Answer by Yakov for Normal abelian subgroups in p-groups
The Sylow subgroup $\Sigma_n$ of the Symmetric group $S_{p^n}$ has only one maximal normal abelian subgroup, say $B$, for $p>2$ and $n>1$. As $\Sigma_n/B$ is isomorphic to$\Sigma_{n-1}$, the...
View ArticleAnswer by Will Sawin for Normal abelian subgroups in p-groups
The answer to question 1 is yes. We will demonstrate a way to add one to $n(G)$, so $n(G)$ can be any natural number.Given a group $G$, and an abelian group $A$, take a faithful action of $G$ on $A$,...
View ArticleAnswer by Marty Isaacs for Normal abelian subgroups in p-groups
The answer to Qustion 2 is that the index of $T(G)$ in $G$ can be unboundedly large. The reason is that if $G = X \times Y$, the direct product, then $T(G) = T(X) \times T(Y)$. Thus, for example, if...
View ArticleNormal abelian subgroups in p-groups
Given a group $G$, we denote by $T(G)$ the subgroup generated by all (maximal) normal abelian subgroups of $G$.Let define the series $(T_i(G))$ by $T_0(G)=1$ and $T_{i+1}(G)/T_i(G)=T(G/T_i(G)$, and...
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