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An inequality on derived length of a solvable group
- Zhengfei Wu^{1},
- Xianhe Zhao^{2}Email author and
- Shirong Li^{3}
https://doi.org/10.1186/s13660-015-0553-3
© Wu et al.; licensee Springer 2015
- Received: 20 October 2014
- Accepted: 7 January 2015
- Published: 21 January 2015
Abstract
Let G be a finite solvable group. Write \(\delta^{*}(G)\) for the number of conjugacy classes of non-abelian 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 \(\delta^{*}(G)\).
Keywords
- solvable group
- non-abelian subgroup
- derived length
MSC
- 20D10
- 20D20
1 Introduction
In this paper by \(\delta^{*}(G)\) denote the number of conjugacy classes of non-abelian 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].
Here, if r satisfies \(G^{{\mathcal{F}}^{r}}=1\) but \(G^{{\mathcal{F}}^{(r-1)}}\neq1\), we say that r is the ℱ-length of G. As G is solvable, r must exist and \(r\geq1\).
- (1)
In the definition of \(M_{i,j}\), \(j\geq i+2\) is required.
- (2)
For every \(i\in\{0, 1,\ldots,{r-4}\}\), the set \(\{y_{i} : y_{i}\in G^{(i)}- G^{{\mathcal{F}}^{(i+1)}}\}\) is non-empty, the \((x_{0}, x_{1},\ldots,x_{r-4})\) may be replaced by \((y_{0}, y_{1}, \ldots, y_{r-4})\) in the note.
- (3)
If for some i, \(G^{(i)}/G^{(i+1)}\) is a non-abelian 2-group, we say that the i is a \(\lambda(G)\). The investigation of the non-ℱ-subgroups of G is an interesting problem.
2 Preliminaries
In this section we list some known results which are needed in the sequel.
Lemma 1
- (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 \(G^{{\mathcal{F}}^{(r-1)}}\neq1\), we can see that \(G^{{\mathcal{F}}^{(r-2)}}\) is not in ℱ. Next, by the definition of \(M_{i,j}\), we have \(r-2\geq j\), so \(G^{{\mathcal {F}}^{(r-2)}}\leq G^{{\mathcal{F}}^{j}}\leq M_{i,j}\). As ℱ is subgroup-closed, 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
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 \(x_{i}=x_{i'}\).
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 \(j\leq r-2\), which is a contradiction (see the definition of \(M_{i,j}\)). □
Lemma 4
([1, Lemma 4.1])
- (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
- (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 non-abelian, and hence every maximal subgroup of G which contains \(G'\) is non-abelian, 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 non-cyclic 2-group for some fixed \(i\in\{0,1,\ldots,r-4\}\). 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
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 \(i\in\{ 0, 1, \ldots, r-4\}\), \(G^{(i)}/G^{(i+2)}\cong Q_{8}\) of order 8. Then \(G^{(i)}\) is non-nilpotent and contains an abnormal maximal subgroup K which is non-abelian such that \(K\in\delta^{*}(G)\).
Proof
The condition that \(d(G)=r\geq 4\) shows that \(G^{(r-4)}\) is non-abelian. 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 \(r\leq i+2\leq r-3 +2= r-1\), a contradiction. Now we find that \(G^{(i)}\) is non-nilpotent. Then there exists an abnormal maximal subgroup \(K_{i}\) of \(G^{(i)}\). If \(K_{i}\) is abelian, then \(G^{(r-3)}K_{i} = G\), this implies that \(G'\leq G^{(r-3)}\), contrary to \(r\geq4\). Now, we conclude that \(K_{i}\) is non-abelian, 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 \(K_{i}\). □
3 Main results
Now, we are able to give the main theorems of this note as follows:
Theorem 8
Proof
- (a)
\(G(=G^{(0)}), G^{(1)}, G^{(2)}, \ldots, G^{(r-2)}\);
- (b)\(M_{0,2}\); \(M_{0,4}, M_{1,4}, M_{2,4}\); \(M_{0,6}, M_{1,6}, M_{2,6}, M_{3,6},M_{4,6}\); … ;$$ M_{0,2k}, M_{1, 2k}, \ldots, M_{2k-2, 2k},\quad r-2\geq2k\geq r-3. $$
- (c)
\(K_{i_{0}}, K_{i_{1}}, \ldots, K_{i_{c}}\).
In this case when G is nilpotent, we have the following.
Theorem 9
Proof
- (a)
\(G\ (=G^{(0)}), G^{(1)}, G^{(2)}, \ldots, G^{(r-2)}\);
- (b)
\(M_{i,j}\), \(i\in\{ 0,1, \ldots, r-4\}\), \(r+2\leq j \leq r-2\).
- (c)
\(K_{0}, K_{1}, \ldots, K_{r-4}\).
Example 10
- (1)By Theorem 8, we have (\(\lambda(G)=c+1=0\))We conclude that \(d(G)=3< 4.47\) and \(4.47-r= 1.47\).$$2\bigl(\delta^{*}(G)- 2- c\bigr)^{1/2}+ 1= 2(4-1)^{1/2} + 1< 4.47. $$
- (2)\(\delta(G)=5 (S_{4}, S_{3}, A_{4}, Q_{8}, C_{2}\times C_{2})\), by [1, Theorem 4.1], we haveWe conclude that \(d(G)=3< 5\) and \(5-r= 2\).$$4.4\leq2\bigl(\delta(G)-1\bigr)^{1/2}+ 1= 2(4)^{1/2} + 1\leq5. $$
Declarations
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).
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Authors’ Affiliations
References
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