| Version 2 |
Version 1 |
| Let $G$ be a finite group whose order is divisible by the prime $p$. Suppose $p^m$ is the highest power of $p$ which is a factor of $|G|$ and set $k = \frac{|G|}{p^m}$. |
Let $G$ be a finite group whose order is divisible by the prime $p$. Suppose $p^m$ is the highest power of $p$ which is a factor of $|G|$ and set $k = \frac{|G|}{p^m}$ |
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| \begin{itemize} |
\begin{itemize} |
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\item The group $G$ contains at least one subgroup of order $p^m$.
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\item The group $G$ contains at least one subgroup of order $p^m$ |
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\item Any two subgroups of $G$ of order $p^m$ are conjugate.
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\item Any two subgroups of $G$ of order $p^m$ are conjugate |
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\item The number of subgroups of $G$ of order $p^m$ is congruent to $1$ modulo $p$ and is a factor of $k$.
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\item The number of subgroups of $G$ of order $p^m$ is congruent to $1$ modulo $p$ and is a factor of $k$ |
| \end{itemize} |
\end{itemize} |
