Maximal subgroup growth of a few polycyclic groups
We give here the exact maximal subgroup growth of two classes of polycyclic groups. Let \(G_k = \langle x_1, x_2, \dots , x_k \mid x_ix_jx_i^{-1}x_j \text{ for all } i < j \rangle\), so \(G_k = \mathbb{Z} \rtimes (\mathbb{Z} \rtimes (\mathbb{Z} \rtimes \dots \rtimes \mathbb{Z}))\). Then for a...
Збережено в:
| Дата: | 2022 |
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| Автори: | , |
| Формат: | Стаття |
| Мова: | English |
| Опубліковано: |
Lugansk National Taras Shevchenko University
2022
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| Теми: | |
| Онлайн доступ: | https://admjournal.luguniv.edu.ua/index.php/adm/article/view/1506 |
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| Назва журналу: | Algebra and Discrete Mathematics |
Репозитарії
Algebra and Discrete Mathematics| Резюме: | We give here the exact maximal subgroup growth of two classes of polycyclic groups. Let \(G_k = \langle x_1, x_2, \dots , x_k \mid x_ix_jx_i^{-1}x_j \text{ for all } i < j \rangle\), so \(G_k = \mathbb{Z} \rtimes (\mathbb{Z} \rtimes (\mathbb{Z} \rtimes \dots \rtimes \mathbb{Z}))\). Then for all integers \(k \geq 2\), we calculate \(m_n(G_k)\), the number of maximal subgroups of \(G_k\) of index \(n\), exactly. Also, for infinitely many groups \(H_k\) of the form \(\mathbb{Z}^2 \rtimes G_2\), we calculate \(m_n(H_k)\) exactly. |
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