Exercise 4.5.19 ( If $|G| = 6545$ then $G$ is not simple)

Question

Prove that if $|G| = 6545$ then $G$ is not simple.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 6545 = 5 \cdot 7 \cdot 11 \cdot 17$.

By Sylow's Theorem,
\begin{align*}
&n_{17} \mid 5 \cdot 7 \cdot 11 = 385 \qquad 	ext{and} \qquad n_{17} \equiv 1 \pmod {17},\
&n_{11} \mid  5\cdot 7 \cdot 17  =  595 \qquad 	ext{and} \qquad n_{11} \equiv 1 \pmod {11},\
&n_7 \mid 5 \cdot 11 \cdot 17 = 935 \qquad 	ext{and} \qquad n_7 \equiv 1 \pmod 7,\
\end{align*}
thus
\begin{align}
n_{17} \in \{1, 35\},\qquad n_{11} \in \{1, 595\}, \qquad n_{7} \in \{1, 85\}.
\end{align}
Assume for the sake of contradiction that $G$ is simple. Then $n_{17} 
e 1,\ n_{11} 
e 1, \ n_{7} 
e 1$, so
$$n_{17} = 35,\qquad n_{11} = 595,\qquad  n_7 = 85.$$
Since distinct Sylow subgroups have trivial intersection, there are
\begin{itemize}
\item $n_{17} \cdot (17 - 1) = 35 \cdot 16 = 560$ elements of order 17,
\item $n_{11} \cdot (11 - 1) = 595 \cdot 10 = 5950$ elements of order 11,
\item $n_7 \cdot (7-1) = 85 \cdot 6 =  510$ elements of order 7.
\end{itemize}
Hence
$$|G| > 560 + 5950 + 510 = 7020.$$
Since $|G| = 6545$, this is a contradiction, which proves that $G$ is not simple.

If $|G| = 6545$ then $G$ is not simple.
\end{proof}
We check (1) with sagemath:
\begin{verbatim}
sage: n = 5 * 7 * 11 * 17; n
6545
sage: p = 17; m = n // p; m
385
sage:  l = [d for d in range(1,m+1) if m % d == 0]; l
[1, 5, 7, 11, 35, 55, 77, 385]
sage: [d for d in l if d % p == 1]
[1, 35]
sage: p = 11; m = n // p; m
595
sage:  l = [d for d in range(1,m+1) if m % d == 0]; l
[1, 5, 7, 17, 35, 85, 119, 595]
sage: [d for d in l if d % p == 1]
[1, 595]
sage: p = 7; m = n // p; m
935
sage:  l = [d for d in range(1,m+1) if m % d == 0]; l
[1, 5, 11, 17, 55, 85, 187, 935]
sage: [d for d in l if d % p == 1]
[1, 85]
\end{verbatim}

Exercise 4.5.19 ( If $|G| = 6545$ then $G$ is not simple)

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Exercise 4.5.19 ( If $|G| = 6545$ then $G$ is not simple)

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