Exercise 4.5.20 (If $|G| = 1365$ then $G$ is not simple)

Question

Prove that if $|G| = 1365$ then $G$ is not simple.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 1365 = 3 \cdot 5 \cdot 7 \cdot 13.$
By Sylow's Theorem,
\begin{align*}
&n_{13} \mid 3 \cdot 5 \cdot 7 =  105 \qquad 	ext{and} \qquad n_{13} \equiv 1 \pmod {13},\
&n_{7} \mid  3\cdot 5 \cdot 13  = 195  \qquad 	ext{and} \qquad n_{7} \equiv 1 \pmod {7},\
&n_5 \mid 3 \cdot 7 \cdot 13 = 273  \qquad 	ext{and} \qquad n_5 \equiv 1 \pmod 5,\
&n_3 \mid 5 \cdot 7 \cdot 13 =  455 \qquad 	ext{and} \qquad n_3 \equiv 1 \pmod 3,
\end{align*}
thus
\begin{align}
n_{13} \in \{1, 105\},\qquad n_{7} \in \{1, 15 \}, \qquad n_{5} \in \{1, 21, 91\}, \qquad n_{3} \in \{1,7,13,91\}.
\end{align}
Assume for the sake of contradiction that $G$ is not simple. Then $n_{13} 
e 1, n_7 
e 1, n_5 
e 1$ and $n_3 
e 1$. Therefore
$$n_{13} = 105,\qquad n_7 = 15,\qquad n_5 \geq 21,\qquad n_3 \geq 7.$$
Since distinct Sylow subgroups have trivial intersection, there are
\begin{itemize}
\item $n_{13} \cdot (13 - 1) = 105 \cdot 12 = 1260$ elements of order 17,
\item $n_{7} \cdot (7 - 1) = 15 \cdot 6 = 90$ elements of order 11,
\item $n_5 \cdot (5-1) \geq  21 \cdot 4 = 84$ elements of order 7,
\item $n_3 \cdot (3-1) \geq 7 \cdot 2 = 14$ elements of order 3,
\item $1$ element of order $1$.
\end{itemize}
Hence
$$|G|\geq  1260 + 90 + 84 + 14 + 1 = 1449.$$
Since $|G| = 1365$, this is a contradiction, which proves that $G$ is not simple.
\end{proof}

Exercise 4.5.20 (If $|G| = 1365$ then $G$ is not simple)

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Exercise 4.5.20 (If $|G| = 1365$ then $G$ is not simple)

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