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| Format: | Preprint |
| Published: |
2024
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| Online Access: | https://arxiv.org/abs/2402.12221 |
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| _version_ | 1866929248250363904 |
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| author | Lewis, Mark L. |
| author_facet | Lewis, Mark L. |
| contents | Let $G$ be a $p$-group for some prime $p$. Let $n$ be the positive integer so that $|G:Z(G)| = p^n$. Suppose $A$ is a maximal abelian subgroup of $G$. Let $$p^l = {\rm max} \{|Z(C_G (g)):Z(G)| : g \in G \setminus Z(G)\},$$ $$p^b = {\rm max} \{|cl(g)| : g \in G \setminus Z(G) \},$$ and $p^a = |A:Z(G)|$. Then we show that $a \ge n/(b+l)$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2402_12221 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | A lower bound on the size of maximal abelian subgroups Lewis, Mark L. Group Theory 20D25 Let $G$ be a $p$-group for some prime $p$. Let $n$ be the positive integer so that $|G:Z(G)| = p^n$. Suppose $A$ is a maximal abelian subgroup of $G$. Let $$p^l = {\rm max} \{|Z(C_G (g)):Z(G)| : g \in G \setminus Z(G)\},$$ $$p^b = {\rm max} \{|cl(g)| : g \in G \setminus Z(G) \},$$ and $p^a = |A:Z(G)|$. Then we show that $a \ge n/(b+l)$. |
| title | A lower bound on the size of maximal abelian subgroups |
| topic | Group Theory 20D25 |
| url | https://arxiv.org/abs/2402.12221 |