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Let $G$ be a finite groupsgroup and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}.$$ Define $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?

Let $G$ be a finite groups and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}.$$ Define $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?

Let $G$ be a finite group and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}.$$ Define $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?
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YCor
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Let $G$ be a finite groups and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}$$$$X_n(G):=\{x\in G: x^n=1\}.$$ DoesDefine $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $$\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}=\mathbb Q\cap[0,1]$$ where$J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?

Let $G$ be a finite groups and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}$$ Does there exist a positive integer $n$, such that $$\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}=\mathbb Q\cap[0,1]$$ where $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?

Let $G$ be a finite groups and $n$ be a positive integer. Define $$X_n(G):=\{x\in G: x^n=1\}.$$ Define $$J_n=\left\{\frac{|X_n(G)|}{|G|}: G\,\text{is a finite group}\right\}.$$ Question. Does there exist a positive integer $n$, such that $J_n=\mathbb Q\cap[0,1]$?

Here $\mathbb Q$ is the set of rational numbers? The answer to the simpler question below is also desirable to me:

Does there exist $n\in\mathbb N$, and a family of finite groups $(G_n)_n$ such that

  1. $|X_n(G_k)|<|G_k|$ for all $k$,
  2. $0<\prod_{k=1}^\infty\frac{|X_n(G_k)|}{|G_k|}$?
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