Exercise 4.5.23 (If $|G| = 462$ then $G$ is not simple)

Question

Prove that if $|G| = 462$ then $G$ is not simple.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 462 = 2 \cdot 3 \cdot 7 \cdot 11$. By Sylow's Theorem,
\begin{align*}
&n_{11} \mid 2\cdot 3 \cdot 7 = 42,\qquad n_{11} \equiv 1 \pmod {11},\
\end{align*}
thus $n_{11} \in \{1, 2, 3 , 6 ,7 ,14,21, 42\}$, where  $n_{11} \equiv 1 \pmod {11}$, thus
\begin{align*}
n_{11} = 1.
\end{align*}
So the unique Sylow $11$-subgroup is normal in $G$, and $G$ is not simple.

\bigskip
 If $|G| = 462$ then $G$ is not simple.
\end{proof}

Exercise 4.5.23 (If $|G| = 462$ then $G$ is not simple)

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Exercise 4.5.23 (If $|G| = 462$ then $G$ is not simple)

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