Exercise 4.5.21 (If $|G| = 2907$ then $G$ is not simple)

Question

Prove that if $|G| = 2907$ then $G$ is not simple.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 2907 = 3^2 \cdot 17 \cdot 19$. By Sylow's Theorem
\begin{align*}
&n_{19} \mid 3^2 \cdot 17 = 153, \qquad n_{19} \equiv 1 \pmod {19},\
&n_{17} \mid 3^2 \cdot 19 = 171, \qquad n_{17} \equiv 1 \pmod {19}.\
\end{align*}
Therefore
$$n_{19} \in \{1,153\},\qquad n_{17} \in \{1,171\}.$$
Assume for the sake of contradiction that $n_{19} 
e 1$ and $n_{17} 
e 1$. Then $n_{19} = 153$ and $n_{17} = 171$. 
Since the intersection of distinct Sylow $p$-subgroups for $p = 17$ or $19$ is trivial, $G$ contains
\begin{itemize}
\item $n_{19} \cdot (19 - 1) = 153 \cdot 18 = 2754$ elements of order $19$,
\item $n_{17} \cdot (17 - 1) = 171 \cdot 16 = 2736$ elements of order $17$.
\end{itemize}
Therefore
$$|G| \geq 2754 + 2736 = 5490.$$
Since $|G| = 2907 $, this is a contradiction, so $G$ is not simple.

\bigskip
If $|G| = 2907$ then $G$ is not simple.
\end{proof}

Exercise 4.5.21 (If $|G| = 2907$ then $G$ is not simple)

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Exercise 4.5.21 (If $|G| = 2907$ then $G$ is not simple)

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