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An inequality on derived length of a solvable group
Journal of Inequalities and Applications volume 2015, Article number: 20 (2015)
Abstract
Let G be a finite solvable group. Write \(\delta^{*}(G)\) for the number of conjugacy classes of nonabelian subgroups of G, and by \(d(G)\) denote the length of derived subgroups. In this paper an upper bound of \(d(G)\) is given in terms of \(\delta^{*}(G)\).
Introduction
In this paper G is a solvable group of finite order and let \(d(G)\) denote the derived length of G. By \(\delta(G)\) denote the number of conjugacy classes of noncyclic subgroups of G. It can be proved (see [1, Theorem 4.2]) that
This gives an upper bound of the derived length of G. Note that this bound is not nice, for which we have an improvement in this note.
In this paper by \(\delta^{*}(G)\) denote the number of conjugacy classes of nonabelian subgroups of G, which will replace \(\delta(G)\).
Recall some information about a formation which is required in this note. A class ℱ of finite groups is called a formation if \({G\in \mathcal{F}}\) and \(N\trianglelefteq G\) then \(G/N\in\mathcal{F}\), and if \(G/N_{i}\ (i=1,2)\in\mathcal{F}\) then \(G/N_{1}\cap N_{2}\in\mathcal{F}\). If, in addition, \(G/\Phi(G)\in\mathcal{F}\) implies \(G\in\mathcal{F}\), we say that ℱ to be saturated. The class of all abelian groups is a formation but not saturated; the class of all nilpotent groups is a formation and saturated [2, 9.5 Formations].
Let ℱ be a subgroupclosed formation. and let \(G^{\mathcal{F}}\) be the ℱresidual of G, that is,
Define
Consider the series of characteristic (normal) subgroups of G:
Here, if r satisfies \(G^{{\mathcal{F}}^{r}}=1\) but \(G^{{\mathcal{F}}^{(r1)}}\neq1\), we say that r is the ℱlength of G. As G is solvable, r must exist and \(r\geq1\).
Note that \(G/G^{\mathcal{F}}\) and \(G^{{\mathcal{F}}^{(r1)}}\) both are in ℱ, and
Suppose that \(r \geq4\). Now, we are able to choose elements \(x_{0}, x_{1}, \ldots, x_{r4}\) in \(G=G^{{\mathcal{F}}^{0}}, G^{{\mathcal{F}}^{1}}, \ldots, G^{{\mathcal{F}}^{(r4)}}\) satisfying the following condition:
Define subgroups as follows:
Note that:

(1)
In the definition of \(M_{i,j}\), \(j\geq i+2\) is required.

(2)
For every \(i\in\{0, 1,\ldots,{r4}\}\), the set \(\{y_{i} : y_{i}\in G^{(i)} G^{{\mathcal{F}}^{(i+1)}}\}\) is nonempty, the \((x_{0}, x_{1},\ldots,x_{r4})\) may be replaced by \((y_{0}, y_{1}, \ldots, y_{r4})\) in the note.

(3)
If for some i, \(G^{(i)}/G^{(i+1)}\) is a nonabelian 2group, we say that the i is a \(\lambda(G)\). The investigation of the nonℱsubgroups of G is an interesting problem.
Preliminaries
In this section we list some known results which are needed in the sequel.
Lemma 1
For all possible i and j we have:

(1)
All \(M_{i,j}\) are subgroups of G.

(2)
No \(M_{i,j}\) is in ℱ.
Proof
(1) As all \(G^{{\mathcal{F}}^{j}}\) are characteristic (normal) subgroups of G, all \(M_{i,j}\) must be subgroups of G.
(2) As \(G^{{\mathcal{F}}^{r}}=1\) but \(G^{{\mathcal{F}}^{(r1)}}\neq1\), we can see that \(G^{{\mathcal{F}}^{(r2)}}\) is not in ℱ. Next, by the definition of \(M_{i,j}\), we have \(r2\geq j\), so \(G^{{\mathcal {F}}^{(r2)}}\leq G^{{\mathcal{F}}^{j}}\leq M_{i,j}\). As ℱ is subgroupclosed, we conclude that \(M_{i,j}\notin\mathcal{F}\). □
In the following part ℱ is assumed to be the class of all abelian groups. Then \(G^{\mathcal{F}}=G'\) and \(G^{{\mathcal {F}}^{(i)}}=G^{(i)}\) for all i. Lemma 2 is valuable for the following proofs.
Lemma 2
Let G be a nilpotent group. Then that \(G/G'\) is cyclic implies G is cyclic.
Proof
In a nilpotent group, the derived subgroup is contained in the Frattini subgroup [2, 5.2.16], so \(G'\leq\Phi(G)\) and hence \(G/\Phi(G)\) is cyclic. Thus G is cyclic. □
Lemma 3
Suppose that G is a nilpotent group. Then no two of \(M_{i,j}\) for all possible i and j are conjugate in G.
Proof
Assume the lemma is false. So that there exist \(M_{i,j}\) with \(j\geq r+ 2\)and \(M_{i',j'}\) with \(j'\geq r+2\) which are conjugate, that is, there exists a \(y\in G\) such that \(M_{i,j}^{y}=M_{i',j'}\). By definition of \(M_{i, j}\) we have \(G^{(j)} \leq G^{(i)}\). Thus, we have \(M_{i,j}(=\langle x_{i}\rangle G^{(j)})\leq G^{(i)}\). It follows that
Hence \(x_{i'}\in G^{(i)}\). By the choice of \(x_{i}\), we have \(i'\geq i\). Similarly, \(i\geq i'\). It follows that \(i=i'\) and \(x_{i}=x_{i'}\).
In order to finish the proof, we also claim that \(j=j'\). Suppose that \(j\neq j'\). Without loss of generality, let \(j < j'\). Then \(j+ 1\leq j'\) and
We thus get
Consequently, \(G^{(j)}/ G^{(j+1)}\) is cyclic.
Now, applying the hypothesis that G is a nilpotent group, by Lemma 2 we find that \(G^{(j)}\) is cyclic, consequently \(j\leq r2\), which is a contradiction (see the definition of \(M_{i,j}\)). □
Lemma 4
([1, Lemma 4.1])
Let G be a solvable group. Then the following statements are true:

(1)
\(M_{i,2j}\) and \(M_{i', 2j'}\) are conjugate if and only if \(i=i'\) and \(j=j'\).

(2)
No \(M_{i,j}\) is conjugate to some \(G^{(k)}\).
Please note Lemmas 6 and 7 below.
Lemma 5
Let G be a nilpotent group and \(G^{(2)}\neq1\). Then there exists a nonabelian subgroup M which is a maximal subgroup in G, and the following statements are true:

(1)
No subgroups \(M_{i,j}\) for all possible i and j are conjugate to M.

(2)
No subgroups \(G^{(k)}\) for all possible k are conjugate to M.
Proof
By the condition that \(G^{(2)}\neq1\), so \(G'\) is nonabelian, and hence every maximal subgroup of G which contains \(G'\) is nonabelian, for which we write M.
As G is nilpotent, it follows that \(G'\leq\Phi(G)\) and hence M is normal. Suppose some \(M_{i,j}\) is conjugate to M. Then \(M_{i,j}=\langle x_{i}\rangle G^{(j)}=M\trianglelefteq G\) and \(G'\leq M= M_{i, j}\). If \(x_{i}\in G'\), as \(G'\) contains \(G^{(j)}\), we see that \(M_{i,j}= G' < M< G\), which is a contradiction. Thus \(x_{i}\notin G'\), and it follows that \(x_{i}=x_{0}\), and \(M_{i,j}=M_{0,j}\), \(j\geq2\). Now, both \(x_{0}\) and \(G''\) are in \(M_{0,j}\), hence \(M_{0, 2}=\langle x_{0}\rangle G''= M_{0,j}\). It follows that \(M_{0,2}/G''\) is cyclic, by applying Lemma 2, \(G'\) is cyclic, a contradiction. □
Lemma 6
Let G be a solvable group with \(d(G)=r\geq4\). Suppose that \(G^{(i)}/ G^{(i+1)}\) is a noncyclic 2group for some fixed \(i\in\{0,1,\ldots,r4\}\). Fix this i and take \(a_{i}\) for an element of \(G^{(i)}\) but not in \(G^{(i+1)}\) and let \(K_{i}= \langle a_{i}\rangle G^{(i+1)}\) (note that \(G^{(i)}> K_{i} > G^{(i+1)}\)). If \(K_{i}\) is conjugate to some \(M_{s,t}\) or some \(G^{(k)}\), then \(G^{(i)}/G^{(i+2)}\cong Q_{8}\), the quaternion group of order 8.
Proof
Fix i and write
where all \(a_{h}\) are 2element, and \(l\geq2\). \(G^{(i)}\) is nonabelian, then \(\langle a_{h},G^{(i+1)}\rangle=\langle a_{h}\rangle G^{(i+1)}=K_{h}\) is nonabelian too.
As \(G^{(i)}> K_{h} >G^{(i+1)}\), there is no \(G^{(k)}\) which is conjugate to \(K_{h}\). By condition, some \(M_{s,t}\) is conjugate to \(K_{h}\). Thus, for some \(y\in G\) we have
As \(K_{h}< G^{(i)}\), it follows that
Now, as \(G^{(i+2)}< G^{(i+1)}< K_{h}\) when \(i< r4\), we have \(G^{(i+2)}< G^{(i+1)}< K_{h}=M_{s,t}\), it follows that \(K_{h}/ G^{(i+2)}\) is a cyclic group of \(M_{s,t}/G^{(i+2)}\) which is generated by \(a_{h} G^{(i+2)}\). Hence \(G^{(i+1)}/ G^{(i+2)}\) is cyclic.
For any element a of \(G^{(i)}\) with order 2 (mould \(G^{(i+2)}\)), if \(a\notin G^{(i+1)}\), then \(\langle a\rangle\cap G^{(i+1)}=1\), contrary to \(G^{(i+1)}/ G^{(i+2)}\) being cyclic. Thus, \(a\in G^{(i+1)}\) and it follows that \(\langle a\rangle\) is a unique subgroup of order 2 (mould \(G^{(i+2)}\)), consequently \(G^{(i)}\cong Q_{8}\) of order 8 [2, 5.3.6], as desired. □
Lemma 7
Let G be a solvable group with \(d(G)=r\geq4\). Suppose that for some fixed \(i\in\{ 0, 1, \ldots, r4\}\), \(G^{(i)}/G^{(i+2)}\cong Q_{8}\) of order 8. Then \(G^{(i)}\) is nonnilpotent and contains an abnormal maximal subgroup K which is nonabelian such that \(K\in\delta^{*}(G)\).
Proof
The condition that \(d(G)=r\geq 4\) shows that \(G^{(r4)}\) is nonabelian. By the condition that \(G^{(i)}/G^{(i+2)}\cong Q_{8}\), \(G^{(i+1)}/G^{(i+2)}\) is cyclic of order 2, by Lemma 2 we see \(G^{(i+1)}\) is cyclic, hence \(G^{(i+2)}=1\). It follows that \(r\leq i+2\leq r3 +2= r1\), a contradiction. Now we find that \(G^{(i)}\) is nonnilpotent. Then there exists an abnormal maximal subgroup \(K_{i}\) of \(G^{(i)}\). If \(K_{i}\) is abelian, then \(G^{(r3)}K_{i} = G\), this implies that \(G'\leq G^{(r3)}\), contrary to \(r\geq4\). Now, we conclude that \(K_{i}\) is nonabelian, as desired.
Obviously, no \(G^{(s)}\) is conjugate to \(K_{i}\) for all possible s. Suppose that some \(M_{s,t}\) is conjugate to \(K_{i}\). So, \(M_{s,t}^{y}= K_{i} < G^{(i)}\) for some \(y\in G\). By definition, \(M_{s,t}=\langle x_{s}\rangle G^{(t)}\) with \(t\geq s+2\). When \(s\geq i+1\), then \(x_{s}\) and \(G^{(t)}\) both are in \(G^{(i+1)}\), so \(K_{i} = M_{s,t}^{y}\leq G^{(i+1)}=(G^{(i)})'\), contrary to \(K_{i}\) being a maximal subgroup of \(G^{(i)}\). Thus \(s=i\) and \(M_{s,t}=M_{i,t}\) with \(t\geq i+2\). Now, \(M_{s,t}=M_{i,t}=\langle x_{i}\rangle G^{(t)}\leq G^{(i)}\), so \(G^{(i+2)}\geq G^{(t)}\). If \(G^{(i+2)}> G^{(t)}\), we have \(\langle x_{i}\rangle G^{(t)}> K\), consequently, \(\langle x_{i}\rangle G^{(i+2)}=G^{(i)}\), and hence \(\langle x_{i}\rangle G^{(i+1)}=G^{(i)}\), contrary to \(Q_{8}\) (\(\cong G^{(i)}/G^{(i+2)}\)). Thus, \(M_{s,t}=M_{i,i+2}=\langle x_{i}\rangle G^{(i+2)}\), which is normal in \(G^{(i)}\), consequently, \(K_{i}\) would be normal in \(G^{(i)}\), a contradiction. We conclude that no \(M_{s,t}\) is conjugate to \(K_{i}\). □
Main results
Now, we are able to give the main theorems of this note as follows:
Theorem 8
Let G be a solvable group with \(\delta ^{*}(G)\geq1\). Write \(\lambda(G)=c+1\). Then
Proof
In this section ℱ denotes the class of all abelian groups, then \(G^{\mathcal{F}}=G'\) and \(G^{{\mathcal{F}}^{i}}=G^{(i)}\). If \(d(G)=1, 2, 3\), then the theorem holds obviously. Let \(d(G)=r \geq4\). By Lemmas 1 and 3, there exist the following nonabelian subgroups in G:

(a)
\(G(=G^{(0)}), G^{(1)}, G^{(2)}, \ldots, G^{(r2)}\);

(b)
\(M_{0,2}\); \(M_{0,4}, M_{1,4}, M_{2,4}\); \(M_{0,6}, M_{1,6}, M_{2,6}, M_{3,6},M_{4,6}\); … ;
$$ M_{0,2k}, M_{1, 2k}, \ldots, M_{2k2, 2k},\quad r2\geq2k\geq r3. $$
By Lemmas 6 and 7 for every \(\lambda(G)\), \(G^{i_{l}}\) contains at least a nonabelian subgroup \(K_{i_{l}}\), which belongs to \(\delta^{*}(G)\), so we have

(c)
\(K_{i_{0}}, K_{i_{1}}, \ldots, K_{i_{c}}\).
No two of these subgroups are conjugate in G, therefore
That is, \(4\delta^{*}(G)\geq(r1)^{2}+4c +9\),
□
In this case when G is nilpotent, we have the following.
Theorem 9
Let G be a nilpotent group with \(d(G)=r\). Then
Proof
If \(d(G)=r=1, 2, 3\), then the theorem holds obviously. Let \(d(G)\geq4\). By Lemmas 1 and 4, there exist the following nonabelian subgroups in G:

(a)
\(G\ (=G^{(0)}), G^{(1)}, G^{(2)}, \ldots, G^{(r2)}\);

(b)
\(M_{i,j}\), \(i\in\{ 0,1, \ldots, r4\}\), \(r+2\leq j \leq r2\).
By Lemma 5, every \(G^{(i)}\) contains a nonabelian subgroup \(K_{i}\) which is in \(\delta^{*}(G)\), so we have

(c)
\(K_{0}, K_{1}, \ldots, K_{r4}\).
No two of these subgroups are conjugate in G, therefore
□
Example 10
Let \(G=S_{4}\). Then \(d(G)=3\) and \(\delta^{*}(G)=4 (S_{4}, S_{3}, A_{4}, Q_{8})\).

(1)
By Theorem 8, we have (\(\lambda(G)=c+1=0\))
$$2\bigl(\delta^{*}(G) 2 c\bigr)^{1/2}+ 1= 2(41)^{1/2} + 1< 4.47. $$We conclude that \(d(G)=3< 4.47\) and \(4.47r= 1.47\).

(2)
\(\delta(G)=5 (S_{4}, S_{3}, A_{4}, Q_{8}, C_{2}\times C_{2})\), by [1, Theorem 4.1], we have
$$4.4\leq2\bigl(\delta(G)1\bigr)^{1/2}+ 1= 2(4)^{1/2} + 1\leq5. $$We conclude that \(d(G)=3< 5\) and \(5r= 2\).
References
Li, S, Zhao, X: Finite groups with few noncyclic subgroups. J. Group Theory 10, 225233 (2007)
Robinson, DJS: A Course in the Theory of Groups. Springer, New York (1982)
Acknowledgements
The authors are grateful to the editors and the referees, who provided their detailed reports. The research of the work was supported by NSFC (Grant Nos. 11171364, 11271301, U1204101, 11471266), Fundamental Research Funds for the Central Universities (No. XDJK2014C163), National Youth Science Foundation (No. 11201385) and the Major Project of Education Department of Henan Province (No. 13B110085).
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Wu, Z., Zhao, X. & Li, S. An inequality on derived length of a solvable group. J Inequal Appl 2015, 20 (2015). https://doi.org/10.1186/s1366001505533
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DOI: https://doi.org/10.1186/s1366001505533
MSC
 20D10
 20D20
Keywords
 solvable group
 nonabelian subgroup
 derived length