Exercise 4.5.14 (A group of order $312$ has a normal Sylow $p$-subgroup)

Question

Prove that a group of order $312$ has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $|G| = 312 = 2^3\cdot  3 \cdot 13$.

By Sylow's Theorem,
$$n_{13} \mid 24,\qquad n_{13} \equiv 1 \pmod {13}.$$
Therefore
\begin{align*}
&n_{13} \in \{1, 2, 3, 4, 6, 8, 12, 24\} \cap \{1, 14\} = \{1\}.
  \end{align*}
This gives $n_{13} = 1$: a group of order $312$ has a normal Sylow $13$-subgroup
\end{proof}
Check with Sagemath:
\begin{verbatim}
 [1 + k*13 for k in range(24) if 312 % (1 + k*13) == 0]
 [1]
\end{verbatim}

Exercise 4.5.14 (A group of order $312$ has a normal Sylow $p$-subgroup)

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Exercise 4.5.14 (A group of order $312$ has a normal Sylow $p$-subgroup)

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