A lower bound on the size of maximal abelian subgroups

Abstract

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 $a \ge n/(b+l)$.

Theorems & Definitions (28)

• Theorem 1.1
• Theorem 1.2
• Lemma 2.1
• Lemma 2.2
• proof
• Lemma 2.3
• proof
• Lemma 2.4
• proof
• Lemma 3.1