I want to prove
Let $G$ be a group of order $385$. Prove that $G$ is not simple.
Can you check my attempt please ?
Since a simple group $G$ is a group in which the only normal subgroups are the trivial ones, namely $\{e_G\}$ and $\{G\}$, we want to show here that the given group $G$ has a normal subgroup, we want to show that there is a unique Sylow $p$-subgroup, $n_p=1$ for some prime $p$.
Let $ord(G)=385=5^1 \times7^1 \times11^1$.
Since $5 \nmid7,5 \nmid11$ and $7 \nmid11$, $G$ has subgroups of order $5,7$ and $11$.
So $G$ has $3$ $p$-Sylow-subgroups with $p=5,7,11$:
P a Sylow $5$-subgroup of order $5$
Q a Sylow $7$-subgroup of order $7$
R a Sylow $11$-subgroup of order $11$
With Sylow it follows that $n_{7}|55$ which means
$n_7 \in \{1,5,11,55\}$ and $n_{7}\equiv 1 \mod 7$ but
for $5$ and $11$ $n_{7}\not\equiv 1 \mod 7$ which results in $n_{7}\in \{1\}$
Again with Sylow it follows that $n_{11}|35$ which means
$n_{11} \in \{1,5,7,35\}$ and $n_{11}\equiv 1 \mod 11$ but
for $5,7$ and $11$ $n_{11}\not\equiv 1 \mod 11$ which results in $n_{11}\in \{1\}$
Since we've found an unique Sylow $p$-subgroup, $n_p=1$ for some prime $p$ this group must be a normal subgroup and tehrefore $G$ is not simple.
Best Answer
You've got the right idea, but I think you made a mistake in applying Sylow's theorems.
For instance, since $n_7\equiv 1\pmod 7$, and since $n_7\mid 55$, we have right away that $n_7=1$.
So your counting argument won't be necessary.