Exercise 4.5.30 (Elements of order $7$ in a simple group of order $168$)

Question

How many elements of order $7$ must there be in a simple group of order $168$?

Richard Ganaye · Accepted Answer

\begin{proof}
Let $G$ be a simple group of order $168 = 2^3\cdot 3$. By Sylow's Theorem,
$$n_7 \mid 2^3\cdot 3 = 24,\qquad n_7 \equiv 1 \pmod 7,$$
therefore $n_7 \in \{1,8\}$.

Since $G$ is simple, $n_7 
e 1$, thus
$$n_7 = 8.$$
Two distinct Sylow $7$-subgroups have trivial intersection, thus there are $n_7 \cdot 6 = 48$ elements of order $7$.
\end{proof}


With Sagemath:
\begin{verbatim}
sage: def sylow(n):
....:     l = []
....:     for p,a in n.factor():
....:         m = n // p^a
....:         divisors = [d for d in range(1,m+1) if m % d == 0]
....:         possible = [d for d in divisors if d % p == 1]   
....:         l.append([p,possible])
....:     return(l)
....: 
sage: n = 168
sage: for p, l in sylow(n):
....:     print p, '=>', l
....:     
2 => [1, 3, 7, 21]
3 => [1, 4, 7, 28]
7 => [1, 8]
\end{verbatim}

Exercise 4.5.30 (Elements of order $7$ in a simple group of order $168$)

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Exercise 4.5.30 (Elements of order $7$ in a simple group of order $168$)

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