to] (G' [intersection] SH).Therefore G/SH is abelian which implies that G' [less than or equal to] SH and hence G = SH which is not the case. So SH/H is a Sylow p-subgroup of G/H which is not normal. Thus we conclude that 1 < [G/H : [N.sub.G/H](SH/H)] [equivalent to] 1 (p).But [G/H : [N.sub.G/H] (SH/H)] divides [G' : H] which is a power of 2. This contradiction completes the proof.

doi: 10.3792/pjaa.89.103

Acknowledgements. We would like to thank to the referee for his/her suggestions which improved our original paper. The statements of Proposition 2.1 and Theorem 2.2 belong essentially to him/her.

The work of the first author was supported by Scientific Research Projects Coordination Unit of Istanbul University. The project number is 4383.

References

[1] T. R. Berger, Characters and derived length in groups of odd order, J. Algebra 39 (1976), no. 1, 199-207.

[2] Y. Berkovich, D. Chillag and M. Herzog, Finite groups in which the degrees of the nonlinear irreducible characters are distinct, Proc. Amer. Math. Soc. 115 (1992), no. 4, 955-959.

[3] S. C. Garrison, III, On groups with a small number of character degrees, Pro-Quest LLC, Ann Arbor, MI, 1973.

[4] I. M. Isaacs, Groups having at most three irreducible character degrees, Proc. Amer. Math. Soc. 21 (1969), 185 188.

[5] I. M. Isaacs and G. Knutson, Irreducible character degrees and normal subgroups, J. Algebra 199 (1998), no. 1, 302 326.

[6] I. M. Isaacs, Character theory of finite groups, corrected reprint of the 1976 original [Academic Press, New York], AMS Chelsea Publishing, Providence, RI, 2006.

[7] M. L. Lewis, Derived lengths of solvable groups having five irreducible character degrees. I, Algebr. Represent. Theory 4 (2001), no. 5, 469 489.

Utku YILMAZTURK, *) Temha ERKOC *) and Ismail S. GULOGLU **)

(Communicated by Masaki Kashiwara, m.j.a., Oct. 15, 2013)

2000 Mathematics Subject Classification. Primary 20C15.

*) Department of Mathematics, Faculty of Science, Istanbul University, 34134, Vezneciler, Istanbul, Turkey.

**) Department of Mathematics, Faculty of Arts and Science, Dogus University, Acibadem, Kadikoy, 34722, Istanbul, Turkey.

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Author: | Yilmazturk, Utku; Erkoc, Temha; Guloglu, Ismail S. |
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Publication: | Japan Academy Proceedings Series A: Mathematical Sciences |

Article Type: | Report |

Geographic Code: | 7TURK |

Date: | Nov 1, 2013 |

Words: | 3367 |

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