Skip to main content
Became Hot Network Question
Link to the last question
Source Link
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Let all groups here be finite $p$--groups.

Given $K<H$, let $r(K,H)$ be the smallest $r$ such that there exists a chain of subgroups $K=L_0 \lhd L_1 \lhd \cdots \lhd L_r = H$, such that each $L_i/L_{i-1}$ is cyclic.

Question: Let $H$ be a subgroup of $G$. If $K_1$ and $K_2$ are subgroups of $H$ that are conjugate in $G$, does it follow that $r(K_1,H) = r(K_2,H)$?

As in the last questionquestion I posed, a minimal counterexample would have $G = \langle H,g\rangle$, and $H$ not normal in $G$.

Let all groups here be finite $p$--groups.

Given $K<H$, let $r(K,H)$ be the smallest $r$ such that there exists a chain of subgroups $K=L_0 \lhd L_1 \lhd \cdots \lhd L_r = H$, such that each $L_i/L_{i-1}$ is cyclic.

Question: Let $H$ be a subgroup of $G$. If $K_1$ and $K_2$ are subgroups of $H$ that are conjugate in $G$, does it follow that $r(K_1,H) = r(K_2,H)$?

As in the last question I posed, a minimal counterexample would have $G = \langle H,g\rangle$, and $H$ not normal in $G$.

Let all groups here be finite $p$--groups.

Given $K<H$, let $r(K,H)$ be the smallest $r$ such that there exists a chain of subgroups $K=L_0 \lhd L_1 \lhd \cdots \lhd L_r = H$, such that each $L_i/L_{i-1}$ is cyclic.

Question: Let $H$ be a subgroup of $G$. If $K_1$ and $K_2$ are subgroups of $H$ that are conjugate in $G$, does it follow that $r(K_1,H) = r(K_2,H)$?

As in the last question I posed, a minimal counterexample would have $G = \langle H,g\rangle$, and $H$ not normal in $G$.

Source Link
Nicholas Kuhn
  • 11.1k
  • 31
  • 58

A finite p-group question: can this happen?

Let all groups here be finite $p$--groups.

Given $K<H$, let $r(K,H)$ be the smallest $r$ such that there exists a chain of subgroups $K=L_0 \lhd L_1 \lhd \cdots \lhd L_r = H$, such that each $L_i/L_{i-1}$ is cyclic.

Question: Let $H$ be a subgroup of $G$. If $K_1$ and $K_2$ are subgroups of $H$ that are conjugate in $G$, does it follow that $r(K_1,H) = r(K_2,H)$?

As in the last question I posed, a minimal counterexample would have $G = \langle H,g\rangle$, and $H$ not normal in $G$.